Question

Find the point on the curve $y=x^3-11x+5$ at which the equation of tangent is $y=x-11$

Answer 1

Let the required point of contact be (x, y)
Given: The equation of the curve is $y=x^3-11x+5$
The equation of the tangent to the circle as $y=x-11$ (which is of the form y = mx + c)
Therefore, Slope of the tangent = 1
Now, the slope of the tangent to the given circle at the point (x, y) is given by $\frac{dy}{dx}=3x^2-11$
Then, we have:
$3x^2-11=1$
$3x^2=12$
$x^2=4$
$x=\pm 2$
When$x=2,y=(2{)}^3-11(2)+5=8-22+5=-9$.
When $x=-2,y=(-2{)}^3-11(-2)+5=-8+22+5=19$.
So, the required point are $(2,-9)$ and $(-2,19)$
But (2, -19) does not satisfy the line y = x - 11
Therefore, (2, -9) is required point of curve at which tangent is $y=x-11.$