# Find the points on the curve $$y=x^3$$ at which the slope of the tangent is equal to the y - coordinate of the point.

Toolbox:
• Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
• Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1) Step 1: Let P(x_1,y_1) be the required point. The given curve is y=x^3----(1) Differentiating w.r.t x we get, \large\frac{dy}{dx}$$=3x^2$
The slope of the tangent at $P(x_1,y_1)$ is
$\large\frac{dy}{dx}_{(x_1,y_1)}$$=3x^2------(2) Step 2: It is given that the slope of the tangent is equal to the y coordinate of the point. Hence \large\frac{dy}{dx}$$=y_1$------(3)
Equating equ(2) and equ(3) we get,
$3x_1^2=y_1$-----(4)
Also $(x_1,y_1)$ lies on equ(1)
Hence $y_1=x_1^3$----(5)
Step 3:
Now equating equ(4) and equ(5)
$3x_1^2=x_1^3$
$\Rightarrow 3x_1^2-x_1^3=0$
$x_1^2(3-x_1)=0$
Therefore $x_1=0$ and $x_1=3$
Step 4:
When $x_1=0,y_1=(0)^3=0$
When $x_1=3,y_1=(3)^3=27$
Hence the required points are $(0,0)$ and $(3,27)$