Question

Find a point on the curve y = (x – 3)², where the tangent is parallel to the chord joining the points (3, 0) and (4, 1)

Answer 1

We have, y =  (x – 3)², which is polynomial function.

So it is continuous and differentiable.

Thus conditions of mean value theorem are satisfied.

Hence, there exists atleast one c ∈ (3, 4) such that,

f'(c) = `("f"(4) - "f"(3))/(4 - 3)`

⇒ 2(c – 3) = `(1 - 0)/1`

⇒ c – 3 = `1/2`

⇒ c = `7/2 ∈ (3, 4)`

⇒ x = `7/2`, where tangent is parallel to the chord joining points (3, 0) and (4, 1).

For x = `7/2`, y = `(7/2 - 3)^2`

= `(1/2)^2`

= `1/4`

So, `(7/2, 1/4)` is the point on the curve, where tangent drawn is parallel to the chord joining the points (3, 0) and (4, 1).