Find the point on the parabola `y=(x-3)^2,` where the tangent is parabola to the line joining (3,0) and (4,1

Answer

N/a

Explanation

We have, `y = (x-3)^(2)`, which is continuous in `x_(1) = 3` and ` x_(2) = 4` i.e. `[3,4]`
Also. `y' =2x(x-3).1 = 2(x-3)` which exists in `(3,4)`
Hence, by mean value theorem there exists a poin ton the curve at which,
Thus, `f'(x) = (f(4) - f(3))/(4-3)`
`rArr 2(c-3) = ((4-3)^(2) - (3-3)^(2))/(4-3)`
`rArr 2c-6 = (1-0)/(1) rArr c = 7/2`
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 at which tangent drawn in parallel to the chord joining the point `(3,0)` and `(4,1)`.