Question

If the normal at the point $P(\Theta)$ to the ellipse $\frac{x^2}{3}+\frac{y^2}{2}=1$ intersects it again at the point $Q(2\Theta)$, then find $\cos\Theta$.

• A. $\frac{3}{2}$
• B. $-1$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

Answer: (B)

$-1$

Answer 2

The equation of the ellipse is $\frac{x^2}{3}+\frac{y^2}{2}=1$, so $a^2=3$ and $b^2=2$. The equation of the normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $(a\cos\Theta, b\sin\Theta)$ is $\frac{ax}{\cos\Theta} - \frac{by}{\sin\Theta} = a^2-b^2$. Substituting $a^2=3$ and $b^2=2$, the normal at $P(\Theta)$ is $\frac{\sqrt{3}x}{\cos\Theta} - \frac{\sqrt{2}y}{\sin\Theta} = 1$. Since the normal intersects the ellipse again at $Q(2\Theta)$, the coordinates of $Q$ satisfy the normal equation: $\frac{\sqrt{3}(\sqrt{3}\cos(2\Theta))}{\cos\Theta} - \frac{\sqrt{2}(\sqrt{2}\sin(2\Theta))}{\sin\Theta} = 1$ $\frac{3\cos(2\Theta)}{\cos\Theta} - \frac{2\sin(2\Theta)}{\sin\Theta} = 1$ Using double angle formulas: $\frac{3(2\cos^2\Theta - 1)}{\cos\Theta} - \frac{2(2\sin\Theta\cos\Theta)}{\sin\Theta} = 1$ $\frac{3(2\cos^2\Theta - 1)}{\cos\Theta} - 4\cos\Theta = 1$ $6\cos^2\Theta - 3 - 4\cos^2\Theta = \cos\Theta$ $2\cos^2\Theta - \cos\Theta - 3 = 0$ $(2\cos\Theta - 3)(\cos\Theta + 1) = 0$ $\cos\Theta = \frac{3}{2}$ or $\cos\Theta = -1$. Since $-1 \le \cos\Theta \le 1$, we have $\cos\Theta = -1$.