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)

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
$(\sqrt{14} \,cos\,2\theta, \sqrt{5}\,sin\,2\theta)$
$\therefore \sqrt{14}\sqrt{14} \,cos\,2\theta \,sec\,\theta$
$- \sqrt{5} \sqrt{5}\,sin\,2\theta \,cosec\,\theta = 9$
$\Rightarrow 14 \frac{cos\,2\theta}{cos\,\theta} - 5 \frac{sin\,\theta}{sin\,\theta} = 9$
$\Rightarrow 14(2\,cos^2\,\theta -1) - 10\,cos^2\,\theta = 9\,cos\,\theta$
$\Rightarrow 18\,cos^2\theta - 9\,cos\,\theta - 14 = 0$
$\Rightarrow (3\,cos\,\theta + 2)(6\,cos\,\theta - 7) = 0$
$\Rightarrow cos\,\theta = -\frac{2}{3}, cos\,\theta \ne \frac{7}{6}$