Question

Find the points on the curve $4x^2+9y^2=1$, where the tangents are perpendicular to the line $2y+x=0$

Answer 1

The equation of curve is $4x^2+9y^2=\mathrm{1......}\left(1\right)$

Differentiating w.r.t x

$8x+18y\frac{dy}{dx}=0$

$\frac{dy}{dx}=-\frac{4}{9}\frac{x}{y}$

Slope of tangent $=-\frac{4x}{9y}$

$m_1=-\frac{4}{9}\frac{x}{y}$

where $m_1$ is slope of tangent

Let $m_2$ be slope of $2y+x=0$

$m_2=-\frac{1}{2}.........\left(2\right)$

Since tangent is perpendicular to line (2)

$\therefore m_1{m}_2=-1$

$(-\frac{4x}{9y})(-\frac{1}{2})=-1$

$y=-\frac{2x}{9}.......\left(3\right)$

Putting this value of y in (1), we get

$4x^2+\frac{4x^2}{9}=1$

$x^2=\frac{9}{40}$

$x=\pm \frac{3}{\sqrt{40}}=\pm \frac{3}{2\sqrt{10}}$

From (3)

$y=-\frac{2}{9}\times \frac{3}{2\sqrt{10}}=-\frac{1}{3\sqrt{40}}$

$y=-\frac{2}{9}\times \frac{-3}{2\sqrt{10}}=\frac{1}{3\sqrt{40}}$

Therefore points are $(\frac{3}{2\sqrt{20}},-\frac{1}{3\sqrt{10}})$ and $(-\frac{3}{2\sqrt{20}},\frac{1}{3\sqrt{10}})$