The point P(-2√6,√3) lies on the hyperbola x^2a^2-y^2b^2=1 having eccentricity √52.

Question

The point $P(-2\sqrt6,\sqrt3)$ lies on the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ having eccentricity $\dfrac{\sqrt5}2$. If the tangent and normal at $P$ to the hyperbola intersect its conjugate axis at the points $Q$ and $R$ respectively, then $QR$ is equal to

Accepted Answer

$$P\left( { - 2\sqrt 6 ,\sqrt 3 } \right)$$ lies on hyperbola

$$ \Rightarrow {{24} \over {{a^2}}} - {3 \over {{b^2}}} = 1$$ ...... (i)

$$e = {{\sqrt 5 } \over 2} \Rightarrow {b^2} = {a^2}\left( {{5 \over 4} - 1} \right) \Rightarrow 4{b^2} = {a^2}$$

Put in (i) $$ \Rightarrow {6 \over {{b^2}}} - {3 \over {{b^2}}} = 1 \Rightarrow b = \sqrt 3 $$

$$ \Rightarrow a = \sqrt {12} $$

Tangent at P :

$${{ - x} \over {\sqrt 6 }} - {y \over {\sqrt 3 }} = 1 \Rightarrow Q(0,\sqrt 3 )$$

Slope of $$T =  - {1 \over {\sqrt 2 }}$$

Normal at P :

$$y - \sqrt 3  = \sqrt 2 (x + 2\sqrt 6 )$$

$$ \Rightarrow R = (0,5\sqrt 3 )$$

$$QR = 6\sqrt 3 $$

Solutions · 1