Question

For the curve $y=4x^3-2x^5$ find all points at which the tangent passes through the origin.

Answer 1

$y={4x}^3-{2x}^5$

Slope at any point $(a,b)$ is

$y^1={12x}^2-{10x}^4$

$m={12a}^2-10a^4$

Eqn of tangent is $y-b=({12a}^2-{10a}^4)(x-a)$

It passes through $(0,0)$

$b={4a}^3-{2a}^5$ $+b=+a({12a}^2-{10a}^4)$

$b=4\left(0\right)-0=0$ ${4a}^3-{2a}^5={12a}^3-{10a}^5$

$a=1$, $b=4-2=2$ ${8a}^3={8a}^5$

$a=-1$, $b=4(-1)-2(-1)$ $a^5-a^3=0$

$=-4+2=-2$ $a^3(a^2-1)=0$

$a=0,1,-1$

$\therefore$ , Points are $(0,0),(1,2);(-1,-2)$