120750
If a directrix of a hyperbola centred at the origin and passing through the point \((4,-2 \sqrt{3})\) is \(5 x=4 \sqrt{5}\) and its eccentricity is e, then
1 \(4 \mathrm{e}^4-12 \mathrm{e}^2-27=0\)
2 \(4 \mathrm{e}^4-24 \mathrm{e}^2+27=0\)
3 \(4 \mathrm{e}^4+8 \mathrm{e}^2-35=0\)
4 \(4 \mathrm{e}^4-24 \mathrm{e}^2+35=0\)
Explanation:
D The given directrix of hyperbola is -
\(5 \mathrm{x}=4 \sqrt{5}\)
\(\mathrm{x}=\frac{4 \sqrt{5}}{5}\)
\(\therefore \frac{\mathrm{a}}{\mathrm{e}}=\frac{4}{\sqrt{5}}\)
Now, the hyperbola
Given,
below passes through \((4,-2 \sqrt{3})\)
Substituting the value of \(\mathrm{x}\) and \(\mathrm{y}\)
\(\frac{4^2}{\mathrm{a}^2}-\frac{(2 \sqrt{3})^2}{\mathrm{~b}^2}=1\)
\(\frac{16}{\mathrm{a}^2}-\frac{12}{\mathrm{a}^2\left(\mathrm{e}^2-1\right)}=1\)
\({\left[\because \mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2} \Rightarrow \mathrm{a}^2\left(\mathrm{e}^2-1\right)=\mathrm{b}^2\right]}\)
\(\frac{16}{\mathrm{a}^2}-\frac{12}{\mathrm{a}^2\left(\mathrm{e}^2-1\right)}=1\)
\(\frac{16\left(\mathrm{e}^2-1\right)-12}{\mathrm{a}^2\left(\mathrm{e}^2-1\right)}=1\)
\(16 \mathrm{e}^2-16-12=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(16 \mathrm{e}^2-28=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
On substituting \(\mathrm{a}=\frac{4 \mathrm{e}}{\sqrt{5}}\) in the above \(\mathrm{eq}^{\mathrm{n}}\), we get-
\(\Rightarrow 16 \mathrm{e}^2-28=\left(\frac{4 \mathrm{e}}{\sqrt{5}}\right)^2\left(\mathrm{e}^2-1\right)\)
\(\Rightarrow 16 \mathrm{e}^2-28=\frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}\)
\(\Rightarrow \frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}-16 \mathrm{e}^2+28=0\)
\(\frac{16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140}{5}=0\)
\(16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140=0\)
\(16 \mathrm{e}^4-96 \mathrm{e}^2+140=0\)
On dividing the above equation with \(4\)
\(\frac{16 \mathrm{e}^2}{4}-\frac{96 \mathrm{e}^2}{4}+\frac{140}{4}=0\)
\(4 \mathrm{e}^2-24 \mathrm{e}^2+35=0\)
\(\Rightarrow 16 \mathrm{e}^2-28=\left(\frac{4 \mathrm{e}}{\sqrt{5}}\right)^2\left(\mathrm{e}^2-1\right)\)
\(\Rightarrow 16 \mathrm{e}^2-28=\frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}\)
\(\Rightarrow \frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}-16 \mathrm{e}^2+28=0\)
\(\frac{16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140}{5}=0\)
\(16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140=0\)
\(16 \mathrm{e}^4-96 \mathrm{e}^2+140=0\)On dividing the above equation with 4
JEE Main 10.04.2019
Hyperbola
120751
If \(5 x+9=0\) is the directirx of the hyperbola \(16 x^2-9 y^2=144\), then its corresponding focus is
1 \(\left(-\frac{5}{3}, 0\right)\)
2 \((-5,0)\)
3 \(\left(\frac{5}{3}, 0\right)\)
4 \((5,0)\)
Explanation:
B Given,
The equation of the hyperbola
\(16 x^2-9 y^2=144\)
On dividing by 144 both sides
\(\frac{16 \mathrm{x}^2}{144}-\frac{9 \mathrm{y}^2}{144}=\frac{144}{144}\)
\(\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{16}=1\)
For eccentricity (e) in hyperbola is -
\(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(16=9\left(\mathrm{e}^2-1\right)\)
\(\frac{16}{9}+1=\mathrm{e}^2\)
\(\frac{16+9}{9}=\mathrm{e}^2\)
\(\frac{25}{9}=\mathrm{e}^2\)
\(\mathrm{e}=\frac{5}{3}\)
And, directirx \(5 \mathrm{x}+9=0\)
\(5 \mathrm{x}+9=0\)
\(5 \mathrm{x}=-9\)
\(\frac{-9}{5}=\frac{-\mathrm{a}}{\mathrm{e}} \text { is }\)
\(a=\frac{9}{5} \times \frac{5}{3}\)
\(a=3\)
Focus of hyperbola (-ae, 0 )
\(=\left(-3 \times \frac{5}{3}, 0\right)\)
\(=(-5,0)\)
JEE Main 10.04.2019
Hyperbola
120752
For some \(\theta \in\left(0, \frac{\pi}{2}\right)\). If the eccentricity of the hyperbola, \(x^2-y^2 \sec ^2 \theta=10\) is \(\sqrt{5}\) times the eccentricity of the ellipse, \(x^2 \sec ^2 \theta+y^2=5\), then the length of the latus rectum of the ellipse, is
120753
A hyperbola passes through the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is
1 \(\frac{x^2}{9}-\frac{y^2}{16}=1\)
2 \(\frac{x^2}{9}-\frac{y^2}{4}=1\)
3 \(\frac{x^2}{9}-\frac{y^2}{25}=1\)
4 \(x^2-y^2=9\)
Explanation:
A Given,
The ellipse, \(\frac{x^2}{25}+\frac{y^2}{16}=1\),
Comparing with equation,
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(a^2=25, b^2=16\)
Let, \(e_1\) is eccentricity of hyperbola-
\(\mathrm{e}_1=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}_1=\sqrt{1-\frac{16}{25}}\)
\(\mathrm{e}_1=\frac{3}{5}\)
Foci \(=( \pm 3,0) \quad[\because\) Foci \(=(\mathrm{ae}, 0)]\)
II- Let the equation of hyperbola be
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \quad[\therefore \text { passes through }( \pm 3,0)]\)
Let, \(\mathrm{e}_2\) is eccentricity of hyperbola
\(a^2=9\)
\(a=3\)
\(\therefore \quad \mathrm{e}_2 =\frac{5}{3}\)
\(\mathrm{e}_2{ }^2 =1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{25}{9} =1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{25}{9} -1 =\frac{\mathrm{b}^2}{9}\)
\(\frac{16}{9} =\frac{\mathrm{b}^2}{9}\)
\(\mathrm{~b}^2 =16\)
\(\therefore\) The equation of the hyperbola is
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Put, the value of \(a\) and \(b\) in \(\mathrm{eq}^{\mathrm{n}}\). (i) we get,
\(\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{16}=1\)
JEE Main 25.02.2021
Hyperbola
120754
If \(e_1\) and \(e_2\) are the eccentricities of the ellipse, \(\frac{x^2}{18}+\frac{y^2}{4}=1\) and the
hyperbola, \(\frac{x^2}{9}-\frac{y^2}{4}=1\) respectively and \(\left(e_1, e_2\right)\) is a point on the ellipse, \(15 x^2+3 y^2=k\), then \(k\) is equal to
1 14
2 15
3 17
4 16
Explanation:
D Given,
Case I: \(-\frac{x^2}{18}+\frac{y^2}{4}=1\)
\(\mathrm{a}^2=18, \mathrm{~b}^2=4\)
Eccentricity \(\left(\mathrm{e}_1\right)=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(=\sqrt{1-\frac{4}{18}}\)
\(=\sqrt{\frac{14}{18}}\)
\(\mathrm{e}_1=\frac{\sqrt{7}}{3}\)
Case II:- \(\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{4}=1\)
\(\mathrm{a}^2=9, \mathrm{~b}^2=4\)
Eccentricity \(\left(\mathrm{e}_2\right)=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}_2=\sqrt{1+\frac{4}{9}}\)
\(\mathrm{e}_2=\frac{\sqrt{13}}{3}\)
The value of \(e_1, e_2\) lies on ellipse,
\(\therefore \quad 15 \mathrm{x}^2+3 \mathrm{y}^2=\mathrm{k} \quad\left[\frac{\sqrt{7}}{3}, \frac{\sqrt{13}}{3}\right]\)
\(15\left(\frac{\sqrt{7}}{3}\right)^2+3\left(\frac{\sqrt{13}}{3}\right)^2=\mathrm{k}\)
\(15 \times \frac{7}{9}+3 \times \frac{13}{9}=\mathrm{k}\)
\(\frac{105}{9}+\frac{39}{9}=\mathrm{k}\)
\(\frac{105+39}{9}=\mathrm{k}\)
\(\frac{144}{9}=\mathrm{k}\)
\(\mathrm{k}=16\)
120750
If a directrix of a hyperbola centred at the origin and passing through the point \((4,-2 \sqrt{3})\) is \(5 x=4 \sqrt{5}\) and its eccentricity is e, then
1 \(4 \mathrm{e}^4-12 \mathrm{e}^2-27=0\)
2 \(4 \mathrm{e}^4-24 \mathrm{e}^2+27=0\)
3 \(4 \mathrm{e}^4+8 \mathrm{e}^2-35=0\)
4 \(4 \mathrm{e}^4-24 \mathrm{e}^2+35=0\)
Explanation:
D The given directrix of hyperbola is -
\(5 \mathrm{x}=4 \sqrt{5}\)
\(\mathrm{x}=\frac{4 \sqrt{5}}{5}\)
\(\therefore \frac{\mathrm{a}}{\mathrm{e}}=\frac{4}{\sqrt{5}}\)
Now, the hyperbola
Given,
below passes through \((4,-2 \sqrt{3})\)
Substituting the value of \(\mathrm{x}\) and \(\mathrm{y}\)
\(\frac{4^2}{\mathrm{a}^2}-\frac{(2 \sqrt{3})^2}{\mathrm{~b}^2}=1\)
\(\frac{16}{\mathrm{a}^2}-\frac{12}{\mathrm{a}^2\left(\mathrm{e}^2-1\right)}=1\)
\({\left[\because \mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2} \Rightarrow \mathrm{a}^2\left(\mathrm{e}^2-1\right)=\mathrm{b}^2\right]}\)
\(\frac{16}{\mathrm{a}^2}-\frac{12}{\mathrm{a}^2\left(\mathrm{e}^2-1\right)}=1\)
\(\frac{16\left(\mathrm{e}^2-1\right)-12}{\mathrm{a}^2\left(\mathrm{e}^2-1\right)}=1\)
\(16 \mathrm{e}^2-16-12=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(16 \mathrm{e}^2-28=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
On substituting \(\mathrm{a}=\frac{4 \mathrm{e}}{\sqrt{5}}\) in the above \(\mathrm{eq}^{\mathrm{n}}\), we get-
\(\Rightarrow 16 \mathrm{e}^2-28=\left(\frac{4 \mathrm{e}}{\sqrt{5}}\right)^2\left(\mathrm{e}^2-1\right)\)
\(\Rightarrow 16 \mathrm{e}^2-28=\frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}\)
\(\Rightarrow \frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}-16 \mathrm{e}^2+28=0\)
\(\frac{16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140}{5}=0\)
\(16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140=0\)
\(16 \mathrm{e}^4-96 \mathrm{e}^2+140=0\)
On dividing the above equation with \(4\)
\(\frac{16 \mathrm{e}^2}{4}-\frac{96 \mathrm{e}^2}{4}+\frac{140}{4}=0\)
\(4 \mathrm{e}^2-24 \mathrm{e}^2+35=0\)
\(\Rightarrow 16 \mathrm{e}^2-28=\left(\frac{4 \mathrm{e}}{\sqrt{5}}\right)^2\left(\mathrm{e}^2-1\right)\)
\(\Rightarrow 16 \mathrm{e}^2-28=\frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}\)
\(\Rightarrow \frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}-16 \mathrm{e}^2+28=0\)
\(\frac{16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140}{5}=0\)
\(16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140=0\)
\(16 \mathrm{e}^4-96 \mathrm{e}^2+140=0\)On dividing the above equation with 4
JEE Main 10.04.2019
Hyperbola
120751
If \(5 x+9=0\) is the directirx of the hyperbola \(16 x^2-9 y^2=144\), then its corresponding focus is
1 \(\left(-\frac{5}{3}, 0\right)\)
2 \((-5,0)\)
3 \(\left(\frac{5}{3}, 0\right)\)
4 \((5,0)\)
Explanation:
B Given,
The equation of the hyperbola
\(16 x^2-9 y^2=144\)
On dividing by 144 both sides
\(\frac{16 \mathrm{x}^2}{144}-\frac{9 \mathrm{y}^2}{144}=\frac{144}{144}\)
\(\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{16}=1\)
For eccentricity (e) in hyperbola is -
\(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(16=9\left(\mathrm{e}^2-1\right)\)
\(\frac{16}{9}+1=\mathrm{e}^2\)
\(\frac{16+9}{9}=\mathrm{e}^2\)
\(\frac{25}{9}=\mathrm{e}^2\)
\(\mathrm{e}=\frac{5}{3}\)
And, directirx \(5 \mathrm{x}+9=0\)
\(5 \mathrm{x}+9=0\)
\(5 \mathrm{x}=-9\)
\(\frac{-9}{5}=\frac{-\mathrm{a}}{\mathrm{e}} \text { is }\)
\(a=\frac{9}{5} \times \frac{5}{3}\)
\(a=3\)
Focus of hyperbola (-ae, 0 )
\(=\left(-3 \times \frac{5}{3}, 0\right)\)
\(=(-5,0)\)
JEE Main 10.04.2019
Hyperbola
120752
For some \(\theta \in\left(0, \frac{\pi}{2}\right)\). If the eccentricity of the hyperbola, \(x^2-y^2 \sec ^2 \theta=10\) is \(\sqrt{5}\) times the eccentricity of the ellipse, \(x^2 \sec ^2 \theta+y^2=5\), then the length of the latus rectum of the ellipse, is
120753
A hyperbola passes through the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is
1 \(\frac{x^2}{9}-\frac{y^2}{16}=1\)
2 \(\frac{x^2}{9}-\frac{y^2}{4}=1\)
3 \(\frac{x^2}{9}-\frac{y^2}{25}=1\)
4 \(x^2-y^2=9\)
Explanation:
A Given,
The ellipse, \(\frac{x^2}{25}+\frac{y^2}{16}=1\),
Comparing with equation,
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(a^2=25, b^2=16\)
Let, \(e_1\) is eccentricity of hyperbola-
\(\mathrm{e}_1=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}_1=\sqrt{1-\frac{16}{25}}\)
\(\mathrm{e}_1=\frac{3}{5}\)
Foci \(=( \pm 3,0) \quad[\because\) Foci \(=(\mathrm{ae}, 0)]\)
II- Let the equation of hyperbola be
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \quad[\therefore \text { passes through }( \pm 3,0)]\)
Let, \(\mathrm{e}_2\) is eccentricity of hyperbola
\(a^2=9\)
\(a=3\)
\(\therefore \quad \mathrm{e}_2 =\frac{5}{3}\)
\(\mathrm{e}_2{ }^2 =1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{25}{9} =1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{25}{9} -1 =\frac{\mathrm{b}^2}{9}\)
\(\frac{16}{9} =\frac{\mathrm{b}^2}{9}\)
\(\mathrm{~b}^2 =16\)
\(\therefore\) The equation of the hyperbola is
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Put, the value of \(a\) and \(b\) in \(\mathrm{eq}^{\mathrm{n}}\). (i) we get,
\(\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{16}=1\)
JEE Main 25.02.2021
Hyperbola
120754
If \(e_1\) and \(e_2\) are the eccentricities of the ellipse, \(\frac{x^2}{18}+\frac{y^2}{4}=1\) and the
hyperbola, \(\frac{x^2}{9}-\frac{y^2}{4}=1\) respectively and \(\left(e_1, e_2\right)\) is a point on the ellipse, \(15 x^2+3 y^2=k\), then \(k\) is equal to
1 14
2 15
3 17
4 16
Explanation:
D Given,
Case I: \(-\frac{x^2}{18}+\frac{y^2}{4}=1\)
\(\mathrm{a}^2=18, \mathrm{~b}^2=4\)
Eccentricity \(\left(\mathrm{e}_1\right)=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(=\sqrt{1-\frac{4}{18}}\)
\(=\sqrt{\frac{14}{18}}\)
\(\mathrm{e}_1=\frac{\sqrt{7}}{3}\)
Case II:- \(\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{4}=1\)
\(\mathrm{a}^2=9, \mathrm{~b}^2=4\)
Eccentricity \(\left(\mathrm{e}_2\right)=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}_2=\sqrt{1+\frac{4}{9}}\)
\(\mathrm{e}_2=\frac{\sqrt{13}}{3}\)
The value of \(e_1, e_2\) lies on ellipse,
\(\therefore \quad 15 \mathrm{x}^2+3 \mathrm{y}^2=\mathrm{k} \quad\left[\frac{\sqrt{7}}{3}, \frac{\sqrt{13}}{3}\right]\)
\(15\left(\frac{\sqrt{7}}{3}\right)^2+3\left(\frac{\sqrt{13}}{3}\right)^2=\mathrm{k}\)
\(15 \times \frac{7}{9}+3 \times \frac{13}{9}=\mathrm{k}\)
\(\frac{105}{9}+\frac{39}{9}=\mathrm{k}\)
\(\frac{105+39}{9}=\mathrm{k}\)
\(\frac{144}{9}=\mathrm{k}\)
\(\mathrm{k}=16\)
120750
If a directrix of a hyperbola centred at the origin and passing through the point \((4,-2 \sqrt{3})\) is \(5 x=4 \sqrt{5}\) and its eccentricity is e, then
1 \(4 \mathrm{e}^4-12 \mathrm{e}^2-27=0\)
2 \(4 \mathrm{e}^4-24 \mathrm{e}^2+27=0\)
3 \(4 \mathrm{e}^4+8 \mathrm{e}^2-35=0\)
4 \(4 \mathrm{e}^4-24 \mathrm{e}^2+35=0\)
Explanation:
D The given directrix of hyperbola is -
\(5 \mathrm{x}=4 \sqrt{5}\)
\(\mathrm{x}=\frac{4 \sqrt{5}}{5}\)
\(\therefore \frac{\mathrm{a}}{\mathrm{e}}=\frac{4}{\sqrt{5}}\)
Now, the hyperbola
Given,
below passes through \((4,-2 \sqrt{3})\)
Substituting the value of \(\mathrm{x}\) and \(\mathrm{y}\)
\(\frac{4^2}{\mathrm{a}^2}-\frac{(2 \sqrt{3})^2}{\mathrm{~b}^2}=1\)
\(\frac{16}{\mathrm{a}^2}-\frac{12}{\mathrm{a}^2\left(\mathrm{e}^2-1\right)}=1\)
\({\left[\because \mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2} \Rightarrow \mathrm{a}^2\left(\mathrm{e}^2-1\right)=\mathrm{b}^2\right]}\)
\(\frac{16}{\mathrm{a}^2}-\frac{12}{\mathrm{a}^2\left(\mathrm{e}^2-1\right)}=1\)
\(\frac{16\left(\mathrm{e}^2-1\right)-12}{\mathrm{a}^2\left(\mathrm{e}^2-1\right)}=1\)
\(16 \mathrm{e}^2-16-12=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(16 \mathrm{e}^2-28=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
On substituting \(\mathrm{a}=\frac{4 \mathrm{e}}{\sqrt{5}}\) in the above \(\mathrm{eq}^{\mathrm{n}}\), we get-
\(\Rightarrow 16 \mathrm{e}^2-28=\left(\frac{4 \mathrm{e}}{\sqrt{5}}\right)^2\left(\mathrm{e}^2-1\right)\)
\(\Rightarrow 16 \mathrm{e}^2-28=\frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}\)
\(\Rightarrow \frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}-16 \mathrm{e}^2+28=0\)
\(\frac{16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140}{5}=0\)
\(16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140=0\)
\(16 \mathrm{e}^4-96 \mathrm{e}^2+140=0\)
On dividing the above equation with \(4\)
\(\frac{16 \mathrm{e}^2}{4}-\frac{96 \mathrm{e}^2}{4}+\frac{140}{4}=0\)
\(4 \mathrm{e}^2-24 \mathrm{e}^2+35=0\)
\(\Rightarrow 16 \mathrm{e}^2-28=\left(\frac{4 \mathrm{e}}{\sqrt{5}}\right)^2\left(\mathrm{e}^2-1\right)\)
\(\Rightarrow 16 \mathrm{e}^2-28=\frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}\)
\(\Rightarrow \frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}-16 \mathrm{e}^2+28=0\)
\(\frac{16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140}{5}=0\)
\(16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140=0\)
\(16 \mathrm{e}^4-96 \mathrm{e}^2+140=0\)On dividing the above equation with 4
JEE Main 10.04.2019
Hyperbola
120751
If \(5 x+9=0\) is the directirx of the hyperbola \(16 x^2-9 y^2=144\), then its corresponding focus is
1 \(\left(-\frac{5}{3}, 0\right)\)
2 \((-5,0)\)
3 \(\left(\frac{5}{3}, 0\right)\)
4 \((5,0)\)
Explanation:
B Given,
The equation of the hyperbola
\(16 x^2-9 y^2=144\)
On dividing by 144 both sides
\(\frac{16 \mathrm{x}^2}{144}-\frac{9 \mathrm{y}^2}{144}=\frac{144}{144}\)
\(\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{16}=1\)
For eccentricity (e) in hyperbola is -
\(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(16=9\left(\mathrm{e}^2-1\right)\)
\(\frac{16}{9}+1=\mathrm{e}^2\)
\(\frac{16+9}{9}=\mathrm{e}^2\)
\(\frac{25}{9}=\mathrm{e}^2\)
\(\mathrm{e}=\frac{5}{3}\)
And, directirx \(5 \mathrm{x}+9=0\)
\(5 \mathrm{x}+9=0\)
\(5 \mathrm{x}=-9\)
\(\frac{-9}{5}=\frac{-\mathrm{a}}{\mathrm{e}} \text { is }\)
\(a=\frac{9}{5} \times \frac{5}{3}\)
\(a=3\)
Focus of hyperbola (-ae, 0 )
\(=\left(-3 \times \frac{5}{3}, 0\right)\)
\(=(-5,0)\)
JEE Main 10.04.2019
Hyperbola
120752
For some \(\theta \in\left(0, \frac{\pi}{2}\right)\). If the eccentricity of the hyperbola, \(x^2-y^2 \sec ^2 \theta=10\) is \(\sqrt{5}\) times the eccentricity of the ellipse, \(x^2 \sec ^2 \theta+y^2=5\), then the length of the latus rectum of the ellipse, is
120753
A hyperbola passes through the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is
1 \(\frac{x^2}{9}-\frac{y^2}{16}=1\)
2 \(\frac{x^2}{9}-\frac{y^2}{4}=1\)
3 \(\frac{x^2}{9}-\frac{y^2}{25}=1\)
4 \(x^2-y^2=9\)
Explanation:
A Given,
The ellipse, \(\frac{x^2}{25}+\frac{y^2}{16}=1\),
Comparing with equation,
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(a^2=25, b^2=16\)
Let, \(e_1\) is eccentricity of hyperbola-
\(\mathrm{e}_1=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}_1=\sqrt{1-\frac{16}{25}}\)
\(\mathrm{e}_1=\frac{3}{5}\)
Foci \(=( \pm 3,0) \quad[\because\) Foci \(=(\mathrm{ae}, 0)]\)
II- Let the equation of hyperbola be
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \quad[\therefore \text { passes through }( \pm 3,0)]\)
Let, \(\mathrm{e}_2\) is eccentricity of hyperbola
\(a^2=9\)
\(a=3\)
\(\therefore \quad \mathrm{e}_2 =\frac{5}{3}\)
\(\mathrm{e}_2{ }^2 =1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{25}{9} =1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{25}{9} -1 =\frac{\mathrm{b}^2}{9}\)
\(\frac{16}{9} =\frac{\mathrm{b}^2}{9}\)
\(\mathrm{~b}^2 =16\)
\(\therefore\) The equation of the hyperbola is
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Put, the value of \(a\) and \(b\) in \(\mathrm{eq}^{\mathrm{n}}\). (i) we get,
\(\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{16}=1\)
JEE Main 25.02.2021
Hyperbola
120754
If \(e_1\) and \(e_2\) are the eccentricities of the ellipse, \(\frac{x^2}{18}+\frac{y^2}{4}=1\) and the
hyperbola, \(\frac{x^2}{9}-\frac{y^2}{4}=1\) respectively and \(\left(e_1, e_2\right)\) is a point on the ellipse, \(15 x^2+3 y^2=k\), then \(k\) is equal to
1 14
2 15
3 17
4 16
Explanation:
D Given,
Case I: \(-\frac{x^2}{18}+\frac{y^2}{4}=1\)
\(\mathrm{a}^2=18, \mathrm{~b}^2=4\)
Eccentricity \(\left(\mathrm{e}_1\right)=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(=\sqrt{1-\frac{4}{18}}\)
\(=\sqrt{\frac{14}{18}}\)
\(\mathrm{e}_1=\frac{\sqrt{7}}{3}\)
Case II:- \(\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{4}=1\)
\(\mathrm{a}^2=9, \mathrm{~b}^2=4\)
Eccentricity \(\left(\mathrm{e}_2\right)=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}_2=\sqrt{1+\frac{4}{9}}\)
\(\mathrm{e}_2=\frac{\sqrt{13}}{3}\)
The value of \(e_1, e_2\) lies on ellipse,
\(\therefore \quad 15 \mathrm{x}^2+3 \mathrm{y}^2=\mathrm{k} \quad\left[\frac{\sqrt{7}}{3}, \frac{\sqrt{13}}{3}\right]\)
\(15\left(\frac{\sqrt{7}}{3}\right)^2+3\left(\frac{\sqrt{13}}{3}\right)^2=\mathrm{k}\)
\(15 \times \frac{7}{9}+3 \times \frac{13}{9}=\mathrm{k}\)
\(\frac{105}{9}+\frac{39}{9}=\mathrm{k}\)
\(\frac{105+39}{9}=\mathrm{k}\)
\(\frac{144}{9}=\mathrm{k}\)
\(\mathrm{k}=16\)
120750
If a directrix of a hyperbola centred at the origin and passing through the point \((4,-2 \sqrt{3})\) is \(5 x=4 \sqrt{5}\) and its eccentricity is e, then
1 \(4 \mathrm{e}^4-12 \mathrm{e}^2-27=0\)
2 \(4 \mathrm{e}^4-24 \mathrm{e}^2+27=0\)
3 \(4 \mathrm{e}^4+8 \mathrm{e}^2-35=0\)
4 \(4 \mathrm{e}^4-24 \mathrm{e}^2+35=0\)
Explanation:
D The given directrix of hyperbola is -
\(5 \mathrm{x}=4 \sqrt{5}\)
\(\mathrm{x}=\frac{4 \sqrt{5}}{5}\)
\(\therefore \frac{\mathrm{a}}{\mathrm{e}}=\frac{4}{\sqrt{5}}\)
Now, the hyperbola
Given,
below passes through \((4,-2 \sqrt{3})\)
Substituting the value of \(\mathrm{x}\) and \(\mathrm{y}\)
\(\frac{4^2}{\mathrm{a}^2}-\frac{(2 \sqrt{3})^2}{\mathrm{~b}^2}=1\)
\(\frac{16}{\mathrm{a}^2}-\frac{12}{\mathrm{a}^2\left(\mathrm{e}^2-1\right)}=1\)
\({\left[\because \mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2} \Rightarrow \mathrm{a}^2\left(\mathrm{e}^2-1\right)=\mathrm{b}^2\right]}\)
\(\frac{16}{\mathrm{a}^2}-\frac{12}{\mathrm{a}^2\left(\mathrm{e}^2-1\right)}=1\)
\(\frac{16\left(\mathrm{e}^2-1\right)-12}{\mathrm{a}^2\left(\mathrm{e}^2-1\right)}=1\)
\(16 \mathrm{e}^2-16-12=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(16 \mathrm{e}^2-28=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
On substituting \(\mathrm{a}=\frac{4 \mathrm{e}}{\sqrt{5}}\) in the above \(\mathrm{eq}^{\mathrm{n}}\), we get-
\(\Rightarrow 16 \mathrm{e}^2-28=\left(\frac{4 \mathrm{e}}{\sqrt{5}}\right)^2\left(\mathrm{e}^2-1\right)\)
\(\Rightarrow 16 \mathrm{e}^2-28=\frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}\)
\(\Rightarrow \frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}-16 \mathrm{e}^2+28=0\)
\(\frac{16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140}{5}=0\)
\(16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140=0\)
\(16 \mathrm{e}^4-96 \mathrm{e}^2+140=0\)
On dividing the above equation with \(4\)
\(\frac{16 \mathrm{e}^2}{4}-\frac{96 \mathrm{e}^2}{4}+\frac{140}{4}=0\)
\(4 \mathrm{e}^2-24 \mathrm{e}^2+35=0\)
\(\Rightarrow 16 \mathrm{e}^2-28=\left(\frac{4 \mathrm{e}}{\sqrt{5}}\right)^2\left(\mathrm{e}^2-1\right)\)
\(\Rightarrow 16 \mathrm{e}^2-28=\frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}\)
\(\Rightarrow \frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}-16 \mathrm{e}^2+28=0\)
\(\frac{16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140}{5}=0\)
\(16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140=0\)
\(16 \mathrm{e}^4-96 \mathrm{e}^2+140=0\)On dividing the above equation with 4
JEE Main 10.04.2019
Hyperbola
120751
If \(5 x+9=0\) is the directirx of the hyperbola \(16 x^2-9 y^2=144\), then its corresponding focus is
1 \(\left(-\frac{5}{3}, 0\right)\)
2 \((-5,0)\)
3 \(\left(\frac{5}{3}, 0\right)\)
4 \((5,0)\)
Explanation:
B Given,
The equation of the hyperbola
\(16 x^2-9 y^2=144\)
On dividing by 144 both sides
\(\frac{16 \mathrm{x}^2}{144}-\frac{9 \mathrm{y}^2}{144}=\frac{144}{144}\)
\(\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{16}=1\)
For eccentricity (e) in hyperbola is -
\(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(16=9\left(\mathrm{e}^2-1\right)\)
\(\frac{16}{9}+1=\mathrm{e}^2\)
\(\frac{16+9}{9}=\mathrm{e}^2\)
\(\frac{25}{9}=\mathrm{e}^2\)
\(\mathrm{e}=\frac{5}{3}\)
And, directirx \(5 \mathrm{x}+9=0\)
\(5 \mathrm{x}+9=0\)
\(5 \mathrm{x}=-9\)
\(\frac{-9}{5}=\frac{-\mathrm{a}}{\mathrm{e}} \text { is }\)
\(a=\frac{9}{5} \times \frac{5}{3}\)
\(a=3\)
Focus of hyperbola (-ae, 0 )
\(=\left(-3 \times \frac{5}{3}, 0\right)\)
\(=(-5,0)\)
JEE Main 10.04.2019
Hyperbola
120752
For some \(\theta \in\left(0, \frac{\pi}{2}\right)\). If the eccentricity of the hyperbola, \(x^2-y^2 \sec ^2 \theta=10\) is \(\sqrt{5}\) times the eccentricity of the ellipse, \(x^2 \sec ^2 \theta+y^2=5\), then the length of the latus rectum of the ellipse, is
120753
A hyperbola passes through the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is
1 \(\frac{x^2}{9}-\frac{y^2}{16}=1\)
2 \(\frac{x^2}{9}-\frac{y^2}{4}=1\)
3 \(\frac{x^2}{9}-\frac{y^2}{25}=1\)
4 \(x^2-y^2=9\)
Explanation:
A Given,
The ellipse, \(\frac{x^2}{25}+\frac{y^2}{16}=1\),
Comparing with equation,
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(a^2=25, b^2=16\)
Let, \(e_1\) is eccentricity of hyperbola-
\(\mathrm{e}_1=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}_1=\sqrt{1-\frac{16}{25}}\)
\(\mathrm{e}_1=\frac{3}{5}\)
Foci \(=( \pm 3,0) \quad[\because\) Foci \(=(\mathrm{ae}, 0)]\)
II- Let the equation of hyperbola be
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \quad[\therefore \text { passes through }( \pm 3,0)]\)
Let, \(\mathrm{e}_2\) is eccentricity of hyperbola
\(a^2=9\)
\(a=3\)
\(\therefore \quad \mathrm{e}_2 =\frac{5}{3}\)
\(\mathrm{e}_2{ }^2 =1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{25}{9} =1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{25}{9} -1 =\frac{\mathrm{b}^2}{9}\)
\(\frac{16}{9} =\frac{\mathrm{b}^2}{9}\)
\(\mathrm{~b}^2 =16\)
\(\therefore\) The equation of the hyperbola is
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Put, the value of \(a\) and \(b\) in \(\mathrm{eq}^{\mathrm{n}}\). (i) we get,
\(\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{16}=1\)
JEE Main 25.02.2021
Hyperbola
120754
If \(e_1\) and \(e_2\) are the eccentricities of the ellipse, \(\frac{x^2}{18}+\frac{y^2}{4}=1\) and the
hyperbola, \(\frac{x^2}{9}-\frac{y^2}{4}=1\) respectively and \(\left(e_1, e_2\right)\) is a point on the ellipse, \(15 x^2+3 y^2=k\), then \(k\) is equal to
1 14
2 15
3 17
4 16
Explanation:
D Given,
Case I: \(-\frac{x^2}{18}+\frac{y^2}{4}=1\)
\(\mathrm{a}^2=18, \mathrm{~b}^2=4\)
Eccentricity \(\left(\mathrm{e}_1\right)=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(=\sqrt{1-\frac{4}{18}}\)
\(=\sqrt{\frac{14}{18}}\)
\(\mathrm{e}_1=\frac{\sqrt{7}}{3}\)
Case II:- \(\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{4}=1\)
\(\mathrm{a}^2=9, \mathrm{~b}^2=4\)
Eccentricity \(\left(\mathrm{e}_2\right)=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}_2=\sqrt{1+\frac{4}{9}}\)
\(\mathrm{e}_2=\frac{\sqrt{13}}{3}\)
The value of \(e_1, e_2\) lies on ellipse,
\(\therefore \quad 15 \mathrm{x}^2+3 \mathrm{y}^2=\mathrm{k} \quad\left[\frac{\sqrt{7}}{3}, \frac{\sqrt{13}}{3}\right]\)
\(15\left(\frac{\sqrt{7}}{3}\right)^2+3\left(\frac{\sqrt{13}}{3}\right)^2=\mathrm{k}\)
\(15 \times \frac{7}{9}+3 \times \frac{13}{9}=\mathrm{k}\)
\(\frac{105}{9}+\frac{39}{9}=\mathrm{k}\)
\(\frac{105+39}{9}=\mathrm{k}\)
\(\frac{144}{9}=\mathrm{k}\)
\(\mathrm{k}=16\)
120750
If a directrix of a hyperbola centred at the origin and passing through the point \((4,-2 \sqrt{3})\) is \(5 x=4 \sqrt{5}\) and its eccentricity is e, then
1 \(4 \mathrm{e}^4-12 \mathrm{e}^2-27=0\)
2 \(4 \mathrm{e}^4-24 \mathrm{e}^2+27=0\)
3 \(4 \mathrm{e}^4+8 \mathrm{e}^2-35=0\)
4 \(4 \mathrm{e}^4-24 \mathrm{e}^2+35=0\)
Explanation:
D The given directrix of hyperbola is -
\(5 \mathrm{x}=4 \sqrt{5}\)
\(\mathrm{x}=\frac{4 \sqrt{5}}{5}\)
\(\therefore \frac{\mathrm{a}}{\mathrm{e}}=\frac{4}{\sqrt{5}}\)
Now, the hyperbola
Given,
below passes through \((4,-2 \sqrt{3})\)
Substituting the value of \(\mathrm{x}\) and \(\mathrm{y}\)
\(\frac{4^2}{\mathrm{a}^2}-\frac{(2 \sqrt{3})^2}{\mathrm{~b}^2}=1\)
\(\frac{16}{\mathrm{a}^2}-\frac{12}{\mathrm{a}^2\left(\mathrm{e}^2-1\right)}=1\)
\({\left[\because \mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2} \Rightarrow \mathrm{a}^2\left(\mathrm{e}^2-1\right)=\mathrm{b}^2\right]}\)
\(\frac{16}{\mathrm{a}^2}-\frac{12}{\mathrm{a}^2\left(\mathrm{e}^2-1\right)}=1\)
\(\frac{16\left(\mathrm{e}^2-1\right)-12}{\mathrm{a}^2\left(\mathrm{e}^2-1\right)}=1\)
\(16 \mathrm{e}^2-16-12=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(16 \mathrm{e}^2-28=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
On substituting \(\mathrm{a}=\frac{4 \mathrm{e}}{\sqrt{5}}\) in the above \(\mathrm{eq}^{\mathrm{n}}\), we get-
\(\Rightarrow 16 \mathrm{e}^2-28=\left(\frac{4 \mathrm{e}}{\sqrt{5}}\right)^2\left(\mathrm{e}^2-1\right)\)
\(\Rightarrow 16 \mathrm{e}^2-28=\frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}\)
\(\Rightarrow \frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}-16 \mathrm{e}^2+28=0\)
\(\frac{16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140}{5}=0\)
\(16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140=0\)
\(16 \mathrm{e}^4-96 \mathrm{e}^2+140=0\)
On dividing the above equation with \(4\)
\(\frac{16 \mathrm{e}^2}{4}-\frac{96 \mathrm{e}^2}{4}+\frac{140}{4}=0\)
\(4 \mathrm{e}^2-24 \mathrm{e}^2+35=0\)
\(\Rightarrow 16 \mathrm{e}^2-28=\left(\frac{4 \mathrm{e}}{\sqrt{5}}\right)^2\left(\mathrm{e}^2-1\right)\)
\(\Rightarrow 16 \mathrm{e}^2-28=\frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}\)
\(\Rightarrow \frac{16 \mathrm{e}^4}{5}-\frac{16 \mathrm{e}^2}{5}-16 \mathrm{e}^2+28=0\)
\(\frac{16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140}{5}=0\)
\(16 \mathrm{e}^4-16 \mathrm{e}^2-80 \mathrm{e}^2+140=0\)
\(16 \mathrm{e}^4-96 \mathrm{e}^2+140=0\)On dividing the above equation with 4
JEE Main 10.04.2019
Hyperbola
120751
If \(5 x+9=0\) is the directirx of the hyperbola \(16 x^2-9 y^2=144\), then its corresponding focus is
1 \(\left(-\frac{5}{3}, 0\right)\)
2 \((-5,0)\)
3 \(\left(\frac{5}{3}, 0\right)\)
4 \((5,0)\)
Explanation:
B Given,
The equation of the hyperbola
\(16 x^2-9 y^2=144\)
On dividing by 144 both sides
\(\frac{16 \mathrm{x}^2}{144}-\frac{9 \mathrm{y}^2}{144}=\frac{144}{144}\)
\(\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{16}=1\)
For eccentricity (e) in hyperbola is -
\(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(16=9\left(\mathrm{e}^2-1\right)\)
\(\frac{16}{9}+1=\mathrm{e}^2\)
\(\frac{16+9}{9}=\mathrm{e}^2\)
\(\frac{25}{9}=\mathrm{e}^2\)
\(\mathrm{e}=\frac{5}{3}\)
And, directirx \(5 \mathrm{x}+9=0\)
\(5 \mathrm{x}+9=0\)
\(5 \mathrm{x}=-9\)
\(\frac{-9}{5}=\frac{-\mathrm{a}}{\mathrm{e}} \text { is }\)
\(a=\frac{9}{5} \times \frac{5}{3}\)
\(a=3\)
Focus of hyperbola (-ae, 0 )
\(=\left(-3 \times \frac{5}{3}, 0\right)\)
\(=(-5,0)\)
JEE Main 10.04.2019
Hyperbola
120752
For some \(\theta \in\left(0, \frac{\pi}{2}\right)\). If the eccentricity of the hyperbola, \(x^2-y^2 \sec ^2 \theta=10\) is \(\sqrt{5}\) times the eccentricity of the ellipse, \(x^2 \sec ^2 \theta+y^2=5\), then the length of the latus rectum of the ellipse, is
120753
A hyperbola passes through the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is
1 \(\frac{x^2}{9}-\frac{y^2}{16}=1\)
2 \(\frac{x^2}{9}-\frac{y^2}{4}=1\)
3 \(\frac{x^2}{9}-\frac{y^2}{25}=1\)
4 \(x^2-y^2=9\)
Explanation:
A Given,
The ellipse, \(\frac{x^2}{25}+\frac{y^2}{16}=1\),
Comparing with equation,
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(a^2=25, b^2=16\)
Let, \(e_1\) is eccentricity of hyperbola-
\(\mathrm{e}_1=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}_1=\sqrt{1-\frac{16}{25}}\)
\(\mathrm{e}_1=\frac{3}{5}\)
Foci \(=( \pm 3,0) \quad[\because\) Foci \(=(\mathrm{ae}, 0)]\)
II- Let the equation of hyperbola be
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \quad[\therefore \text { passes through }( \pm 3,0)]\)
Let, \(\mathrm{e}_2\) is eccentricity of hyperbola
\(a^2=9\)
\(a=3\)
\(\therefore \quad \mathrm{e}_2 =\frac{5}{3}\)
\(\mathrm{e}_2{ }^2 =1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{25}{9} =1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{25}{9} -1 =\frac{\mathrm{b}^2}{9}\)
\(\frac{16}{9} =\frac{\mathrm{b}^2}{9}\)
\(\mathrm{~b}^2 =16\)
\(\therefore\) The equation of the hyperbola is
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Put, the value of \(a\) and \(b\) in \(\mathrm{eq}^{\mathrm{n}}\). (i) we get,
\(\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{16}=1\)
JEE Main 25.02.2021
Hyperbola
120754
If \(e_1\) and \(e_2\) are the eccentricities of the ellipse, \(\frac{x^2}{18}+\frac{y^2}{4}=1\) and the
hyperbola, \(\frac{x^2}{9}-\frac{y^2}{4}=1\) respectively and \(\left(e_1, e_2\right)\) is a point on the ellipse, \(15 x^2+3 y^2=k\), then \(k\) is equal to
1 14
2 15
3 17
4 16
Explanation:
D Given,
Case I: \(-\frac{x^2}{18}+\frac{y^2}{4}=1\)
\(\mathrm{a}^2=18, \mathrm{~b}^2=4\)
Eccentricity \(\left(\mathrm{e}_1\right)=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(=\sqrt{1-\frac{4}{18}}\)
\(=\sqrt{\frac{14}{18}}\)
\(\mathrm{e}_1=\frac{\sqrt{7}}{3}\)
Case II:- \(\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{4}=1\)
\(\mathrm{a}^2=9, \mathrm{~b}^2=4\)
Eccentricity \(\left(\mathrm{e}_2\right)=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}_2=\sqrt{1+\frac{4}{9}}\)
\(\mathrm{e}_2=\frac{\sqrt{13}}{3}\)
The value of \(e_1, e_2\) lies on ellipse,
\(\therefore \quad 15 \mathrm{x}^2+3 \mathrm{y}^2=\mathrm{k} \quad\left[\frac{\sqrt{7}}{3}, \frac{\sqrt{13}}{3}\right]\)
\(15\left(\frac{\sqrt{7}}{3}\right)^2+3\left(\frac{\sqrt{13}}{3}\right)^2=\mathrm{k}\)
\(15 \times \frac{7}{9}+3 \times \frac{13}{9}=\mathrm{k}\)
\(\frac{105}{9}+\frac{39}{9}=\mathrm{k}\)
\(\frac{105+39}{9}=\mathrm{k}\)
\(\frac{144}{9}=\mathrm{k}\)
\(\mathrm{k}=16\)
