First slide

Parabola

Question

If the normal at the point $P (θ)$ to the ellipse $\frac{x^{2}}{14} + \frac{y^{2}}{5} = 1$
intersect it again at the point $Q (2 θ),$ then $\cos θ =$

Moderate

A

$\frac{- 2}{3}$
B

$\frac{2}{3}$
C

$\frac{- 6}{7}$
D

$\frac{6}{7}$

Solution

Normal at $P (\sqrt{14} \cos θ, \sqrt{5} \sin θ)$ to the ellipse is
$\frac{\sqrt{14} x}{\cos θ} - \frac{\sqrt{5} y}{\sin θ} = 14 - 5 = 9$
Since it passes through $Q (2 θ)$
$\frac{\sqrt{14} \times \sqrt{14} \cos 2 θ}{\cos θ} - \frac{\sqrt{5} \times \sqrt{5} \sin 2 θ}{\sin θ} = 9$
$\begin{array}{l} \Rightarrow \frac{14 (2 \cos^{2} θ - 1)}{\cos θ} - \frac{5 \times 2 \sin θ \cos θ}{\sin θ} = 9 \\ \Rightarrow 18 \cos^{2} θ - 9 \cos θ - 14 = 0 \\ \Rightarrow (3 \cos θ + 2) (6 \cos θ - 7) = 0 \\ \Rightarrow \cos θ = - 2 / 3 . \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App