Question

Find the points on the curve y = x3 - 11x + 5at which the tangent has the equation y = x- 1​

Answer 1

Answer:

Slope of the tangent y = x - 11 is 1, y = x3 - 11x + 5 Since (-2, 19) does not lie on the tangent y = x - 11 .. .  Requred point is (2, -9).Read more on Sarthaks.com - https://www.sarthaks.com/201300/find-the-point-on-the-curve-y-x-3-11x-5-at-which-the-equation-of-tangent-is-y-x-11

Answer 2

Answer:

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

dx

dy

=3x

2

−11

Then, we have:

3x

2

−11=1

3x

2

=12

x

2

=4

x=±2

Whenx=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.

Step-by-step explanation:

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