Hyperbola – JEE Main Previous Year Question with Solutions

Class 9-10, JEE & NEET

JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects. Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more. Download eSaral app for free study material and video tutorials.
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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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 $\frac{x}{2} \sec \theta-\frac{y}{\sqrt{2}} \tan \theta=1$ $P\left(\frac{2}{\sec \theta}, 0\right), Q\left(0,-\frac{\sqrt{2}}{\tan \theta}\right)$ $R\left(\frac{2}{\sec \theta},-\frac{\sqrt{2}}{\tan \theta}\right)=(h, k)$ 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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) Equation of tangent for $\frac{x^{2}}{(9 / 2)}-\frac{y^{2}}{(9 / 4)}=1$ 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]

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 $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ 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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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

• April 26, 2021 at 11:01 am

It’s not enough for us?

100
• April 26, 2021 at 11:04 am

Correct ?

5
• April 26, 2021 at 11:06 am

Right it is not enough we are doing more questions

2
• February 6, 2021 at 8:24 pm

thanks for the precious questions 🙂

3
• December 18, 2020 at 5:17 am

🤟🤟🤟

0
• October 30, 2020 at 7:56 pm

Emmett dddddddddd Jenna am keen is add we nnaa all manner nee d see we let me lengthens meet meeru key kk lily

To GM

0
• August 20, 2020 at 10:41 pm

Interesting and also helpful to all

0
• August 13, 2020 at 11:34 am

Plz give us some more problems

0
• July 25, 2020 at 12:19 am

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

0
• June 19, 2020 at 12:47 pm

Do u have some more questions ?…

0
• October 30, 2020 at 11:38 am

No,these are not helpful

0
• May 22, 2020 at 10:29 am

Chill bro

0
• March 5, 2020 at 8:59 am

Solution of 4 question is not understood

0
• October 23, 2020 at 8:06 pm

i’m getting multiple answers in Q4

0
• 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

1