# Find the point on the curve

Question:

Find the point on the curve $y=x^{3}-11 x+5$ at which the tangent is $y=x-11$.

Solution:

The equation of the given curve is $y=x^{3}-11 x+5$.

The equation of the tangent to the given curve is given as $y=x-11$ (which is of the form $y=m x+c$ ).

$\therefore$ Slope of the tangent $=1$

Now, the slope of the tangent to the given curve at the point $(x, y)$ is given by, $\frac{d y}{d x}=3 x^{2}-11$

Then, we have:

$3 x^{2}-11=1$

$\Rightarrow 3 x^{2}=12$

$\Rightarrow x^{2}=4$

$\Rightarrow x=\pm 2$

When $x=2, y=(2)^{3}-11(2)+5=8-22+5=-9$

When $x=-2, y=(-2)^{3}-11(-2)+5=-8+22+5=19$

Hence, the required points are (2, −9) and (−2, 19).

But, both these points should satisfy the equation of the tangent as there would be point of contact between tangent and the curve.

∴ (2, −9) is the required point as (−2, 19) is not satisfying the given equation of tangent.