Question

How many tangents are possible from origin on the curve $y=(x+1{)}^3$.Also find the equation of these tangents.

Answer 1

Let the tangent on the curve $y=(x+1{)}^3$ passing through $(0,0)$ be at $(x_1,y_1)$

Then,

$\frac{dy}{dx}=3(x+1{)}^2$

$(\frac{dy}{dx}{)}_{x=x_1}=3(x_1+1{)}^2$

Equation of tangent at $(x_1,y_1)$ is

$y-y_1={\left(\frac{dy}{dx}\right)}_{x=x_1}(x-x_1)$

$y-y_1=3(x_1+1{)}^2(x-x_1)$

Since this curve passes through $(0,0)$

So, $y_1=3x_1(x_1+1{)}^2$ ---- (i)

Since $(x_1,y_1)$ lies on $y=(x+1{)}^3$

So, $y_1=(x_1+1{)}^3$ ----- (ii)

From (i) and (ii)

$3x_1(x_1+1{)}^2=(x_1+1{)}^3$

$3x_1(x_1+1{)}^2-(x_1+1{)}^3=0$

$(x_1+1{)}^2\left[3x_1-x_1-1\right]=0$

$(x_1+1{)}^2\left[2x_1-1\right]=0$

$x_1=-1,\frac{1}{2}$

Therefore, two tangents are possible from origin to the curve $y=(x+1{)}^3$