Question

Let P(6,3) be a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

If the normal at the point P intersects the x-axis at (9,0), then the eccentricity of hyperbola is,

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Given hyperbola is,

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

We know that slope of normal at $(x_1,y_1)=-\frac{a^2{y}_1}{b^2{x}_1}$

and equation of normal at $(x_1,y_1)$ is,

$(y-y_1)=-\frac{a^2{y}_1}{b^2{x}_1}(x-x_1)$

Given that $(x_1,y_1)\equiv (6,3)$

$\therefore$ Equation of normal becomes,

$y-3=-\frac{a^{2}3}{b^2{6}_2}(x-6)$

$2b^2.y-6b^2+a^{2}x-6a^2=0$

Its given that this line passes through the point (9,0)

$\therefore 0-6b^2+9a^2-6a^2=0$

$6b^2=3a^2$

$\frac{b^2}{a^2}=\left[\frac{1}{2}\right]$ .............(1)

Eccentricity of the hyperbola is given by the relation

$e^2=1+\frac{b^2}{a^2}=1+\frac{1}{2}\left(from\right(1\left)\right)$

$=\frac{3}{2}$

$\therefore e=\sqrt{\frac{3}{2}}$