The tangent to the curve
Question:

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

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

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

3. $\left(-\frac{1}{8}, 7\right)$

4. $\left(\frac{1}{8},-7\right)$

Correct Option: , 4

Solution:

$y=x^{2}-5 x+5$

$\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}-5=2 \Rightarrow \mathrm{x}=\frac{7}{2}$

at $\mathrm{x}=\frac{7}{2}, \mathrm{y}=\frac{-1}{4}$

Equation of tangent at $\left(\frac{7}{2}, \frac{-1}{4}\right)$ is $2 \mathrm{x}-\mathrm{y}-\frac{29}{4}=0$

Now check options

$x=\frac{1}{8}, y=-7$