Question

The point on the curve $y-(x-3{)}^2$ where the tangent is parallel to the chord joining $(3,0)$ and $(4,1)$ is

• A. $\left(-\frac{7}{2},\frac{1}{4}\right)$
• B. $\left(\frac{5}{2},\frac{1}{4}\right)$
• C. $\left(-\frac{5}{2},\frac{1}{4}\right)$
• D. $\left(\frac{7}{2},\frac{1}{4}\right)$

Answer 1

Answer: (D)

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 $\left(\frac{7}{2},\frac{1}{4}\right)$.