Equation of trajectory?

Using the formula for equation of trajectory:-
$\begin{aligned} y & = x\tan \theta - \dfrac{gx^2}{2u^2{\cos}^2\theta}\\ \\ y & = x\tan 45 - \dfrac{10x^2}{2×{20}^2 {\cos}^2 45}\\ \\ y & = x - \dfrac{10x^2}{800 × \frac{1}{2}}\\ \\ y & = x - \dfrac{x^2}{40}\\ \\ y & = \dfrac{40x - x^2}{40}\\ \\ 40y & = 40x - x^2\\ x^2 - 40x + 40y & = 0\\ \therefore a + b & = 1 + 40\\ & = \color{#3D99F6}{\boxed{41}} \end{aligned}$

So, here is the proof of the formula:-
The velocity along the x-axis would the cosine component of the velocity of the projectile and the velocity along the y-axis would be its sin component.
So, $v_x = u\cos\theta$
Along this axis the acceleration is $0$ because there is no displacement along x-axis.

Now, $v_y = u\sin\theta$ and since it is projected upwards $a = -g$

Now, distance covered(x) = speed × time = $u\cos\theta t\\ t = \dfrac{x}{u\cos\theta}$

Let $y$ be the maximum height obtained.

Applying the formula:-
$S = u_yt + \dfrac{1}{2}a_yt^2\\ y = u\sin\theta × \dfrac{x}{u\cos\theta} - \dfrac{1}{2}×g×\dfrac{x^2}{u^2{\cos}^2\theta}\\ \color{#3D99F6}{\boxed{y = x\tan\theta - \dfrac{gx^2}{2u^2{\cos}^2\theta}}}$

Before we begin, let us define our variables:

Let $x$ be the horizontal displacement, $y$ be the vertical displacement

$u$ be the initial velocity, and $u_x,u_y$ be the respective velocity components

$a_x,a_y$ be the acceleration along the $x$ -axis and $y$ -axis

$t$ be the time that passed after the projectile is thrown

Now, we know these values:

$u=20\\ u_x=20\cos45^{\circ} = 10\sqrt{2}\\ u_y=20\sin45^{\circ} = 10\sqrt{2}\\ a_x=0\\ a_y=-10$

Clarification: We neglect air resistance, so there is no horizontal acceleration. Gravity works downwards, which is negative on the Cartesian plane.

Using the equations of linear motion with constant acceleration:

$s=ut+\dfrac{1}{2}at^2$

$x=10\sqrt{2}t+\dfrac{1}{2}(0)t^2\\ x=10\sqrt{2}t$

$t=\dfrac{x}{10\sqrt{2}}\implies$ Eq.(1)

$y=10\sqrt{2}t+\dfrac{1}{2}(-10)t^2$

$y=10\sqrt{2}t-5t^2\implies$ Eq.(2)

Substitute Eq.(1) into Eq.(2):

$y=10\sqrt{2}\left(\dfrac{x}{10\sqrt{2}}\right)-5\left(\dfrac{x}{10\sqrt{2}}\right)^2\\ y=x-5\left(\dfrac{x^2}{100(2)}\right)\\ y=x-\dfrac{x^2}{40}\\ 40y=40x-x^2\\ x^2-40x+40y=0$

$a=1,\;b=40,\;a+b=1+40=\boxed{41}$

Although Ashish's solution is great ( +1 from my side;-) ) I would like to present a different (not so clever) solution .

We can see that the trajectory ought to be a parabola thus its two solution must represent the position of launch and final position (or position of crash!! XD) thus if position of launch is the origin then solving for y=0 and neglecting x=0 since it is the origin we get x= b/a that would equal its range which is u^2/g=40. Thus a=1 , b= 40 , a+b=41.