Ellipse
Locus of Midpoint
IIT-JEE 2004 Screening
Grade 11
Question:
Tangents are drawn to the ellipse $x^2 + 2y^2 = 2$. The locus of the midpoint of the intercept made by the tangents between the coordinate axes is:
(1) $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$
(2) $\frac{1}{4x^2} + \frac{1}{2y^2} = 1$
(3) $\frac{1}{x^2} + \frac{1}{2y^2} = 2$
(4) $\frac{1}{2x^2} + \frac{1}{y^2} = 2$
Step-by-Step Solution
Key Concept: Tangent at $(x_1, y_1)$: $xx_1 + 2yy_1 = 2$. Intercepts: $(2/x_1, 0)$ and $(0, 1/y_1)$. Midpoint $(h, k)$: $x_1 = 1/h$, $y_1 = 1/(2k)$. Substituting into the ellipse equation gives $1/h^2 + 1/(2k^2) = 2$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (3)