Question

If tangents are drawn to the ellipse $x^2 + 2 y^2 = 2$, then the locus of the midpoints of the intercepts made by the tangents between the coordinate axes is:

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

Hint:

The locus of midpoints of intercepts of tangents to conics is found by expressing intercepts in terms of tangent parameters and eliminating parameters via tangent equation.

Answer 1

The Correct Option is D

Step 1: Equation of ellipse and tangent
Ellipse: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \] where $a^2 = 2$, $b^2 = 1$. Equation of tangent at $(x_1,y_1)$: \[ \frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1 \Rightarrow \frac{x x_1}{2} + y y_1 = 1. \] Step 2: Intercepts on axes
$x$-intercept ($y=0$): \[ \frac{x x_1}{2} = 1 \implies x = \frac{2}{x_1}. \] $y$-intercept ($x=0$): \[ y y_1 = 1 \implies y = \frac{1}{y_1}. \] Step 3: Midpoint of intercepts
\[ M = \left(\frac{1}{2} \times \frac{2}{x_1}, \frac{1}{2} \times \frac{1}{y_1}\right) = \left(\frac{1}{x_1}, \frac{1}{2 y_1}\right). \] Step 4: Find locus of $M$
From tangent condition: \[ \frac{x_1^2}{2} + y_1^2 = 1. \] Express in terms of $M=(X,Y)$: \[ x_1 = \frac{1}{X}, \quad y_1 = \frac{1}{2 Y}. \] Substitute: \[ \frac{1}{2 X^2} + \frac{1}{4 Y^2} = 1. \]