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] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2)Tangent is \frac{x}{2} \sec \theta-\frac{y}{\sqrt{2}} \tan \theta=1P\left(\frac{2}{\sec \theta}, 0\right), Q\left(0,-\frac{\sqrt{2}}{\tan \theta}\right)$$R\left(\frac{2}{\sec \theta},-\frac{\sqrt{2}}{\tan \theta}\right)=(h, k)Locus of R \Rightarrow \quad \frac{4}{h^{2}}-\frac{9}{k^{2}}=1=\frac{4}{x^{2}}-\frac{9}{y^{2}}=1 Q. A common tangent to the conics x^{2}=6 y and 2 x^{2}-4 y^{2}=9 is :(1) x+y=\frac{9}{2}(2) x+y=1(3) x-y=\frac{3}{2}(4) x – y = 1 [JEE-Main (On line)-2013] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3)Equation of tangent for \frac{x^{2}}{(9 / 2)}-\frac{y^{2}}{(9 / 4)}=1is y=m x \pm \sqrt{\frac{9}{2} m^{2}-\frac{9}{4}}and tangent to x^{2}=6 y is y=m x-\frac{3}{2} m^{2} ….(2)Now equated ( 1) and ( 2) Q. The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is :(1) \sqrt{3}(2) \frac{4}{3}(3) \frac{4}{\sqrt{3}}(4) \frac{2}{\sqrt{3}} [JEE-Main 2016] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4)Given\frac{2 \mathrm{b}^{2}}{\mathrm{a}}=8$$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}$$\Rightarrow \mathrm{e}=\frac{2}{\sqrt{3}} Q. A hyperbola passes through the point \mathrm{P}(\sqrt{2}, \sqrt{3}) and has foci at (\pm 2,0) . Then the tangent to this hyperbola at \mathrm{P} also passes through the point:(1) (-\sqrt{2},-\sqrt{3})(2) (3 \sqrt{2}, 2 \sqrt{3})(3) (2 \sqrt{2}, 3 \sqrt{3})(4) (\sqrt{3}, \sqrt{2}) [JEE-Main 2017] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3)Equation of hyperbola is \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1foci is (\pm 2,0) hence ae =2, \Rightarrow a^{2} e^{2}=4$$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 = 0y = – 12

• August 31, 2021 at 7:41 pm

solutions are not at all clear please try to give clear solutions

• April 26, 2021 at 11:01 am

It’s not enough for us?

• April 26, 2021 at 11:04 am

Correct ?

• April 26, 2021 at 11:06 am

Right it is not enough we are doing more questions

• February 6, 2021 at 8:24 pm

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• December 18, 2020 at 5:17 am

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• October 30, 2020 at 7:56 pm

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To GM

• August 20, 2020 at 10:41 pm

Interesting and also helpful to all

• August 13, 2020 at 11:34 am

Plz give us some more problems

• July 25, 2020 at 12:19 am

𝕐𝕖𝕤 𝕥𝕙𝕚𝕤 𝕚𝕤 𝕣𝕖𝕒𝕝𝕪 𝕙𝕖𝕝𝕡𝕗𝕦𝕝 𝕗𝕠𝕣 𝕒𝕟𝕪 𝕒𝕤𝕡𝕣𝕚𝕟𝕥 𝕨𝕙𝕠 𝕨𝕒𝕟𝕥 𝕥𝕠 𝕕𝕠 𝕓𝕖𝕥𝕥𝕖𝕣 𝕤𝕔𝕠𝕣𝕖 𝕚𝕟 𝕖𝕟𝕥𝕣𝕒𝕟𝕔𝕖 𝕖𝕩𝕒𝕞

• June 19, 2020 at 12:47 pm

Do u have some more questions ?…

• October 30, 2020 at 11:38 am

• May 22, 2020 at 10:29 am

Chill bro

• March 5, 2020 at 8:59 am

Solution of 4 question is not understood

• October 23, 2020 at 8:06 pm

i’m getting multiple answers in Q4

• March 26, 2021 at 10:12 pm

HI BRO
WE DONT NEED TO SEE THE SOLUTION FOR QUESTION NUMBER-4
ITS A PRETTY SIMPLE QUESTION
TRY ONCE MORE U WILL DEFINETLY GET IT