Question

The locus of the middle point of the intercept of the tangents drawn from an external point to the ellipse $x^{2} + 2y^{2} = 2$between the coordinate axes, is

• A. $\frac{1}{x^{2}} + \frac{1}{2y^{2}} = 1$
• B. $\frac{1}{4x^{2}} + \frac{1}{2y^{2}} = 1$
• C. $\frac{1}{2x^{2}} + \frac{1}{4y^{2}} = 1$
• D. $\frac{1}{2x^{2}} + \frac{1}{y^{2}} = 1$

Answer 1

Answer: (C)

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

Answer 2

Let the point of contact be $R \equiv (\sqrt{2}\cos\theta,\sin\theta)$

Equation of tangent $AB$is $\frac{x}{\sqrt{2}}\cos\theta + y\sin\theta = 1$

⇒ $A \equiv (\sqrt{2}\sec\theta,0);B \equiv (0,\text{coses}\theta)$ Let the middle point $Q$of AB be (h, k).

⇒ $h = \frac{\sec\theta}{\sqrt{2}},k = \frac{\cos ⥂ ec\theta}{2}$⇒ $\cos\theta = \frac{1}{h\sqrt{2}},\sin\theta = \frac{1}{2k}$⇒

$\frac{1}{2h^{2}} + \frac{1}{4k^{2}} = 1$

Thus required locus is $\frac{1}{2x^{2}} + \frac{1}{4y^{2}} = 1$