Question

The normal to the curve $5x^5-10x^3+x+2y+6=0$ at the point $P(0,-3)$ is tangent to the curve at some other points. Find those points?

Answer 1

$Y=5x^5+10x^3+x+2y+6$

let y=$\frac{5x^5}{2}+5x^3-\frac{x}{2}-3$

Differentiating w.r.tx

$y=\frac{-25x^4}{2}+15x^2-\frac{1}{2}$

substituting (0,-3)

$y={\frac{25\left(0\right)}{2}}^4+15\left(0\right)\frac{-1}{2}$

$=\frac{-1}{2}$

slope of tangent is $\frac{-1}{2}$

equation is

$y=\frac{-1}{2}x+c$

It passes through$(0,-3)$

$\therefore c=-3$

$\Rightarrow y$=\frac-1}{2}$x-3$

solving equation with the carve, we obtain

$5x^5-10x^3=0$

$\Rightarrow x^3\left(x^2\right)=0$

$so,x=0,\sqrt{2},\sqrt{-2}$

points are

$(0,-3,)(\sqrt{2},\frac{-1}{\sqrt{2}}-3),(\sqrt{-2},\frac{1}{\sqrt{2}}-3)$