Question

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

• A. $-\frac{2}{3}$
• B. $\frac{2}{3}$
• C. $\frac{3}{2}$
• D. $-\frac{3}{2}$

Answer 1

Answer: (A)

$-\frac{2}{3}$

Answer 2

Any point on the ellipse is
$(\sqrt{14} cos\,\theta, \sqrt{5} sin\,\theta)$
$\therefore$ Equation of normal at
$(\sqrt{14}\,cos\,\theta, \sqrt{5}\,sin\,\theta)$ is
$\sqrt{14} x\,sec\,\theta, \sqrt{5}y\,cosec\,\theta = 9$
Since, it passes through
$(\sqrt{14} \,cos\,2\theta, \sqrt{5}\,sin\,2\theta)$
$\therefore \sqrt{14}\sqrt{14} \,cos\,2\theta \,sec\,\theta$
$- \sqrt{5} \sqrt{5}\,sin\,2\theta \,cosec\,\theta = 9$
$\Rightarrow 14 \frac{cos\,2\theta}{cos\,\theta} - 5 \frac{sin\,\theta}{sin\,\theta} = 9$
$\Rightarrow 14(2\,cos^2\,\theta -1) - 10\,cos^2\,\theta = 9\,cos\,\theta$
$\Rightarrow 18\,cos^2\theta - 9\,cos\,\theta - 14 = 0$
$\Rightarrow (3\,cos\,\theta + 2)(6\,cos\,\theta - 7) = 0$
$\Rightarrow cos\,\theta = -\frac{2}{3}, cos\,\theta \ne \frac{7}{6}$