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Q.
The point on the curve $y - (x - 3)^2$ where the tangent is parallel to the chord joining $(3, 0)$ and $(4, 1)$ is
Application of Derivatives
Solution:
We have, $y = (x - 3)^2$
Slope $= \frac{dy}{dx}= 2\left(x - 3\right)$
Slope of chord joining $\left(3, 0\right)$ and $\left(4, 1\right) = \frac{1-0}{4-3}=1$
For parallel lines, slopes to be equal
$\therefore 2\left(x-3\right) = 1$
$\Rightarrow x = \frac{7}{2}$ and $y = \left(\frac{7}{2}-3\right)^{2} = \frac{1}{4}$
Hence, point is $\left(\frac{7}{2}, \frac{1}{4}\right)$.