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 point(s). Find those point(s)?

Answer 1

Given ,

$5x^2-10x^3+x+2y+6=0$

Differentiate both side we get

$25x^4-30x^2+1+2\frac{dy}{dx}=0$

$-2\frac{dy}{dx}=25x^4-30x^2+1$

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

Given point, (0,-3)

${\left(\frac{dy}{dx}\right)}_{(0,-3)}=-\frac{1}{2}$

Now equation of tangent

$\because y-y_1=\frac{dy}{dx}(x-x_1)$

$\because y+3=-\frac{1}{2}(x-0)$

$y=-\frac{x}{2}-3$ …..(1)

Equation of normal

$\because y-y_1=\frac{1}{{}^{-\frac{dy}{dx}}}(x-x_1)$

$y+3=\frac{1}{\frac{1}{2}}(x-0)$

$y+3=2x$

$y=2x-3$ ……(2)

By solving equation (1) and (2) , we get

X=0 ,y = - 3

These are required point