Question

The tangent to the curve $y=x^2-5x+5$, parallel to the line $2y=4x+1$, also passes through the point :

• A. $(\frac{7}{2},\frac{1}{4})$
• B. $(\frac{1}{4},\frac{7}{2})$
• C. $(-\frac{1}{8},7)$
• D. $(\frac{1}{8},-7)$

Answer 1

The correct option is D $(\frac{1}{8},-7)$

$y=x^2-5x+5\cdots \left(1\right)$
$\Rightarrow \frac{dy}{dx}=2x-5$
Slope of the line $2y=4x+1$ is $2$
$\therefore 2x-5=2\Rightarrow x=\frac{7}{2}$
Putting this value in $\left(1\right)$, we get :
$y=\frac{49}{4}-\frac{35}{2}+5=-\frac{1}{4}$

Equation of tangent :
$y+\frac{1}{4}=2(x-\frac{7}{2})$
$\Rightarrow 8x-28=4y+1$
$\Rightarrow 8x-4y-29=0$
$(\frac{1}{8},-7)$ satisfies above equation.