Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

If from any point $\mathrm{P\left(x_1, y_1\right)}$ on the hyperbola $\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1}$ , tangents are drawn to the hyperbola $\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,}$ then the corresponding chord of contact touches the another branch of the hyperbola $\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1}$ at the point
Option: 1
$\left(x_1,-y_1\right)$

Option: 2
$\left(-x_1, y_1\right)$

Option: 3
$\left(x_1, y_1\right)$

Option: 4
$\left(-x_1,-y_1\right)$

Answers (1)

best_answer

Chord of contact of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ w.r.t. point $P\left(x_1, y_1\right)$ is is

$\frac{x x_1}{a^2}-\frac{y y_1}{b^2}=1$ $...(1)$

Eq. (1) can be written as $\frac{x\left(-x_1\right)}{a^2}-\frac{y\left(-y_1\right)}{b^2}=-1$ ,

which is tangent to the hyperbola

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$ at point $\left(-x_1,-y_1\right).$

Obviously, point $\left(x_1, y_1\right)$ and $\left(-x_1,-y_1\right)$ lie on the different branches of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1.$

Posted by

Ritika Harsh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Trending Articles/News

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs

anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question