Hyperbola – JEE Main Previous Year Question with Solutions


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Q. The equation of the hyperbola whose foci are (–2,0) and (2, 0) and eccentricity is 2 is given by : (1) $-3 x^{2}+y^{2}=3$ (2) $x^{2}-3 y^{2}=3$ (3) $3 x^{2}-y^{2}=3$ (4) $-x^{2}+3 y^{2}=3$ [AIEEE-2011]

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Sol. (3) $2 \mathrm{ae}=4$ $\mathrm{ae}=2$ $\mathrm{a}(2)=2$ $\mathrm{a}=1$ $\mathrm{b}^{2}=\mathrm{a}^{2}\left(\mathrm{e}^{2}-1\right)$ $\begin{aligned}=& 1(4-1)=3 \\ \text { equation } & \frac{\mathrm{x}^{2}}{1}-\frac{\mathrm{y}^{2}}{3}=1 \\ & 3 \mathrm{x}^{2}-\mathrm{y}^{2}=3 \end{aligned}$

Q. A tangent to the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$ meets $x$ -axis at $P$ and $y$ -axis at Q. Lines PR and QR are drawn such that OPRQ is a rectangle (where O is the origin). Then R lies on : (1) $\frac{2}{x^{2}}-\frac{4}{y^{2}}=1$ (2) $\frac{4}{x^{2}}-\frac{2}{y^{2}}=1$ (3) $\frac{4}{x^{2}}+\frac{2}{y^{2}}=1$ (4) $\frac{2}{x^{2}}+\frac{4}{y^{2}}=1$ [JEE-Main (On line)-2013]

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Sol. (2) Tangent is $\frac{x}{2} \sec \theta-\frac{y}{\sqrt{2}} \tan \theta=1$ $P\left(\frac{2}{\sec \theta}, 0\right), Q\left(0,-\frac{\sqrt{2}}{\tan \theta}\right)$ $R\left(\frac{2}{\sec \theta},-\frac{\sqrt{2}}{\tan \theta}\right)=(h, k)$ Locus of $R \Rightarrow \quad \frac{4}{h^{2}}-\frac{9}{k^{2}}=1=\frac{4}{x^{2}}-\frac{9}{y^{2}}=1$

Q. A common tangent to the conics $x^{2}=6 y$ and $2 x^{2}-4 y^{2}=9$ is : (1) $x+y=\frac{9}{2}$ (2) $x+y=1$ (3) $x-y=\frac{3}{2}$ (4) x – y = 1 [JEE-Main (On line)-2013]

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Sol. (3) Equation of tangent for $\frac{x^{2}}{(9 / 2)}-\frac{y^{2}}{(9 / 4)}=1$ is $y=m x \pm \sqrt{\frac{9}{2} m^{2}-\frac{9}{4}}$ and tangent to $x^{2}=6 y$ is $y=m x-\frac{3}{2} m^{2}$ ….(2) Now equated ( 1) and ( 2)

Q. The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is : (1) $\sqrt{3}$ (2) $\frac{4}{3}$ (3) $\frac{4}{\sqrt{3}}$ (4) $\frac{2}{\sqrt{3}}$ [JEE-Main 2016]

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Sol. (4) Given $\frac{2 \mathrm{b}^{2}}{\mathrm{a}}=8$ $2 \mathrm{b}=\mathrm{ae}$ we know $\mathrm{b}^{2}=\mathrm{a}^{2}\left(\mathrm{e}^{2}-1\right)$ substitute $\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{e}}{2}$ from ( 2) in ( 3) $\Rightarrow \frac{\mathrm{e}^{2}}{4}=\mathrm{e}^{2}-1 \Rightarrow 4=3 \mathrm{e}^{2}$ $\Rightarrow \mathrm{e}=\frac{2}{\sqrt{3}}$

Q. A hyperbola passes through the point $\mathrm{P}(\sqrt{2}, \sqrt{3})$ and has foci at $(\pm 2,0) .$ Then the tangent to this hyperbola at $\mathrm{P}$ also passes through the point: (1) $(-\sqrt{2},-\sqrt{3})$ (2) $(3 \sqrt{2}, 2 \sqrt{3})$ (3) $(2 \sqrt{2}, 3 \sqrt{3})$ (4) $(\sqrt{3}, \sqrt{2})$ [JEE-Main 2017]

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Sol. (3) Equation of hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ foci is $(\pm 2,0)$ hence ae $=2, \Rightarrow a^{2} e^{2}=4$ $b^{2}=a^{2}\left(e^{2}-1\right)$ $\therefore a^{2}+b^{2}=4$ …(1) Hyperbola passes through $(\sqrt{2}, \sqrt{3})$ $\therefore \frac{2}{\mathrm{a}^{2}}-\frac{3}{\mathrm{b}^{2}}=1$ …(2) On solving ( 1) and ( 2) $\mathrm{a}^{2}=8(\text { is rejected })$ and $\mathrm{a}^{2}=1$ and $\mathrm{b}^{2}=3$ $\therefore \frac{\mathrm{x}^{2}}{1}-\frac{\mathrm{y}^{2}}{3}=1$ Equation of tangent is $\frac{\sqrt{2} \mathrm{x}}{1}-\frac{\sqrt{3} \mathrm{y}}{3}=1$ Hence $(2 \sqrt{2}, 3 \sqrt{3})$ satisfy it.

Q. Tangents are drawn to the hyperbola $4 \mathrm{x}^{2}-\mathrm{y}^{2}=36$ at the point $\mathrm{P}$ and $\mathrm{Q}$. If these tangents intersect at the point $\mathrm{T}(0,3)$ then the area (in sq. units) of $\Delta \mathrm{PTQ}$ is – (1) $54 \sqrt{3}$ (2) $60 \sqrt{3}$ (3) $36 \sqrt{5}$ (4) $45 \sqrt{5}$ [JEE-Main 2018]

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Sol. (4) Equation PQ : chord of contact T = 0 y = – 12

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Comments
  • February 6, 2021 at 8:24 pm

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  • August 20, 2020 at 10:41 pm

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  • August 13, 2020 at 11:34 am

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  • July 25, 2020 at 12:19 am

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  • June 19, 2020 at 12:47 pm

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    • October 30, 2020 at 11:38 am

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  • May 22, 2020 at 10:29 am

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  • March 5, 2020 at 8:59 am

    Solution of 4 question is not understood

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    • October 23, 2020 at 8:06 pm

      i’m getting multiple answers in Q4

      0