Solving for the points of intersection of two parabola of different types

Question

In a test I had to solve these as part of a problem: $y^2=4x $ and $x^2=4y $.It seemed easy but on trying I couldn't figure out how to solve it.

I know that by trial and error we can get the correct solutions, $(4,4) $ & $(0,0) $ but this would not be very convenient for other slightly harder problems.

So I want to know how to solve it by simplifying, factorising, etc. Here's what I did:

From $y^2=4x $ we get $x=y^2/4$; and so, $x^2=4y \implies y^4/16=4y$. Then cancelling out a $y $ from both sides and simplifying we get- $y= \pm 4$. But since y is one-fourth of $x^2$ so it is positive, so $y=4$. Then, putting in this value in either of the equations we get $x=4$ too.

But clearly, I couldn't get the other solution by solving thus (because I cancelled out a y in the beginning I think). But I don't know how to solve to get the other solution too.

So I want to know of a general method to solve such problems easily and accurately.

Answer 1

you will get $$\frac{x^4}{16}=4x$$ this is equivalent to $$1/16\,x \left( x-4 \right) \left( {x}^{2}+4\,x+16 \right) =0$$ from here we get $$x=0,y=0$$ or $$x=4,y=4$$ it Comes from $$x(x^3-4^3)$$ and then use $$a^3-b^3=(a-b)(a^2+ab+b^2)$$

Answer 2

What to expect:

1) $y^2=4x$ is a parabola with symmetry axis $y=0$ $(x-$axis), opening upward.

2) $y=(1/4)x^2$ is a parabola with symmetry axis $x=0$ $(y-$axis), opening to the right.

$2$ points of intersection of 1) and 2) in the first quadrant . That's it.

By inspection $x=y=0$ is a common point.

Equating:

$(x^2)^2 =16 y^2= 4×16x;$

$x(x^3-64)=0.$

A) $x_1=0;$

B) $x^3=64$, or $x_2=4.$

The coordinates of the points of Intersection $P,Q$ are?

Answer 3

$y^2=4x \implies y=\pm2\sqrt x$ after square rooting both sides

$x^2=4y \implies y=\cfrac {x^2}4$ after dividing both sides by $4$

Find the intersection points of $$2\sqrt{x}=\cfrac {x^2}4$$ and $$-2\sqrt{x}=\cfrac {x^2}4$$

Answer (do not hover unless stuck):

Squaring both sides: $$4x=\cfrac {x^4}{16}$$ $$\cfrac {x^4}{16}-4x=0$$ $$x^4-64x=0$$ $$x(x^3-64)=0$$ $$x(x-4)(x^2+4x+16)$$ Since $x^2+4x+16$ yields complex roots, the intersection points are at $x=0,4$.