EllipseHardBloom L3
Question
On the ellipse $4x^{2} + 9y^{2} = 1$, the points at which the tangents are parallel to line $8x = 9y$ are:
Options
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.$\left( \frac{2}{5},\frac{- 1}{5} \right)$
Solution
Sol. Equation of tangents are
$$\begin{matrix} y & \ = \frac{8}{9}x \pm \sqrt{\frac{1}{4}\left( \frac{8}{9} \right)^{2} + \frac{1}{9}} \\ & \ = \frac{8}{9}x \pm \frac{5}{9} \\ 9y & \ = 8x + 5,\ 9y = 8x - 5 \end{matrix}$$
Tangent at $\left( x_{1},y_{1} \right)$ is $4xx_{1} + 9yy_{1} = 1$
$$\Rightarrow \ \frac{9y_{1}}{9} = \frac{4x_{1}}{- 8} = \frac{1}{5}$$
$$\begin{matrix} & & \ \frac{9y_{1}}{9} = \frac{4x_{1}}{- 8} = - \frac{1}{5} \\ \left( x_{1},y_{1} \right) & \ = \left( - \frac{2}{5},\frac{1}{5} \right)\text{~}\text{or}\text{~}\left( \frac{2}{5}, - \frac{1}{5} \right) & \end{matrix}$$
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