SATHEE: Ellipse 2 Question 21

Ellipse 2 Question 21

21. A tangent to the ellipse $x^{2} + 4 y^{2} = 4$ meets the ellipse $x^{2} + 2 y^{2} = 6$ at $P$ and $Q$ . Prove that the tangents at $P$ and $Q$ of the ellipse $x^{2} + 2 y^{2} = 6$ are at right angles.

(1997, 5M)

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Answer:

Correct Answer: 21. (a)

Solution:

Given, $x^{2} + 4 y^{2} = 4$ or $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$

Equation of any tangent to the ellipse on (i) can be written as

$\frac{x}{2} \cos θ + y \sin θ = 1$

Equation of second ellipse is

$\Rightarrow \begin{aligned} x^{2} + 2 y^{2} = 6 \\ \Rightarrow \frac{x^{2}}{6} + \frac{y^{2}}{3} = 1 \end{aligned}$

Suppose the tangents at $P$ and $Q$ meets at $A (h, k)$ . Equation of the chord of contact of the tangents through $A (h, k)$ is

$\frac{h x}{6} + \frac{k y}{3} = 1$

But Eqs. (iv) and (ii) represent the same straight line, so comparing Eqs. (iv) and (ii), we get

$\begin{aligned} \Rightarrow & \frac{h / 6}{\cos θ / 2} & = \frac{k / 3}{\sin θ} = \frac{1}{1} \\ h & = 3 \cos θ and k = 3 \sin θ \end{aligned}$

Therefore, coordinates of $A$ are $(3 \cos θ, 3 \sin θ)$ .

Now, the joint equation of the tangents at $A$ is given by $T^{2} = S S_{1}$ , i.e. $\frac{h x}{6} + \frac{k y}{3} - 1 = \frac{x^{2}}{6} + \frac{y^{2}}{3} - 1 \frac{h^{2}}{6} + \frac{k^{2}}{3} - 1$

In Eq. (v), coefficient of $x^{2} = \frac{h^{2}}{36} - \frac{1}{6} \frac{h^{2}}{6} + \frac{k^{2}}{3} - 1$

$= \frac{h^{2}}{36} - \frac{h^{2}}{36} - \frac{k^{2}}{18} + \frac{1}{6} = \frac{1}{6} - \frac{k^{2}}{18}$

and coefficient of $y^{2} = \frac{k^{2}}{9} - \frac{1}{3} \frac{h^{2}}{6} + \frac{k^{2}}{3} - 1$

$= \frac{k^{2}}{9} - \frac{h^{2}}{18} - \frac{k^{2}}{9} + \frac{1}{3} = - \frac{h^{2}}{18} + \frac{1}{3}$

Again, coefficient of $x^{2} +$ coefficient of $y^{2}$

$\begin{aligned} = - \frac{1}{18} (h^{2} + k^{2}) + \frac{1}{6} + \frac{1}{3} \\ = - \frac{1}{18} (9 \cos^{2} θ + 9 \sin^{2} θ) + \frac{1}{2} \\ = - \frac{9}{18} + \frac{1}{2} = 0 \end{aligned}$