At what points on the curve

Solution:

Given:

The curve $y=x^{2}-4 x+5$ and line is $2 y+x=7$

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

Differentiating the above w.r.t $x$,

we get the Slope of the tangent,

$\Rightarrow \frac{d y}{d x}=2 x^{2}-1-4+0$

$\Rightarrow \frac{d y}{d x}=2 x-4 \ldots(1)$

Since, line is $2 y+x=7$

$\Rightarrow 2 y=-x+7$

$\Rightarrow y=\frac{-x+7}{2}$

$\Rightarrow y=\frac{-x}{2}+\frac{7}{2}$

$\therefore$ The equation of a straight line is $y=m x+c$, where $m$ is the The Slope of the line.

Thus, the The Slope of the line is $\frac{-1}{2} \ldots(2)$

Since, tangent is perpendicular to the line,

$\therefore$ The Slope of the normal $=\frac{-\mathbf{1}}{\text { The Slope of the tangent }}$

From (1) \& (2), we get

i. $e, \frac{-1}{2}=\frac{-1}{2 x-4}$

$\Rightarrow 1=\frac{1}{x-2}$

$\Rightarrow x-2=1$

$\Rightarrow x=3$

Substituting $x=3$ in $y=x^{2}-4 x+5$

$\Rightarrow y=y=3^{2}-4 \times 3+5$

$\Rightarrow y=9-12+5$

$\Rightarrow y=2$

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