Question

Find the distance between the point $(1,1)$ and the tangent to the curve $y=e^{2x}+x^2$ drawn from the point where the curve cuts $y$-axis

• A. $\frac{\sqrt{3}}{\sqrt{5}}$
• B. $\frac{3}{\sqrt{5}}$
• C. $\frac{2}{\sqrt{5}}$
• D. $\frac{\sqrt{2}}{\sqrt{5}}$

Answer 1

The correct option is C $\frac{2}{\sqrt{5}}$

Clearly the the point on the $y$-axis is $(0,1)\equiv P$ (say)
Now $\frac{dy}{dx}=2e^{2x}+2x$
Thus slope at $P$ is $m={\left(\frac{dy}{dx}\right)}_{(0,1)}=2$
Thus required tangent is $(y-1)=2(x-0)\Rightarrow 2x-y+1=0$
Hence, distance of $(1,1)$ from this line is $=\left|\frac{2\times 1-1+1}{\sqrt{2^2+1^2}}\right|=\frac{2}{\sqrt{5}}$