At what points on the curve
Question:

At what points on the curve $y=x^{2}-4 x+5$ is the tangent perpendicular to the line $2 y+x=7 ?$

Solution:

Let $\left(x_{1}, y_{1}\right)$ be the required point.

Slope of the given line $=\frac{-1}{2}$

Slope of the line perpendicular to this line $=2$

Since, the point lies on the curve.

Hence, $y_{1}=x_{1}{ }^{2}-4 x_{1}+5 \quad \ldots(1)$

Now, $y=x^{2}-4 x+5$

$\therefore \frac{d y}{d x}=2 x-4$

Now,

Slope of the tangent at $\left(x_{1}, y_{1}\right)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=2 x_{1}-4$

Slope of the tangent at $\left(x_{1}, y_{1}\right)=$ Slope of the given line [Given]

$\therefore 2 x_{1}-4=2$

$\Rightarrow 2 x_{1}=6$

$\Rightarrow x_{1}=3$

Also,

$y_{1}=9-12+5=2$      [From eq. (1)]

Thus, the required point is $(3,2)$.