Question

A tangent to the ellipse $x^{2} + 4y^{2} = 4$meets the ellipse$x^{2} + 2y^{2} = 6$at P and Q. The angle between the tangents at P and Q of the ellipse $x^{2} + 2y^{2} = 6$is

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{4}$
• D. $\frac{\pi}{6}$

Answer 1

Answer: (A)

$\frac{\pi}{2}$

Answer 2

The given ellipse $x^{2} + 4y^{2} = 4$can be written as $\frac{\mathbf{x}^{\mathbf{2}}}{\mathbf{4}}\mathbf{+}\frac{\mathbf{y}^{\mathbf{2}}}{\mathbf{1}}\mathbf{= 1}$ .....(i)

Any tangent to ellipse (i) is $\frac{x}{2}\cos\theta + y\sin\theta = 1$ .....(ii)

Second ellipse is $x^{2} + 2y^{2} = 6$ , i.e. $\frac{x^{2}}{6} + \frac{y^{2}}{3} = 1$.....(iii)

Let the tangents at $P,Q$meet at $(h,k)$.

∴Equation of PQ, i.e. chord of contact is $\frac{hx}{6} + \frac{ky}{3} = 1$ .....(iv)

Since (ii) and (iv) represent the same line,

∴$\frac{h/6}{(\cos\theta)/2} = \frac{k/3}{\sin\theta} = \frac{1}{1}$⇒ $h = 3\cos\theta$and $k = 3\sin\theta$

So, $h^{2} + k^{2} = 9$or $x^{2} + y^{2} = 9$is the locus of $(h,k)$ which is the director circle of the ellipse $\frac{x^{2}}{6} + \frac{y^{2}}{3} = 1$

∴ The angle between the tangents at P and Q will be $\pi/2$