Question

Derive the equation of parabola in the standard from $y^2=4ax$ with diagram.

Answer 1

Consider the above diagram, we take a point $P(x,y)$ on the parabola such that, $PF=PB$ … (1) … where $PB$ is perpendicular to $l$. The coordinates of the point $B$ are $(-a,y)$. Also, by the distance formula, we know that

$PF=\sqrt{(x–a{)}^2+y^2}$

Also, $PB=\sqrt{(x+a{)}^2}$

Since, $PF=PB$ [from eq. (1)], we get

$\sqrt{(x–a{)}^2+y^2}=\sqrt{(x+a{)}^2}$

So, by squaring both sides, we have $(x–a{)}^2+y^2=(x+a{)}^2$ or, $x^2–2ax+a^2+y^2=x^2+2ax+a^2$ . Further, by the solving the equation, we have $y^2=4ax$ … where $a>0$. Therefore, we can say that any point on the parabola satisfies the equation:

$y^2=4ax$ … (2)

Let’s look at the converse situation now. If $P(x,y)$ satisfies equation (2), then

$PF=\sqrt{(x–a{)}^2+y^2}$

$=\sqrt{(x–a{)}^2+4ax}$ … using the RHS of equation (2)

$=\sqrt{(x^2–2ax+a^2)+4ax}=\sqrt{x^2+2ax+a^2}$

$=\sqrt{(x+a{)}^2}=PB$ … (3)

Hence, we conclude that the point $P(x,y)$ satisfying eq. (2) lies on the parabola. Also, equations (2) and (3) prove that the equation to the parabola with vertex at the origin, focus at $(a,0)$ and directrix $x=–a$, is $y^2=4ax$.