Question

On the ellipse 4x² + 9y² = 1, the points at which the tangents are parallel to the line 8x = 9y are

• A. $\left( \frac{2}{5},\frac{1}{5} \right)$
• B. $\left( - \frac{2}{5},\frac{1}{5} \right)$
• C. $\left( - \frac{2}{5}, - \frac{1}{5} \right)$
• D. $\left( \frac{2}{5}, - \frac{1}{5} \right)$

Answer 1

Answer: (B)

$\left( - \frac{2}{5},\frac{1}{5} \right)$

Answer 2

Ellipse is $\frac{x^{2}}{\frac{1}{4}} + \frac{y^{2}}{\frac{1}{9}}$ = 1 ⇒ a² = $\frac{1}{4}$, b² = $\frac{1}{9}$

The equation of its tangent is 4xx' + 9yy' = 1

∴ m = - $\frac{4x'}{9y'} = \frac{8}{9}$∴ x' = - 2y'

and 4x^'2 + 9y'² = 1 ⇒ 4x'² + 9$\frac{x'^{2}}{4}$ = 1 ⇒ x' = ± $\frac{2}{5}$

When x' = $\frac{2}{5}$, y' = - $\frac{1}{5}$ and when x' = -$\frac{2}{5}$, y' = $\frac{1}{5}$

Hence points are $\left( \frac{2}{5}, - \frac{1}{5} \right)$ and $\left( - \frac{2}{5},\frac{1}{5} \right)$.