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Q.
The distance between the point $(1,1)$ and the tangent to the curve $y= e ^{2 x }+ x ^{2}$ drawn at the point $x =0$ is
Application of Derivatives
Solution:
Putting $x =0$ in $y = e ^{2 x }+ x ^{2}$ we get $y =1$
$\therefore $ The given point is $P (0,1)$
$y = e ^{2 x }+ x ^{2} \frac{ dy }{ dx }=2 e ^{2 x }+2 x $
$\Rightarrow \left[\frac{ dy }{ dx }\right]_{ P }=2$
$\therefore $ Equation of tangent at $P$ to equation (i)
is $y-1=2(x-0) $
$\Rightarrow 2 x-y+1=0$
$\therefore $ Required distance $=$ Length of $\perp$ from $(1,1)$ to
equation (ii). $=\frac{2-1+1}{\sqrt{4+1}}=\frac{2}{\sqrt{5}} $.