NEET Test Series from KOTA - 10 Papers In MS WORD
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Hyperbola
120693
If the eccentricity of the hyperbola \(x^2-y^2 \operatorname{cosec}^2 \alpha=25\) is \(\sqrt{5}\) times the eccentricity of the ellipse \(x^2 \operatorname{cosec}^2 \alpha+y^2=5\), then \(\alpha\) is equal to :
1 \(\tan ^{-1} \sqrt{2}\)
2 \(\sin ^{-1} \sqrt{\frac{3}{4}}\)
3 \(\tan ^{-1} \sqrt{\frac{2}{5}}\)
4 \(\sin ^{-1} \sqrt{\frac{2}{5}}\)
Explanation:
A Eccentricity of \(\frac{x^2}{25}-\frac{y^2}{25 \sin ^2 \alpha}=1\) is
\(\mathrm{a}=5, \mathrm{~b}=5 \sin \alpha\)
So, \(\mathrm{e}_1=\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{\mathrm{a}} \mathrm{e}_1=\frac{\sqrt{25+25 \sin ^2 \alpha}}{5}\)
\(e_1=\sqrt{1+\sin ^2 \alpha}\)
Now for the given ellipse,
\(x^2 \operatorname{cosec}^2 \alpha+y^2=5\)
\(\frac{x^2}{5 \sin ^2 \alpha}+\frac{y^2}{5}=1\)
Here,
\(a=\sqrt{5} \sin \alpha, b=\sqrt{5}\)
So, \(\quad e_2=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{5 \sin ^2 \alpha}{5}}=\cos \alpha\)
Given, \(\sqrt{1+\sin ^2 \alpha}=\sqrt{5} \cos \alpha\)
Squaring on both sides-
\(1+\sin ^2 \alpha=5 \cos ^2 \alpha\)
\(1+\sin ^2 \alpha-5\left(1-\sin ^2 \alpha\right)=0\)
\(1+\sin ^2 \alpha-5+5 \sin ^2 \alpha=0\)
\(6 \sin ^2 \alpha=4\)
\(\sin ^2 \alpha=\frac{4}{6}\)
\(\sin \alpha=\frac{2}{3}\)
\(\alpha=\sin ^{-1} \sqrt{\frac{2}{3}}=\tan ^{-1} \sqrt{2}\)
VITEEE-2017
Hyperbola
120694
Equation of the chord of the hyperbola \(25 x^2-\) \(16 y^2=400\) which is bisected at the point \((6,2)\) is
1 \(6 x-7 y=418\)
2 \(75 x-16 y=418\)
3 \(25 x-4 y=400\)
4 None of these
Explanation:
B Equation of hyperbola is
\(25 \mathrm{x}^2-16 \mathrm{y}^2 =400\)
\(\frac{25 \mathrm{x}^2}{400}-\frac{16 \mathrm{y}^2}{400} =1\)
\(\frac{\mathrm{x}^2}{16}-\frac{y^2}{25} =1\)
Chord is bisected at \((6,2)\) then the equation of chord
\(\mathrm{T}=\mathrm{S}_1\)
\(\therefore \quad \frac{6 \mathrm{x}}{16}-\frac{2 \mathrm{y}}{25}=\frac{6^2}{16}-\frac{2^2}{25}\)
\(\frac{150 \mathrm{x}-32 \mathrm{y}}{400}=\frac{36 \times 25-4 \times 16}{400}\)
\(150 \mathrm{x}-32 \mathrm{y}=900-64\)
\(150 \mathrm{x}-32 \mathrm{y}=836\)
\(75 \mathrm{x}-16 \mathrm{y}=418\)
VITEEE-2014
Hyperbola
120695
If \(P Q\) is a double ordinate of hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), such that OPQ is an equilateral triangle, \(O\) being the centre of the hyperbola, then the eccentricity ' \(e\) ' of the hyperbola satisfies
1 \(1\lt \mathrm{e}\lt \frac{2}{\sqrt{3}}\)
2 \(\mathrm{e}=\frac{2}{\sqrt{3}}\)
3 \(\mathrm{e}=\frac{\sqrt{3}}{2}\)
4 \(\mathrm{e} \geq \frac{2}{\sqrt{3}}\)
Explanation:
D Given, hyperbola be \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\).
Let double ordinate PQ be \((a \sec \theta, b \tan \theta),(a \sec \theta,-b\) \(\tan \theta)\) and \(O\) is centre \((0,0)\).
\(\triangle \mathrm{OPQ}\) being equilateral
\(\text { Thus } \tan 30^{\circ}=\frac{\mathrm{b} \tan \theta}{\mathrm{a} \sec \theta}\)
\(\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{b}}{\mathrm{a}} \sin \theta\)
\(3 \cdot \frac{\mathrm{b}^2}{\mathrm{a}^2}=\operatorname{cosec}^2 \theta\)
\(3\left(\mathrm{e}^2-1\right)=\operatorname{cosec}^2 \theta\)
Since, \(\operatorname{cosec}^2 \theta \geq 1\)
Therefore, \(3\left(\mathrm{e}^2-1\right) \geq 1\)
\(\mathrm{e}^2 \geq \frac{4}{3}\)
\(\mathrm{e} \geq \frac{2}{\sqrt{3}}\)
WB JEE -2016
Hyperbola
120696
Taking axes of hyperbola as coordinate axes, find its equation when the distance between the foci is 16 and eccentricity is \(\sqrt{2}\)
1 \(x^2-y^2=32\)
2 \(x^2-y^2=64\)
3 \(x^2-y^2=8\)
4 \(x^2-y^2=16\)
Explanation:
A Given that,
\(2 \mathrm{ae}=16 \text { and } \mathrm{e}=\sqrt{2}\)
\(\mathrm{a}=\frac{16}{2 \sqrt{2}}\)
\(\mathrm{a}=4 \sqrt{2}\)
Eccentricity, \(\mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(2=1+\frac{\mathrm{b}^2}{32}\)
\(\frac{\mathrm{b}^2}{32}=1\)
\(\mathrm{~b}^2=32 \Rightarrow \mathrm{b}=4 \sqrt{2}\)
Hence, required equation of hyperbola is
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(\frac{x^2}{32}-\frac{y^2}{32}=1\)
\(x^2-y^2=32\)
120693
If the eccentricity of the hyperbola \(x^2-y^2 \operatorname{cosec}^2 \alpha=25\) is \(\sqrt{5}\) times the eccentricity of the ellipse \(x^2 \operatorname{cosec}^2 \alpha+y^2=5\), then \(\alpha\) is equal to :
1 \(\tan ^{-1} \sqrt{2}\)
2 \(\sin ^{-1} \sqrt{\frac{3}{4}}\)
3 \(\tan ^{-1} \sqrt{\frac{2}{5}}\)
4 \(\sin ^{-1} \sqrt{\frac{2}{5}}\)
Explanation:
A Eccentricity of \(\frac{x^2}{25}-\frac{y^2}{25 \sin ^2 \alpha}=1\) is
\(\mathrm{a}=5, \mathrm{~b}=5 \sin \alpha\)
So, \(\mathrm{e}_1=\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{\mathrm{a}} \mathrm{e}_1=\frac{\sqrt{25+25 \sin ^2 \alpha}}{5}\)
\(e_1=\sqrt{1+\sin ^2 \alpha}\)
Now for the given ellipse,
\(x^2 \operatorname{cosec}^2 \alpha+y^2=5\)
\(\frac{x^2}{5 \sin ^2 \alpha}+\frac{y^2}{5}=1\)
Here,
\(a=\sqrt{5} \sin \alpha, b=\sqrt{5}\)
So, \(\quad e_2=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{5 \sin ^2 \alpha}{5}}=\cos \alpha\)
Given, \(\sqrt{1+\sin ^2 \alpha}=\sqrt{5} \cos \alpha\)
Squaring on both sides-
\(1+\sin ^2 \alpha=5 \cos ^2 \alpha\)
\(1+\sin ^2 \alpha-5\left(1-\sin ^2 \alpha\right)=0\)
\(1+\sin ^2 \alpha-5+5 \sin ^2 \alpha=0\)
\(6 \sin ^2 \alpha=4\)
\(\sin ^2 \alpha=\frac{4}{6}\)
\(\sin \alpha=\frac{2}{3}\)
\(\alpha=\sin ^{-1} \sqrt{\frac{2}{3}}=\tan ^{-1} \sqrt{2}\)
VITEEE-2017
Hyperbola
120694
Equation of the chord of the hyperbola \(25 x^2-\) \(16 y^2=400\) which is bisected at the point \((6,2)\) is
1 \(6 x-7 y=418\)
2 \(75 x-16 y=418\)
3 \(25 x-4 y=400\)
4 None of these
Explanation:
B Equation of hyperbola is
\(25 \mathrm{x}^2-16 \mathrm{y}^2 =400\)
\(\frac{25 \mathrm{x}^2}{400}-\frac{16 \mathrm{y}^2}{400} =1\)
\(\frac{\mathrm{x}^2}{16}-\frac{y^2}{25} =1\)
Chord is bisected at \((6,2)\) then the equation of chord
\(\mathrm{T}=\mathrm{S}_1\)
\(\therefore \quad \frac{6 \mathrm{x}}{16}-\frac{2 \mathrm{y}}{25}=\frac{6^2}{16}-\frac{2^2}{25}\)
\(\frac{150 \mathrm{x}-32 \mathrm{y}}{400}=\frac{36 \times 25-4 \times 16}{400}\)
\(150 \mathrm{x}-32 \mathrm{y}=900-64\)
\(150 \mathrm{x}-32 \mathrm{y}=836\)
\(75 \mathrm{x}-16 \mathrm{y}=418\)
VITEEE-2014
Hyperbola
120695
If \(P Q\) is a double ordinate of hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), such that OPQ is an equilateral triangle, \(O\) being the centre of the hyperbola, then the eccentricity ' \(e\) ' of the hyperbola satisfies
1 \(1\lt \mathrm{e}\lt \frac{2}{\sqrt{3}}\)
2 \(\mathrm{e}=\frac{2}{\sqrt{3}}\)
3 \(\mathrm{e}=\frac{\sqrt{3}}{2}\)
4 \(\mathrm{e} \geq \frac{2}{\sqrt{3}}\)
Explanation:
D Given, hyperbola be \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\).
Let double ordinate PQ be \((a \sec \theta, b \tan \theta),(a \sec \theta,-b\) \(\tan \theta)\) and \(O\) is centre \((0,0)\).
\(\triangle \mathrm{OPQ}\) being equilateral
\(\text { Thus } \tan 30^{\circ}=\frac{\mathrm{b} \tan \theta}{\mathrm{a} \sec \theta}\)
\(\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{b}}{\mathrm{a}} \sin \theta\)
\(3 \cdot \frac{\mathrm{b}^2}{\mathrm{a}^2}=\operatorname{cosec}^2 \theta\)
\(3\left(\mathrm{e}^2-1\right)=\operatorname{cosec}^2 \theta\)
Since, \(\operatorname{cosec}^2 \theta \geq 1\)
Therefore, \(3\left(\mathrm{e}^2-1\right) \geq 1\)
\(\mathrm{e}^2 \geq \frac{4}{3}\)
\(\mathrm{e} \geq \frac{2}{\sqrt{3}}\)
WB JEE -2016
Hyperbola
120696
Taking axes of hyperbola as coordinate axes, find its equation when the distance between the foci is 16 and eccentricity is \(\sqrt{2}\)
1 \(x^2-y^2=32\)
2 \(x^2-y^2=64\)
3 \(x^2-y^2=8\)
4 \(x^2-y^2=16\)
Explanation:
A Given that,
\(2 \mathrm{ae}=16 \text { and } \mathrm{e}=\sqrt{2}\)
\(\mathrm{a}=\frac{16}{2 \sqrt{2}}\)
\(\mathrm{a}=4 \sqrt{2}\)
Eccentricity, \(\mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(2=1+\frac{\mathrm{b}^2}{32}\)
\(\frac{\mathrm{b}^2}{32}=1\)
\(\mathrm{~b}^2=32 \Rightarrow \mathrm{b}=4 \sqrt{2}\)
Hence, required equation of hyperbola is
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(\frac{x^2}{32}-\frac{y^2}{32}=1\)
\(x^2-y^2=32\)
120693
If the eccentricity of the hyperbola \(x^2-y^2 \operatorname{cosec}^2 \alpha=25\) is \(\sqrt{5}\) times the eccentricity of the ellipse \(x^2 \operatorname{cosec}^2 \alpha+y^2=5\), then \(\alpha\) is equal to :
1 \(\tan ^{-1} \sqrt{2}\)
2 \(\sin ^{-1} \sqrt{\frac{3}{4}}\)
3 \(\tan ^{-1} \sqrt{\frac{2}{5}}\)
4 \(\sin ^{-1} \sqrt{\frac{2}{5}}\)
Explanation:
A Eccentricity of \(\frac{x^2}{25}-\frac{y^2}{25 \sin ^2 \alpha}=1\) is
\(\mathrm{a}=5, \mathrm{~b}=5 \sin \alpha\)
So, \(\mathrm{e}_1=\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{\mathrm{a}} \mathrm{e}_1=\frac{\sqrt{25+25 \sin ^2 \alpha}}{5}\)
\(e_1=\sqrt{1+\sin ^2 \alpha}\)
Now for the given ellipse,
\(x^2 \operatorname{cosec}^2 \alpha+y^2=5\)
\(\frac{x^2}{5 \sin ^2 \alpha}+\frac{y^2}{5}=1\)
Here,
\(a=\sqrt{5} \sin \alpha, b=\sqrt{5}\)
So, \(\quad e_2=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{5 \sin ^2 \alpha}{5}}=\cos \alpha\)
Given, \(\sqrt{1+\sin ^2 \alpha}=\sqrt{5} \cos \alpha\)
Squaring on both sides-
\(1+\sin ^2 \alpha=5 \cos ^2 \alpha\)
\(1+\sin ^2 \alpha-5\left(1-\sin ^2 \alpha\right)=0\)
\(1+\sin ^2 \alpha-5+5 \sin ^2 \alpha=0\)
\(6 \sin ^2 \alpha=4\)
\(\sin ^2 \alpha=\frac{4}{6}\)
\(\sin \alpha=\frac{2}{3}\)
\(\alpha=\sin ^{-1} \sqrt{\frac{2}{3}}=\tan ^{-1} \sqrt{2}\)
VITEEE-2017
Hyperbola
120694
Equation of the chord of the hyperbola \(25 x^2-\) \(16 y^2=400\) which is bisected at the point \((6,2)\) is
1 \(6 x-7 y=418\)
2 \(75 x-16 y=418\)
3 \(25 x-4 y=400\)
4 None of these
Explanation:
B Equation of hyperbola is
\(25 \mathrm{x}^2-16 \mathrm{y}^2 =400\)
\(\frac{25 \mathrm{x}^2}{400}-\frac{16 \mathrm{y}^2}{400} =1\)
\(\frac{\mathrm{x}^2}{16}-\frac{y^2}{25} =1\)
Chord is bisected at \((6,2)\) then the equation of chord
\(\mathrm{T}=\mathrm{S}_1\)
\(\therefore \quad \frac{6 \mathrm{x}}{16}-\frac{2 \mathrm{y}}{25}=\frac{6^2}{16}-\frac{2^2}{25}\)
\(\frac{150 \mathrm{x}-32 \mathrm{y}}{400}=\frac{36 \times 25-4 \times 16}{400}\)
\(150 \mathrm{x}-32 \mathrm{y}=900-64\)
\(150 \mathrm{x}-32 \mathrm{y}=836\)
\(75 \mathrm{x}-16 \mathrm{y}=418\)
VITEEE-2014
Hyperbola
120695
If \(P Q\) is a double ordinate of hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), such that OPQ is an equilateral triangle, \(O\) being the centre of the hyperbola, then the eccentricity ' \(e\) ' of the hyperbola satisfies
1 \(1\lt \mathrm{e}\lt \frac{2}{\sqrt{3}}\)
2 \(\mathrm{e}=\frac{2}{\sqrt{3}}\)
3 \(\mathrm{e}=\frac{\sqrt{3}}{2}\)
4 \(\mathrm{e} \geq \frac{2}{\sqrt{3}}\)
Explanation:
D Given, hyperbola be \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\).
Let double ordinate PQ be \((a \sec \theta, b \tan \theta),(a \sec \theta,-b\) \(\tan \theta)\) and \(O\) is centre \((0,0)\).
\(\triangle \mathrm{OPQ}\) being equilateral
\(\text { Thus } \tan 30^{\circ}=\frac{\mathrm{b} \tan \theta}{\mathrm{a} \sec \theta}\)
\(\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{b}}{\mathrm{a}} \sin \theta\)
\(3 \cdot \frac{\mathrm{b}^2}{\mathrm{a}^2}=\operatorname{cosec}^2 \theta\)
\(3\left(\mathrm{e}^2-1\right)=\operatorname{cosec}^2 \theta\)
Since, \(\operatorname{cosec}^2 \theta \geq 1\)
Therefore, \(3\left(\mathrm{e}^2-1\right) \geq 1\)
\(\mathrm{e}^2 \geq \frac{4}{3}\)
\(\mathrm{e} \geq \frac{2}{\sqrt{3}}\)
WB JEE -2016
Hyperbola
120696
Taking axes of hyperbola as coordinate axes, find its equation when the distance between the foci is 16 and eccentricity is \(\sqrt{2}\)
1 \(x^2-y^2=32\)
2 \(x^2-y^2=64\)
3 \(x^2-y^2=8\)
4 \(x^2-y^2=16\)
Explanation:
A Given that,
\(2 \mathrm{ae}=16 \text { and } \mathrm{e}=\sqrt{2}\)
\(\mathrm{a}=\frac{16}{2 \sqrt{2}}\)
\(\mathrm{a}=4 \sqrt{2}\)
Eccentricity, \(\mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(2=1+\frac{\mathrm{b}^2}{32}\)
\(\frac{\mathrm{b}^2}{32}=1\)
\(\mathrm{~b}^2=32 \Rightarrow \mathrm{b}=4 \sqrt{2}\)
Hence, required equation of hyperbola is
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(\frac{x^2}{32}-\frac{y^2}{32}=1\)
\(x^2-y^2=32\)
120693
If the eccentricity of the hyperbola \(x^2-y^2 \operatorname{cosec}^2 \alpha=25\) is \(\sqrt{5}\) times the eccentricity of the ellipse \(x^2 \operatorname{cosec}^2 \alpha+y^2=5\), then \(\alpha\) is equal to :
1 \(\tan ^{-1} \sqrt{2}\)
2 \(\sin ^{-1} \sqrt{\frac{3}{4}}\)
3 \(\tan ^{-1} \sqrt{\frac{2}{5}}\)
4 \(\sin ^{-1} \sqrt{\frac{2}{5}}\)
Explanation:
A Eccentricity of \(\frac{x^2}{25}-\frac{y^2}{25 \sin ^2 \alpha}=1\) is
\(\mathrm{a}=5, \mathrm{~b}=5 \sin \alpha\)
So, \(\mathrm{e}_1=\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{\mathrm{a}} \mathrm{e}_1=\frac{\sqrt{25+25 \sin ^2 \alpha}}{5}\)
\(e_1=\sqrt{1+\sin ^2 \alpha}\)
Now for the given ellipse,
\(x^2 \operatorname{cosec}^2 \alpha+y^2=5\)
\(\frac{x^2}{5 \sin ^2 \alpha}+\frac{y^2}{5}=1\)
Here,
\(a=\sqrt{5} \sin \alpha, b=\sqrt{5}\)
So, \(\quad e_2=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{5 \sin ^2 \alpha}{5}}=\cos \alpha\)
Given, \(\sqrt{1+\sin ^2 \alpha}=\sqrt{5} \cos \alpha\)
Squaring on both sides-
\(1+\sin ^2 \alpha=5 \cos ^2 \alpha\)
\(1+\sin ^2 \alpha-5\left(1-\sin ^2 \alpha\right)=0\)
\(1+\sin ^2 \alpha-5+5 \sin ^2 \alpha=0\)
\(6 \sin ^2 \alpha=4\)
\(\sin ^2 \alpha=\frac{4}{6}\)
\(\sin \alpha=\frac{2}{3}\)
\(\alpha=\sin ^{-1} \sqrt{\frac{2}{3}}=\tan ^{-1} \sqrt{2}\)
VITEEE-2017
Hyperbola
120694
Equation of the chord of the hyperbola \(25 x^2-\) \(16 y^2=400\) which is bisected at the point \((6,2)\) is
1 \(6 x-7 y=418\)
2 \(75 x-16 y=418\)
3 \(25 x-4 y=400\)
4 None of these
Explanation:
B Equation of hyperbola is
\(25 \mathrm{x}^2-16 \mathrm{y}^2 =400\)
\(\frac{25 \mathrm{x}^2}{400}-\frac{16 \mathrm{y}^2}{400} =1\)
\(\frac{\mathrm{x}^2}{16}-\frac{y^2}{25} =1\)
Chord is bisected at \((6,2)\) then the equation of chord
\(\mathrm{T}=\mathrm{S}_1\)
\(\therefore \quad \frac{6 \mathrm{x}}{16}-\frac{2 \mathrm{y}}{25}=\frac{6^2}{16}-\frac{2^2}{25}\)
\(\frac{150 \mathrm{x}-32 \mathrm{y}}{400}=\frac{36 \times 25-4 \times 16}{400}\)
\(150 \mathrm{x}-32 \mathrm{y}=900-64\)
\(150 \mathrm{x}-32 \mathrm{y}=836\)
\(75 \mathrm{x}-16 \mathrm{y}=418\)
VITEEE-2014
Hyperbola
120695
If \(P Q\) is a double ordinate of hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), such that OPQ is an equilateral triangle, \(O\) being the centre of the hyperbola, then the eccentricity ' \(e\) ' of the hyperbola satisfies
1 \(1\lt \mathrm{e}\lt \frac{2}{\sqrt{3}}\)
2 \(\mathrm{e}=\frac{2}{\sqrt{3}}\)
3 \(\mathrm{e}=\frac{\sqrt{3}}{2}\)
4 \(\mathrm{e} \geq \frac{2}{\sqrt{3}}\)
Explanation:
D Given, hyperbola be \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\).
Let double ordinate PQ be \((a \sec \theta, b \tan \theta),(a \sec \theta,-b\) \(\tan \theta)\) and \(O\) is centre \((0,0)\).
\(\triangle \mathrm{OPQ}\) being equilateral
\(\text { Thus } \tan 30^{\circ}=\frac{\mathrm{b} \tan \theta}{\mathrm{a} \sec \theta}\)
\(\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{b}}{\mathrm{a}} \sin \theta\)
\(3 \cdot \frac{\mathrm{b}^2}{\mathrm{a}^2}=\operatorname{cosec}^2 \theta\)
\(3\left(\mathrm{e}^2-1\right)=\operatorname{cosec}^2 \theta\)
Since, \(\operatorname{cosec}^2 \theta \geq 1\)
Therefore, \(3\left(\mathrm{e}^2-1\right) \geq 1\)
\(\mathrm{e}^2 \geq \frac{4}{3}\)
\(\mathrm{e} \geq \frac{2}{\sqrt{3}}\)
WB JEE -2016
Hyperbola
120696
Taking axes of hyperbola as coordinate axes, find its equation when the distance between the foci is 16 and eccentricity is \(\sqrt{2}\)
1 \(x^2-y^2=32\)
2 \(x^2-y^2=64\)
3 \(x^2-y^2=8\)
4 \(x^2-y^2=16\)
Explanation:
A Given that,
\(2 \mathrm{ae}=16 \text { and } \mathrm{e}=\sqrt{2}\)
\(\mathrm{a}=\frac{16}{2 \sqrt{2}}\)
\(\mathrm{a}=4 \sqrt{2}\)
Eccentricity, \(\mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(2=1+\frac{\mathrm{b}^2}{32}\)
\(\frac{\mathrm{b}^2}{32}=1\)
\(\mathrm{~b}^2=32 \Rightarrow \mathrm{b}=4 \sqrt{2}\)
Hence, required equation of hyperbola is
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(\frac{x^2}{32}-\frac{y^2}{32}=1\)
\(x^2-y^2=32\)
