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

Answer

Point is ` (frac{7}{2}, frac{1}{4})`

Explanation

ATQ, `y=(x-3)^{2}`


Slope `=\frac{d y}{d x}=2(x-3)`


Given slope of chord joining `(3,0)` and `(4,1)=\frac{1-0}{4-3}=1`


For parallel lines, slopes to be equal


`\therefore 2(x-3)=1`


`\Rightarrow x=\frac{7}{2}` and `y=(\frac{7}{2}-3)^{2}=\frac{1}{4}`


Hence, point is ` (\frac{7}{2}, \frac{1}{4})`.