How do I show that this equation is a parabola?

Question

A question I'm doing asks me to prove that the boundary of a region is a parabola. I have managed to show that the equation of the boundary is given by $$y=a(d^2+x^2)+b\sqrt{d^2+x^2},$$ for some constants $a$ and $b.$

Does this equation represent a parabola?

EDIT

Original question:

A tall building stands on level ground. The nozzle of a water sprinkler is positioned at a point $P$ on the ground at a distance $d$ from a wall of the building. Water sprays from the nozzle with speed $V$ and the nozzle can be pointed in any direction from $P.$

(i) If $V>\sqrt{gd},$ prove that the water can reach the wall above ground level.

(ii) Suppose that $V=2\sqrt{gd}.$ Show that the portion of the wall that can be sprayed with water is a parabolic segment of height $\frac{15d}{8}$ and area $\frac{5}{2}d^2\sqrt{15}.$

I have done part (i).

My attempt at part (ii):

Let $(X,Y)$ be an arbitrary point on the boundary of the portion, and then this shows that $(\sqrt{d^2+X^2},Y)$ is a point on the trajectory of the water sprayed from the nozzle. And that's how I ended up with $$Y=a(d^2+X^2)+b\sqrt{d^2+X^2}.$$

EDIT 2:

It turns out the portion of the wall that can be sprayed with water is in fact a parabola; what I had thought was $a$ and $b$ were constants, they turned out to be functions of $X,$ and so the equation simplifies down to $Y=-\frac{X^2}{8d}+\frac{15d}{8}.$

So my mistake was the assumption that these $a$ and $b$ were constants, in which case the equation $y=a(d^2+x^2)+b\sqrt{d^2+x^2}$ is not a parabola (see the brilliant answers below).

Answer 1

This is not a parabola. It is not too far from one. You can write $y=x^2+\sqrt{1+x^2}\approx x^2+1+\frac {x^2}2-\frac {x^4}8=\frac 32x^2+1-\frac {x^4}8$ The approximation is the first three terms of the Taylor series near $x=0$. The $x^4$ term spoils the parabola.

Answer 2

It's not a parabola (assuming $a\neq 0$). Indeed, if $(x,y)$ satisfies your equation, then $$y-x^2=\sqrt{x^2+a}$$ so $$y^2-2x^2y+(x^4-x^2-a)=0.$$ Considering this as a quadratic in $y$, its discriminant is $4x^4-4(x^4-x^2-a)=4x^2+4a$. If $a\neq 0$, this has no polynomial square root. It follows that $y^2-2x^2y+(x^4-x^2-a)$ is irreducible as a polynomial in two variables. If its vanishing set contained a parabola, then the equation of the parabola would give a degree $2$ factor, contradicting irreducibility. So the vanishing set cannot contain a parabola, and so the graph of your equation cannot be a parabola.