Ellipse
Tangents
IIT-JEE 1999
Grade 11
Question:
On the ellipse $4x^2 + 9y^2 = 1$, the points at which the tangents are parallel to the line $8x = 9y$ are:
(1) $(\frac{2}{5}, \frac{1}{5})$
(2) $(-\frac{2}{5}, -\frac{1}{5})$
(3) $(\frac{2}{5}, -\frac{1}{5})$ and $(-\frac{2}{5}, \frac{1}{5})$
(4) $(\frac{2}{5}, \frac{1}{5})$ and $(-\frac{2}{5}, -\frac{1}{5})$
Step-by-Step Solution
Key Concept: Tangent at $(x_1, y_1)$ to $4x^2 + 9y^2 = 1$: slope $= -4x_1/(9y_1)$. Set equal to slope of $8x = 9y$ (i.e., $8/9$): $x_1 = -2y_1$. Substitute into ellipse to get $y_1 = \pm 1/5$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (3)