Question

find a point on the curve y= (x- 2)^2 at which tangent is parallel to the chord joining the points (2, 0) and (4, 4) find the equation of tangent at this point​

Answer 1

Answer:

(3,1)

Step-by-step explanation:

y=(x−2)2

dxdy=dxd((x−2)2)=2(x−2)

∴ slope of tangent =2x−4

Slope of line joining (2,0) and (4,4) =4−24−0=2

The tangent is parallel to this line

∴ their slopes are equal

2x−4=2 ⇒2x=6

∴x=3

and y=(3−2)2=1

Thus the point is (3,1)