HyperbolaHardBloom L3

Question

The normal at one extremity of latus rectum (in 1st quadrant) of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,a > b > 0$ meets the rectangular hyperbola $xy = 9$ at points $P$ and $Q$, then:

Options

A.If $P$ is $\left( 6,\frac{3}{2} \right) \Rightarrow Q$ is $\left( - \frac{3\sqrt{2}}{2}, - 3\sqrt{2} \right)$
B.Eccentricity of hyperbola is $\sqrt{2}$
C.If $P$ is $\left( 6,\frac{3}{2} \right) \Rightarrow Q$ is $\left( - \frac{3e}{2}, - \frac{6}{e} \right)$ where $e$ is eccentricity of the given ellipse
D.If $O$ is origin, then product of slopes of $OP$ and $OQ$ is positive

Solution

Sol. (b, c, d)

Equation of normal at $\left( ae,\frac{b^{2}}{a} \right)$

$$\begin{matrix} \frac{a}{e}x - ay & \ = a^{2}e^{2} \\ x - ey & \ = ae^{3} \end{matrix}$$

Put $y = \frac{9}{x},x - 9\frac{e}{x} = ae^{3}$

If

$$\begin{matrix} & x_{1} = 6,x_{1}x_{2} = \ - 9e,\ \\ & x_{2} = - \frac{3e}{2} \\ & y_{2} = \frac{9}{x_{2}} = - \frac{6}{e} \end{matrix}$$

$\Rightarrow \ $ (slope of $OP$ ) (slope of $QQ$ )

$$= \left( \frac{y_{1}}{x_{1}} \right)\left( \frac{y_{2}}{x_{2}} \right) = \frac{9}{x_{1}^{2}}\frac{9}{x_{2}^{2}} = \frac{81}{x_{1}^{2}x_{2}^{2}} > 0.$$

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