Understanding that $\sqrt{x} + a\sqrt{y} = 2$ is a branch of a parabola

Question

Is there a simple (or simpler) way to understand that the following curve
$$\sqrt{x} + a\sqrt{y} = 2 \tag{1}$$
is a branch of a parabola?

When I say "simpler" I mean simpler with respect to turning the above curve into the form

$$Ax^2 + By^2 + Cxy + Dx + Ey + F = 0 \tag{2}$$

and then studying the conic as "usual"?

Square it : Isolate $2a^2\sqrt{xy}$ : Square that to eliminate all $\sqrt{\cdots}$ : Move these terms to get $0$ : You will get that form. — Prem, Apr 15 '24 at 16:19
Related: "Is $\sqrt{x/a}+\sqrt{y/b}=1$ the equation of a parabola tangent to the coordinate axes?". In particular, see my answer, as well as an extensive discussion by Hypergeometricx. — Blue, Apr 15 '24 at 16:48
You can check by yourself that every point on that curve has the same distance from the point $\displaystyle\left({4\over1+a^4},{4a^2\over1+a^4}\right)$ and from the line of equation $y=-a^2 x$. — Intelligenti pauca, Apr 15 '24 at 17:26

score 2 · Answer 1 · answered Apr 15 '24 at 16:40

Algebraically, the question is equivalent to the following elimination process. Let $X,Y$ be the square roots of $x,y$. Then we want to eliminate $X,Y$ from the system of algebraic equations: $$ \left\{ \begin{aligned} x &= X^2\ ,\\ y &= Y^2\ ,\\ 2 &= X +aY \ . \end{aligned} \right. $$ So the question is or may be how can we quickly eliminate $X,Y$ from the three equations to obtain only one. Well, here is one way we can do it: $$ \begin{aligned} (x-a^2y)^2 &=(X^2-a^2Y^2)^2 =(X+aY)^2(X-aY)^2 =4(X-aY)^2 \\ &=+4X^2-8aXY+4a^2Y^2\\ &=-4X^2-8aXY-4a^2Y^2+8x+8a^2y=-4\underbrace{(X+aY)^2}_{=2^2=4}+8x+8a^2y\ , \end{aligned} $$ and we get an equation $$ \tag{P} 0=(x-a^2y)^2-8x-8a^2y+16\ , $$ which is the equation of a parabola $(P)$. So any solution of the initial equation involving radicals leads to a point $(x,y)$ on the parabola $(P)$ with $x,y\ge 0$. (It is not a brach, but an intersection with the first quarter of the plane.)

Any other way is equivalent to an elimination in its essence. So we can get more "simplicity" only by minimizing the elimination process. Squaring the given equation is equally simple or complicated.

score 0 · Answer 2 · answered Apr 15 '24 at 19:34

You just need to square twice.

$$\sqrt x =2-a\sqrt y \implies x=4+a^2y-4a\sqrt y\implies4a\sqrt y= 4+a^2y-x \implies 16a^2y=x^2+a^4y^2-2a^2xy-8x+8a^2y+16$$

Final equation:- $$x^2+a^4y^2-2a^2xy-8x-8a^2y+16=0$$ Let coefficient of $x^2=i,\text{of} \ y^2=j , \text{and of} \ 2xy=k$.

By properties of equation of a parabola, $i.j=k^2$

As $1×a^4=(a^2)^2$ is true, it's proven that this is a parabola.

Understanding that $\sqrt{x} + a\sqrt{y} = 2$ is a branch of a parabola

2 Answers2