Question

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

Answer 1

The equation of the given curve is $y=4x^3-2x^5$.

Differentiating w.r.t. $x$, we get
$\Rightarrow \frac{dy}{dx}=12x^2-10x^4$

Let $P(x_1,y_1)$ be the point on the curve at which tangent passes through the orgin.
Slope of the tangent at $P(x_1,y_1)$ is ${\left(\frac{dy}{dx}\right)}_{x_1,y_1}=12{x_1}^2-10{x_1}^4$.

Equation of the tangent at $(x_1,y_1)$ is given by,
$y-y_1=(12x_1^2-10x_1^4)(x-x_1)$ ....(1)
Since, the tangent passes through the origin (0, 0)
So, equation (1) becomes
$-y_1=(12x_1^2-10x_1^4)(-x_1)$
$y_1=12x_1^3-10x_1^5$ .....(2)

Since, $P(x_1,y_1)$ lies on the curve
So ,$y_1=4x_1^3-2x_1^5$ ......(3)

From eqn (2) and (3), we get
$12x_1^3-10x_1^5=4x_1^3-2x_1^5$
$\Rightarrow 8x_1^5-8x_1^3=0$
$\Rightarrow x_1^5-x_1^3=0$
$\Rightarrow x_1^3(x_1^2-1)$
$\Rightarrow x_1=0,\pm 1$
When $x_1=0,y_1=4(0{)}^3-2(0{)}^5=0$.
When $x_1=1,y_1=4(1{)}^3-2(1{)}^5=2$.
When $x_1=-1,y_1=4(-1{)}^3-2(-1{)}^5=-2$.
Hence, the required points are $(0,0),(1,2)$ and $(-1,-2)$.