B Given, equation of hyperbola is -
\(9 x^2-16 y^2=144 \quad \text { or } \frac{x^2}{16}-\frac{y^2}{9}=1\)
The equation of tangent to hyperbola having slope \(\mathrm{m}\) is \(\mathrm{y}=\mathrm{mx} \pm \sqrt{16 \mathrm{~m}^2-9}\)
It touches the circle, then the distance of this line from the centre of circle is the radius of the circle, hence,
\(\Rightarrow \frac{\sqrt{16 \mathrm{~m}^2-9}}{\sqrt{\mathrm{m}^2+1}}=3\)
\(\Rightarrow 9 \mathrm{~m}^2+9=16 \mathrm{~m}^2-9\)
\(\Rightarrow 7 \mathrm{~m}^2=18\)
\(\Rightarrow \mathrm{m}= \pm 3 \sqrt{\frac{2}{7}}\)
So, equation of the tangent is
\(\mathrm{y}= \pm 3 \sqrt{\frac{2}{7}} \mathrm{x}+\frac{15}{\sqrt{7}}\)Hence, option (b) is correct.
JEEE-2013
Hyperbola
120772
If the distance between the foci and distance between the directrices of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) are in the ratio \(3: 2\) then \(a: b\) is
1 \(\sqrt{2}: 1\)
2 \(1: 2\)
3 \(\sqrt{3}: \sqrt{2}\)
4 \(2: 1\)
Explanation:
A Given,
Equation of hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
Distance between directrices \(=\frac{2 \mathrm{a}}{\mathrm{e}}\)
and distance between foci \(=2 \mathrm{ae}\)
According to the question, we have
\(\frac{2 \mathrm{ae}}{\frac{2 \mathrm{a}}{\mathrm{e}}}=\frac{3}{2}\)
\(\Rightarrow \quad \mathrm{e}^2=\frac{3}{2}\)
\(\therefore \quad \mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\Rightarrow \quad \frac{\mathrm{b}}{\mathrm{a}}=\frac{1}{\sqrt{2}}\)Hence, the ratio of \(a: b\) is \(\sqrt{2}: 1\)
Karnataka CET-2013
Hyperbola
120773
The value of \(m\) for which \(y=m x+6\) is a tangent to the hyperbola \(\frac{x^2}{100}-\frac{y^2}{49}=1\) is
1 \(\sqrt{\frac{20}{30}}\)
2 \(\sqrt{\frac{20}{17}}\)
3 \(\sqrt{\frac{17}{20}}\)
4 \(\sqrt{\frac{3}{20}}\)
Explanation:
C Given, equation of the hyperbola-
\(\frac{x^2}{100}-\frac{y^2}{9}=1\)
and tangent is, \(\mathrm{y}=\mathrm{mx}+6\)
For this to be tangent, \(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\)
\(\Rightarrow 6^2= (100) \mathrm{m}^2-49\)
\(\Rightarrow 36+49 =100 \mathrm{~m}^2\)
\(\Rightarrow 85 =100 \mathrm{~m}^2\)
\(\Rightarrow \mathrm{m}^2 =\frac{85}{100}\)
\(\Rightarrow \mathrm{m} =\sqrt{\frac{17}{20}}\)
COMEDK-2016
Hyperbola
120774
Chords of the circle \(x^2+y^2=r^2\) touch the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{16}=1\). The locus of the midpoints of the chords is
1 \(\left(x^2+y^2\right)^2=a^2 x^2-b^2 y^2\)
2 \(\left(x^2+y^2\right)^2=a^2 x^2+b^2 y^2\)
3 \(x^2+y^2=a^2 x^2-b^2 y^2\)
4 \(x^2+y^2=a^2 x^2+b^2 y^2\)
Explanation:
A Let \(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be the midpoint of a chord of \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{r}^2\).
The equation of the chord is given by
\(S_1=S\)
\(\Rightarrow x_1 x+y_1 y=x_1^2+y_1^2\)
\(\text { Or } y=\left(-\frac{x_1}{y_1}\right) x+\frac{x_1^2+y_1^2}{y_1}\)
The line \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) is a tangent to the hyperbola only if \(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\).
\(\therefore \frac{\left(\mathrm{x}_1^2+\mathrm{y}_1^2\right)^2}{\mathrm{y}_1^2}=\frac{\mathrm{a}^2 \mathrm{x}_1^2}{\mathrm{y}_1^2}-\mathrm{b}^2\)
\(\therefore\) The locus of \(\mathrm{P}\) is \(\left(\mathrm{x}^2+\mathrm{y}^2\right)^2=\mathrm{a}^2 \mathrm{x}^2-\mathrm{b}^2 \mathrm{y}^2\).
COMEDK-2017
Hyperbola
120775
The line \(21 x+5 y=116\) is a tangent to the hyperbola \(7 x^2-5 y^2=232\). Its point of contact is
1 \((-6,-2)\)
2 \((6,2)\)
3 \((6,-2)\)
4 \((-6,2)\)
Explanation:
C The line \(21 x+5 y=116\) is a tangent to \(7 x^2-5 y^2\) \(=232\).
If \(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is the point of contact, then \(\mathrm{S}_1=0\).
\(\Rightarrow 7 \mathrm{x}_1 \mathrm{x}-5 \mathrm{y}_1 \mathrm{y}=232 \text {. }\)
It is same as the equation of the given tangent
\(21 x+5 y=116\)
On comparing both the equations,
\(\therefore \frac{7 \mathrm{x}_1}{21}=-\frac{5 \mathrm{y}_1}{5}=\frac{232}{116}=2\)
\(\therefore \mathrm{x}_1=6, \mathrm{y}_1=-2 .\)So, \(\mathrm{P}\) is \((6,-2)\).
B Given, equation of hyperbola is -
\(9 x^2-16 y^2=144 \quad \text { or } \frac{x^2}{16}-\frac{y^2}{9}=1\)
The equation of tangent to hyperbola having slope \(\mathrm{m}\) is \(\mathrm{y}=\mathrm{mx} \pm \sqrt{16 \mathrm{~m}^2-9}\)
It touches the circle, then the distance of this line from the centre of circle is the radius of the circle, hence,
\(\Rightarrow \frac{\sqrt{16 \mathrm{~m}^2-9}}{\sqrt{\mathrm{m}^2+1}}=3\)
\(\Rightarrow 9 \mathrm{~m}^2+9=16 \mathrm{~m}^2-9\)
\(\Rightarrow 7 \mathrm{~m}^2=18\)
\(\Rightarrow \mathrm{m}= \pm 3 \sqrt{\frac{2}{7}}\)
So, equation of the tangent is
\(\mathrm{y}= \pm 3 \sqrt{\frac{2}{7}} \mathrm{x}+\frac{15}{\sqrt{7}}\)Hence, option (b) is correct.
JEEE-2013
Hyperbola
120772
If the distance between the foci and distance between the directrices of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) are in the ratio \(3: 2\) then \(a: b\) is
1 \(\sqrt{2}: 1\)
2 \(1: 2\)
3 \(\sqrt{3}: \sqrt{2}\)
4 \(2: 1\)
Explanation:
A Given,
Equation of hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
Distance between directrices \(=\frac{2 \mathrm{a}}{\mathrm{e}}\)
and distance between foci \(=2 \mathrm{ae}\)
According to the question, we have
\(\frac{2 \mathrm{ae}}{\frac{2 \mathrm{a}}{\mathrm{e}}}=\frac{3}{2}\)
\(\Rightarrow \quad \mathrm{e}^2=\frac{3}{2}\)
\(\therefore \quad \mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\Rightarrow \quad \frac{\mathrm{b}}{\mathrm{a}}=\frac{1}{\sqrt{2}}\)Hence, the ratio of \(a: b\) is \(\sqrt{2}: 1\)
Karnataka CET-2013
Hyperbola
120773
The value of \(m\) for which \(y=m x+6\) is a tangent to the hyperbola \(\frac{x^2}{100}-\frac{y^2}{49}=1\) is
1 \(\sqrt{\frac{20}{30}}\)
2 \(\sqrt{\frac{20}{17}}\)
3 \(\sqrt{\frac{17}{20}}\)
4 \(\sqrt{\frac{3}{20}}\)
Explanation:
C Given, equation of the hyperbola-
\(\frac{x^2}{100}-\frac{y^2}{9}=1\)
and tangent is, \(\mathrm{y}=\mathrm{mx}+6\)
For this to be tangent, \(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\)
\(\Rightarrow 6^2= (100) \mathrm{m}^2-49\)
\(\Rightarrow 36+49 =100 \mathrm{~m}^2\)
\(\Rightarrow 85 =100 \mathrm{~m}^2\)
\(\Rightarrow \mathrm{m}^2 =\frac{85}{100}\)
\(\Rightarrow \mathrm{m} =\sqrt{\frac{17}{20}}\)
COMEDK-2016
Hyperbola
120774
Chords of the circle \(x^2+y^2=r^2\) touch the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{16}=1\). The locus of the midpoints of the chords is
1 \(\left(x^2+y^2\right)^2=a^2 x^2-b^2 y^2\)
2 \(\left(x^2+y^2\right)^2=a^2 x^2+b^2 y^2\)
3 \(x^2+y^2=a^2 x^2-b^2 y^2\)
4 \(x^2+y^2=a^2 x^2+b^2 y^2\)
Explanation:
A Let \(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be the midpoint of a chord of \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{r}^2\).
The equation of the chord is given by
\(S_1=S\)
\(\Rightarrow x_1 x+y_1 y=x_1^2+y_1^2\)
\(\text { Or } y=\left(-\frac{x_1}{y_1}\right) x+\frac{x_1^2+y_1^2}{y_1}\)
The line \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) is a tangent to the hyperbola only if \(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\).
\(\therefore \frac{\left(\mathrm{x}_1^2+\mathrm{y}_1^2\right)^2}{\mathrm{y}_1^2}=\frac{\mathrm{a}^2 \mathrm{x}_1^2}{\mathrm{y}_1^2}-\mathrm{b}^2\)
\(\therefore\) The locus of \(\mathrm{P}\) is \(\left(\mathrm{x}^2+\mathrm{y}^2\right)^2=\mathrm{a}^2 \mathrm{x}^2-\mathrm{b}^2 \mathrm{y}^2\).
COMEDK-2017
Hyperbola
120775
The line \(21 x+5 y=116\) is a tangent to the hyperbola \(7 x^2-5 y^2=232\). Its point of contact is
1 \((-6,-2)\)
2 \((6,2)\)
3 \((6,-2)\)
4 \((-6,2)\)
Explanation:
C The line \(21 x+5 y=116\) is a tangent to \(7 x^2-5 y^2\) \(=232\).
If \(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is the point of contact, then \(\mathrm{S}_1=0\).
\(\Rightarrow 7 \mathrm{x}_1 \mathrm{x}-5 \mathrm{y}_1 \mathrm{y}=232 \text {. }\)
It is same as the equation of the given tangent
\(21 x+5 y=116\)
On comparing both the equations,
\(\therefore \frac{7 \mathrm{x}_1}{21}=-\frac{5 \mathrm{y}_1}{5}=\frac{232}{116}=2\)
\(\therefore \mathrm{x}_1=6, \mathrm{y}_1=-2 .\)So, \(\mathrm{P}\) is \((6,-2)\).
B Given, equation of hyperbola is -
\(9 x^2-16 y^2=144 \quad \text { or } \frac{x^2}{16}-\frac{y^2}{9}=1\)
The equation of tangent to hyperbola having slope \(\mathrm{m}\) is \(\mathrm{y}=\mathrm{mx} \pm \sqrt{16 \mathrm{~m}^2-9}\)
It touches the circle, then the distance of this line from the centre of circle is the radius of the circle, hence,
\(\Rightarrow \frac{\sqrt{16 \mathrm{~m}^2-9}}{\sqrt{\mathrm{m}^2+1}}=3\)
\(\Rightarrow 9 \mathrm{~m}^2+9=16 \mathrm{~m}^2-9\)
\(\Rightarrow 7 \mathrm{~m}^2=18\)
\(\Rightarrow \mathrm{m}= \pm 3 \sqrt{\frac{2}{7}}\)
So, equation of the tangent is
\(\mathrm{y}= \pm 3 \sqrt{\frac{2}{7}} \mathrm{x}+\frac{15}{\sqrt{7}}\)Hence, option (b) is correct.
JEEE-2013
Hyperbola
120772
If the distance between the foci and distance between the directrices of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) are in the ratio \(3: 2\) then \(a: b\) is
1 \(\sqrt{2}: 1\)
2 \(1: 2\)
3 \(\sqrt{3}: \sqrt{2}\)
4 \(2: 1\)
Explanation:
A Given,
Equation of hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
Distance between directrices \(=\frac{2 \mathrm{a}}{\mathrm{e}}\)
and distance between foci \(=2 \mathrm{ae}\)
According to the question, we have
\(\frac{2 \mathrm{ae}}{\frac{2 \mathrm{a}}{\mathrm{e}}}=\frac{3}{2}\)
\(\Rightarrow \quad \mathrm{e}^2=\frac{3}{2}\)
\(\therefore \quad \mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\Rightarrow \quad \frac{\mathrm{b}}{\mathrm{a}}=\frac{1}{\sqrt{2}}\)Hence, the ratio of \(a: b\) is \(\sqrt{2}: 1\)
Karnataka CET-2013
Hyperbola
120773
The value of \(m\) for which \(y=m x+6\) is a tangent to the hyperbola \(\frac{x^2}{100}-\frac{y^2}{49}=1\) is
1 \(\sqrt{\frac{20}{30}}\)
2 \(\sqrt{\frac{20}{17}}\)
3 \(\sqrt{\frac{17}{20}}\)
4 \(\sqrt{\frac{3}{20}}\)
Explanation:
C Given, equation of the hyperbola-
\(\frac{x^2}{100}-\frac{y^2}{9}=1\)
and tangent is, \(\mathrm{y}=\mathrm{mx}+6\)
For this to be tangent, \(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\)
\(\Rightarrow 6^2= (100) \mathrm{m}^2-49\)
\(\Rightarrow 36+49 =100 \mathrm{~m}^2\)
\(\Rightarrow 85 =100 \mathrm{~m}^2\)
\(\Rightarrow \mathrm{m}^2 =\frac{85}{100}\)
\(\Rightarrow \mathrm{m} =\sqrt{\frac{17}{20}}\)
COMEDK-2016
Hyperbola
120774
Chords of the circle \(x^2+y^2=r^2\) touch the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{16}=1\). The locus of the midpoints of the chords is
1 \(\left(x^2+y^2\right)^2=a^2 x^2-b^2 y^2\)
2 \(\left(x^2+y^2\right)^2=a^2 x^2+b^2 y^2\)
3 \(x^2+y^2=a^2 x^2-b^2 y^2\)
4 \(x^2+y^2=a^2 x^2+b^2 y^2\)
Explanation:
A Let \(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be the midpoint of a chord of \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{r}^2\).
The equation of the chord is given by
\(S_1=S\)
\(\Rightarrow x_1 x+y_1 y=x_1^2+y_1^2\)
\(\text { Or } y=\left(-\frac{x_1}{y_1}\right) x+\frac{x_1^2+y_1^2}{y_1}\)
The line \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) is a tangent to the hyperbola only if \(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\).
\(\therefore \frac{\left(\mathrm{x}_1^2+\mathrm{y}_1^2\right)^2}{\mathrm{y}_1^2}=\frac{\mathrm{a}^2 \mathrm{x}_1^2}{\mathrm{y}_1^2}-\mathrm{b}^2\)
\(\therefore\) The locus of \(\mathrm{P}\) is \(\left(\mathrm{x}^2+\mathrm{y}^2\right)^2=\mathrm{a}^2 \mathrm{x}^2-\mathrm{b}^2 \mathrm{y}^2\).
COMEDK-2017
Hyperbola
120775
The line \(21 x+5 y=116\) is a tangent to the hyperbola \(7 x^2-5 y^2=232\). Its point of contact is
1 \((-6,-2)\)
2 \((6,2)\)
3 \((6,-2)\)
4 \((-6,2)\)
Explanation:
C The line \(21 x+5 y=116\) is a tangent to \(7 x^2-5 y^2\) \(=232\).
If \(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is the point of contact, then \(\mathrm{S}_1=0\).
\(\Rightarrow 7 \mathrm{x}_1 \mathrm{x}-5 \mathrm{y}_1 \mathrm{y}=232 \text {. }\)
It is same as the equation of the given tangent
\(21 x+5 y=116\)
On comparing both the equations,
\(\therefore \frac{7 \mathrm{x}_1}{21}=-\frac{5 \mathrm{y}_1}{5}=\frac{232}{116}=2\)
\(\therefore \mathrm{x}_1=6, \mathrm{y}_1=-2 .\)So, \(\mathrm{P}\) is \((6,-2)\).
B Given, equation of hyperbola is -
\(9 x^2-16 y^2=144 \quad \text { or } \frac{x^2}{16}-\frac{y^2}{9}=1\)
The equation of tangent to hyperbola having slope \(\mathrm{m}\) is \(\mathrm{y}=\mathrm{mx} \pm \sqrt{16 \mathrm{~m}^2-9}\)
It touches the circle, then the distance of this line from the centre of circle is the radius of the circle, hence,
\(\Rightarrow \frac{\sqrt{16 \mathrm{~m}^2-9}}{\sqrt{\mathrm{m}^2+1}}=3\)
\(\Rightarrow 9 \mathrm{~m}^2+9=16 \mathrm{~m}^2-9\)
\(\Rightarrow 7 \mathrm{~m}^2=18\)
\(\Rightarrow \mathrm{m}= \pm 3 \sqrt{\frac{2}{7}}\)
So, equation of the tangent is
\(\mathrm{y}= \pm 3 \sqrt{\frac{2}{7}} \mathrm{x}+\frac{15}{\sqrt{7}}\)Hence, option (b) is correct.
JEEE-2013
Hyperbola
120772
If the distance between the foci and distance between the directrices of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) are in the ratio \(3: 2\) then \(a: b\) is
1 \(\sqrt{2}: 1\)
2 \(1: 2\)
3 \(\sqrt{3}: \sqrt{2}\)
4 \(2: 1\)
Explanation:
A Given,
Equation of hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
Distance between directrices \(=\frac{2 \mathrm{a}}{\mathrm{e}}\)
and distance between foci \(=2 \mathrm{ae}\)
According to the question, we have
\(\frac{2 \mathrm{ae}}{\frac{2 \mathrm{a}}{\mathrm{e}}}=\frac{3}{2}\)
\(\Rightarrow \quad \mathrm{e}^2=\frac{3}{2}\)
\(\therefore \quad \mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\Rightarrow \quad \frac{\mathrm{b}}{\mathrm{a}}=\frac{1}{\sqrt{2}}\)Hence, the ratio of \(a: b\) is \(\sqrt{2}: 1\)
Karnataka CET-2013
Hyperbola
120773
The value of \(m\) for which \(y=m x+6\) is a tangent to the hyperbola \(\frac{x^2}{100}-\frac{y^2}{49}=1\) is
1 \(\sqrt{\frac{20}{30}}\)
2 \(\sqrt{\frac{20}{17}}\)
3 \(\sqrt{\frac{17}{20}}\)
4 \(\sqrt{\frac{3}{20}}\)
Explanation:
C Given, equation of the hyperbola-
\(\frac{x^2}{100}-\frac{y^2}{9}=1\)
and tangent is, \(\mathrm{y}=\mathrm{mx}+6\)
For this to be tangent, \(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\)
\(\Rightarrow 6^2= (100) \mathrm{m}^2-49\)
\(\Rightarrow 36+49 =100 \mathrm{~m}^2\)
\(\Rightarrow 85 =100 \mathrm{~m}^2\)
\(\Rightarrow \mathrm{m}^2 =\frac{85}{100}\)
\(\Rightarrow \mathrm{m} =\sqrt{\frac{17}{20}}\)
COMEDK-2016
Hyperbola
120774
Chords of the circle \(x^2+y^2=r^2\) touch the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{16}=1\). The locus of the midpoints of the chords is
1 \(\left(x^2+y^2\right)^2=a^2 x^2-b^2 y^2\)
2 \(\left(x^2+y^2\right)^2=a^2 x^2+b^2 y^2\)
3 \(x^2+y^2=a^2 x^2-b^2 y^2\)
4 \(x^2+y^2=a^2 x^2+b^2 y^2\)
Explanation:
A Let \(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be the midpoint of a chord of \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{r}^2\).
The equation of the chord is given by
\(S_1=S\)
\(\Rightarrow x_1 x+y_1 y=x_1^2+y_1^2\)
\(\text { Or } y=\left(-\frac{x_1}{y_1}\right) x+\frac{x_1^2+y_1^2}{y_1}\)
The line \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) is a tangent to the hyperbola only if \(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\).
\(\therefore \frac{\left(\mathrm{x}_1^2+\mathrm{y}_1^2\right)^2}{\mathrm{y}_1^2}=\frac{\mathrm{a}^2 \mathrm{x}_1^2}{\mathrm{y}_1^2}-\mathrm{b}^2\)
\(\therefore\) The locus of \(\mathrm{P}\) is \(\left(\mathrm{x}^2+\mathrm{y}^2\right)^2=\mathrm{a}^2 \mathrm{x}^2-\mathrm{b}^2 \mathrm{y}^2\).
COMEDK-2017
Hyperbola
120775
The line \(21 x+5 y=116\) is a tangent to the hyperbola \(7 x^2-5 y^2=232\). Its point of contact is
1 \((-6,-2)\)
2 \((6,2)\)
3 \((6,-2)\)
4 \((-6,2)\)
Explanation:
C The line \(21 x+5 y=116\) is a tangent to \(7 x^2-5 y^2\) \(=232\).
If \(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is the point of contact, then \(\mathrm{S}_1=0\).
\(\Rightarrow 7 \mathrm{x}_1 \mathrm{x}-5 \mathrm{y}_1 \mathrm{y}=232 \text {. }\)
It is same as the equation of the given tangent
\(21 x+5 y=116\)
On comparing both the equations,
\(\therefore \frac{7 \mathrm{x}_1}{21}=-\frac{5 \mathrm{y}_1}{5}=\frac{232}{116}=2\)
\(\therefore \mathrm{x}_1=6, \mathrm{y}_1=-2 .\)So, \(\mathrm{P}\) is \((6,-2)\).
B Given, equation of hyperbola is -
\(9 x^2-16 y^2=144 \quad \text { or } \frac{x^2}{16}-\frac{y^2}{9}=1\)
The equation of tangent to hyperbola having slope \(\mathrm{m}\) is \(\mathrm{y}=\mathrm{mx} \pm \sqrt{16 \mathrm{~m}^2-9}\)
It touches the circle, then the distance of this line from the centre of circle is the radius of the circle, hence,
\(\Rightarrow \frac{\sqrt{16 \mathrm{~m}^2-9}}{\sqrt{\mathrm{m}^2+1}}=3\)
\(\Rightarrow 9 \mathrm{~m}^2+9=16 \mathrm{~m}^2-9\)
\(\Rightarrow 7 \mathrm{~m}^2=18\)
\(\Rightarrow \mathrm{m}= \pm 3 \sqrt{\frac{2}{7}}\)
So, equation of the tangent is
\(\mathrm{y}= \pm 3 \sqrt{\frac{2}{7}} \mathrm{x}+\frac{15}{\sqrt{7}}\)Hence, option (b) is correct.
JEEE-2013
Hyperbola
120772
If the distance between the foci and distance between the directrices of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) are in the ratio \(3: 2\) then \(a: b\) is
1 \(\sqrt{2}: 1\)
2 \(1: 2\)
3 \(\sqrt{3}: \sqrt{2}\)
4 \(2: 1\)
Explanation:
A Given,
Equation of hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
Distance between directrices \(=\frac{2 \mathrm{a}}{\mathrm{e}}\)
and distance between foci \(=2 \mathrm{ae}\)
According to the question, we have
\(\frac{2 \mathrm{ae}}{\frac{2 \mathrm{a}}{\mathrm{e}}}=\frac{3}{2}\)
\(\Rightarrow \quad \mathrm{e}^2=\frac{3}{2}\)
\(\therefore \quad \mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\Rightarrow \quad \frac{\mathrm{b}}{\mathrm{a}}=\frac{1}{\sqrt{2}}\)Hence, the ratio of \(a: b\) is \(\sqrt{2}: 1\)
Karnataka CET-2013
Hyperbola
120773
The value of \(m\) for which \(y=m x+6\) is a tangent to the hyperbola \(\frac{x^2}{100}-\frac{y^2}{49}=1\) is
1 \(\sqrt{\frac{20}{30}}\)
2 \(\sqrt{\frac{20}{17}}\)
3 \(\sqrt{\frac{17}{20}}\)
4 \(\sqrt{\frac{3}{20}}\)
Explanation:
C Given, equation of the hyperbola-
\(\frac{x^2}{100}-\frac{y^2}{9}=1\)
and tangent is, \(\mathrm{y}=\mathrm{mx}+6\)
For this to be tangent, \(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\)
\(\Rightarrow 6^2= (100) \mathrm{m}^2-49\)
\(\Rightarrow 36+49 =100 \mathrm{~m}^2\)
\(\Rightarrow 85 =100 \mathrm{~m}^2\)
\(\Rightarrow \mathrm{m}^2 =\frac{85}{100}\)
\(\Rightarrow \mathrm{m} =\sqrt{\frac{17}{20}}\)
COMEDK-2016
Hyperbola
120774
Chords of the circle \(x^2+y^2=r^2\) touch the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{16}=1\). The locus of the midpoints of the chords is
1 \(\left(x^2+y^2\right)^2=a^2 x^2-b^2 y^2\)
2 \(\left(x^2+y^2\right)^2=a^2 x^2+b^2 y^2\)
3 \(x^2+y^2=a^2 x^2-b^2 y^2\)
4 \(x^2+y^2=a^2 x^2+b^2 y^2\)
Explanation:
A Let \(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be the midpoint of a chord of \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{r}^2\).
The equation of the chord is given by
\(S_1=S\)
\(\Rightarrow x_1 x+y_1 y=x_1^2+y_1^2\)
\(\text { Or } y=\left(-\frac{x_1}{y_1}\right) x+\frac{x_1^2+y_1^2}{y_1}\)
The line \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) is a tangent to the hyperbola only if \(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\).
\(\therefore \frac{\left(\mathrm{x}_1^2+\mathrm{y}_1^2\right)^2}{\mathrm{y}_1^2}=\frac{\mathrm{a}^2 \mathrm{x}_1^2}{\mathrm{y}_1^2}-\mathrm{b}^2\)
\(\therefore\) The locus of \(\mathrm{P}\) is \(\left(\mathrm{x}^2+\mathrm{y}^2\right)^2=\mathrm{a}^2 \mathrm{x}^2-\mathrm{b}^2 \mathrm{y}^2\).
COMEDK-2017
Hyperbola
120775
The line \(21 x+5 y=116\) is a tangent to the hyperbola \(7 x^2-5 y^2=232\). Its point of contact is
1 \((-6,-2)\)
2 \((6,2)\)
3 \((6,-2)\)
4 \((-6,2)\)
Explanation:
C The line \(21 x+5 y=116\) is a tangent to \(7 x^2-5 y^2\) \(=232\).
If \(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is the point of contact, then \(\mathrm{S}_1=0\).
\(\Rightarrow 7 \mathrm{x}_1 \mathrm{x}-5 \mathrm{y}_1 \mathrm{y}=232 \text {. }\)
It is same as the equation of the given tangent
\(21 x+5 y=116\)
On comparing both the equations,
\(\therefore \frac{7 \mathrm{x}_1}{21}=-\frac{5 \mathrm{y}_1}{5}=\frac{232}{116}=2\)
\(\therefore \mathrm{x}_1=6, \mathrm{y}_1=-2 .\)So, \(\mathrm{P}\) is \((6,-2)\).
