Get Answers to all your Questions

header-bg qa

Semi transverse axis of hyperbola is 5. Tangent at point P and normal at P meet conjugate axis at A and B respectively. The circle on AB as diameter passes through two fixed points, the distance between which is 20 . The eccentricity of hyperbola is ________.

Option: 1

2


Option: 2

8


Option: 3

9


Option: 4

7


Answers (1)

best_answer

Consider hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.}

Tangent to it any point \mathrm{P(a \sec \theta, b \tan \theta)} is

\mathrm{ \frac{x}{a} \sec \theta-\frac{y}{b} \tan \theta=1 }

It meets y-axis at \mathrm{ A(0,-b \cot \theta).}

Normal at point P is \mathrm{a x \cos \theta+b y \cot \theta=a^2 e^2}

It meets y-axis at \mathrm{B\left(0, \frac{a^2 e^2}{b} \tan \theta\right)}

Now, circle with diameter as AB is

\mathrm{ \begin{aligned} & x^2+(y+b \cot \theta)\left(y-\frac{a^2 e^2}{b} \tan \theta\right)=0 \\ & \Rightarrow x^2+y^2-\left(a^2 e^2\right)+\left(b \cot \theta-\frac{a^2 e^2}{b} \tan \theta\right) y=0 \end{aligned} }
This circle passes through fixed points \mathrm{( \pm a e, 0),} distance between which is 2ae.

\mathrm{ \begin{aligned} & 2 a e=20 \\ \therefore \quad & e=2 \end{aligned} }                           (given)

Posted by

SANGALDEEP SINGH

View full answer

JEE Main high-scoring chapters and topics

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