Fun With Calculus 2

When $x = 0, y = 1$
Differentiating,
$\dfrac{dy}{dx} = 2e^{2x} + 2x$
$\dfrac{dy}{dx} \left. \right|_{x=0} = 2$
Slope of normal = $\dfrac{-dx}{dy} = -\dfrac{1}{2}$
Equation of normal using slope-intercept form,
$y - 1 = -\dfrac{1}{2}(x-0) \to 2y+x = 2$
Dividing by $\sqrt{5}$
$\dfrac{2y}{\sqrt{5}} + \dfrac{x}{\sqrt{5}} = \dfrac{2}{\sqrt{5}}$
This is of the form,
$x\cos(\alpha) + y\sin(\alpha) = p$ where p is perpendicular distance from the origin.

$a + b = 2+5 = 7$

Side note :
Distance(d) of the point $(x_{1},y_{1})$ from the line $ax + by + c$ is given by ,
$d = \dfrac{|ax_{1} + by_{1} + c |}{\sqrt{a^{2} + b^{2}}}$

Let $f$ defined by $\forall x \in \mathbb{R}, f(x)=e^{2x}+x^2$

First find the equation of the normal to the curve $y=f(x)$ at $x=0$ :

The slope is $-\frac{1}{f'(0)}=-\frac{1}{2}$ . The normal passes trough the point $(0,f(0))=(0,1)$ .

So the equation is $y=-\frac{1}{2}x+1$ .

And we have that the normal passes through the point $(0,2)$ by solving $0=-\frac{1}{2}x+1$ .

We have this:

red: $y=f(x)$ , orange: the tangent at $x=0$ , purple: the normal, black: the altitude of the triangle $OAB$ that passes through $O$ .

So finally we search the distance $d=OH$ .

Use inverse Pythagorean theorem in triangle $OAB$ which is a right-angled triangle with the right angle at $O$ :

$\frac{1}{d^2}=\frac{1}{(OA)^2} + \frac{1}{(OB)^2}$

This gives us $d=\frac{2}{\sqrt{5}}$

So the answer is $\boxed{2+5=7}$