Question

The locus of the middle point of the intercept of the tangents drawn from an external point to the ellipse x² + 2y² = 2 between the co-ordinates 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 ≡ ($\sqrt{2}$ cos θ, sin θ)

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

⇒ A ≡ ($\sqrt{2}$sec θ, 0); B ≡ (0, cosec θ)

Let the middle point Q of AB be (h, k)

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

⇒ $\frac{1}{2h^{2}} + \frac{1}{4kh2}$ =1,

∴Required locus is $\frac{1}{2x^{2}} + \frac{1}{4y^{2}}$ = 1.

Trick: The locus of mid-points of the portion of tangents to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ = 1 intercepted between axes is

a² y² + b² x² = 4x² y²

i.e., $\frac{a^{2}}{4x^{2}} + \frac{b^{2}}{4^{2}}$ = 1 or $\frac{1}{2x^{2}} + \frac{1}{4y^{2}}$ = 1.