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Equation of Circle with Origin as Center

The normal at...

Question

The normal at a point $P$ to the parabola $y^{2} = 4 a x$ meets axis at $G$ . $Q$ is another point on the parabola such that $Q G$ is perpendicular to the axis of the parabola. Prove that ${Q G}^{2} - {P G}^{2} =$ constant

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Solution

Let us take any arbitrary point $P (a t^{2}, 2 a t)$ on parabola $y^{2} = 4 a x .$
So,
Equation of normal at P on parabola: $t x + y = 2 a t + a t^{3}$
this normal intersects x-axis at point $G$ when $y = 0$ (At x-axis, y coords.=0 ), so on substituting $y = 0$ in equation of normal we get:
$x = 2 a + a t^{2}$

as, x-coordinates of point $Q$ and $G$ will be same as $Q G$ is perpendicular to axis(x-axis) of parabola.
Thus,
$y - c o o r d i n a t e$ of Q can be determined using $y^{2} = 4 a x$ as Q is on parabola, we get:
$y = 2 a \sqrt{2 + t^{2}}$

Now,
length $Q G$ can be determined using distance formula: $\sqrt{(0) + (8 a^{2} + 4 a^{2} t^{2})}$
length $P G$ = $\sqrt{(4 a^{2}) + (4 a^{2} t^{2})}$

so, $Q G^{2} - P G^{2} = (8 a^{2} + 4 a^{2} t^{2}) - (4 a^{2} + 4 a^{2} t^{2})$
= $4 a^{2}$
= $c o n s t a n t$
$H e n c e, p r o v e d!$

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The normal at a point P to the parabola y2=4ax meets axis at G. Q is another point on the parabola such that QG is perpendicular to the axis of the parabola. Prove that QG2−PG2= constant

The normal at a point $P$ to the parabola $y^{2} = 4 a x$ meets axis at $G$ . $Q$ is another point on the parabola such that $Q G$ is perpendicular to the axis of the parabola. Prove that ${Q G}^{2} - {P G}^{2} =$ constant