$y^2=4 a x ;$

$y^2=-4 a x ;$

$x^2=4 a y ;$

$x^2=-4 a y$

Consider the general quadratic $y=a x^2+b x+c$, where $a \neq 0$.

$y =a\left[x^2+\frac{b}{a} x+\frac{c}{a}\right]$

$=a\left[\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4 a^2}+\frac{c}{a}\right]$

$=a\left(x+\frac{b}{2 a}\right)^2-\frac{b^2}{4 a}+c$

$y =a\left(x+\frac{b}{2 a}\right)^2-\left(\frac{b^2-4 a c}{4 a}\right)$

i.e., $\quad y+\frac{b^2-4 a c}{4 a}=a\left(x+\frac{b}{2 a}\right)^2$

This is the form $y-k=a(x-h)^2$

where, $$k=\frac{4ac-b^2}{4a} \quad and \quad h=\frac{-b}{2a}\quad a > 0$$

$\alpha-k=|l-k|=k-l$

$\Rightarrow \alpha+l=2 k$

$P \equiv(x, \alpha)$

$y-k=a(x-h)^2$
$\alpha-k=a(x-h)^2$
$$\Rightarrow(x-h)^{2}=\frac{\alpha-k}{a} \Rightarrow x=h \pm \sqrt{\frac{\alpha-k}{a}}.$$

$P \equiv\left(h+\sqrt{\frac{\alpha-h}{a}}, \alpha\right)$

$P F=\sqrt{\frac{\alpha-h}{a}}$

$P l=\alpha-l$

$P F=P l \Rightarrow \frac{\alpha-k}{a}=(\alpha-l)^2-(i)$

(i) $\rightarrow l=2 k-\alpha$

Putting in (ii) $\Rightarrow \frac{\alpha-k}{a}=[\alpha-(3 k-\alpha)]^2$

$\Rightarrow \alpha-k=a \cdot 4(\alpha-k)^2$

$\Rightarrow \alpha-k=\frac{1}{4 a} \quad[\because \alpha-k \neq 0]$

$\Rightarrow \alpha=k+\frac{1}{4 a}$

$\therefore l=2 k-\left(k+\frac{1}{4 a}\right)=k-\frac{1}{4 a}$

So, for the parabola $y-k=a(x-h)^2$,

the vertex is at $(h,k)$

Focus is at $(h,k)$ = $(h,k-\frac{1}{4a}$

and directrix is the line $y=l$

i.e., $y= (k-\frac{1}{4a})$

Ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$a>b, c^2=a^2-b^2$

Consider the line:

$x=\frac{a^2}{c}>a \quad\left(\because \frac{a}{c}>1\right)$

Distance of $P(x, y)$ on the ellipse

to $F_1 \equiv(c, 0)$

$P F_1 =\sqrt{(x-c)^2+y^2}$

$\Rightarrow P F_1^2 =(x-c)^2+y^2$

$=x^2-2 c x+c^2+y^2 \quad\left[y^2=b^2\left(1-\frac{x^2}{a^2}\right)\right]$

$=x^2-2 c x+c^2+b^2\left(1-\frac{x^2}{a^2}\right)$

$=\left(1-\frac{b^2}{a^2}\right) x^2-2 c x+c^2+b^2$

$=\frac{a^2-b^2}{a^2} x^2-2 c x+a^2 \quad\left(\because c^2+b^2=a^2\right)$

$=\frac{c^2}{a^2} x^2-2 c x+a^2$

$=\left(\frac{c}{a} x-a\right)^2=\frac{c^2}{a^2}\left(x-\frac{a^2}{c}\right)^2$

$\therefore \quad P F_1^2=\left(\frac{c}{a}\right)^2 P l^2$

$\Rightarrow P F_1=\frac{c}{a} P l_1=e P l_1$

$\therefore \quad \frac{P F_1}{P l}=e$

Similarly for te line $x=-\frac{-a^2}{c}$, the ratio $=e$

Then lines $x= \pm \frac{a^2}{c}$ are called the directrices of the ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a>b$.

we can define ellipse (hyperbola) using directory Lets take a conic section having a vetex at $(0,0)$, focus on the x-axis and directrix a line parallel to to y-axis.

Let’s find the locus of points $P$ such that $\frac{P F}{P l}=e \leftarrow$ constant

Since $V=(0,0)$ lies on the curve,

$\frac{V f}{V R}=e^{\prime} \Rightarrow V e=\frac{f}{e}$

$\therefore$ the line l is $x=\frac{-f}{e}$

$P F^2=e^2 P l^2$

$(x-f)^2+y^2=e^2\left(x+\frac{f}{e}\right)^2$

$\Rightarrow x^2-2 f x+f^2+y^2$

$$=(e x+f)^2 =e^2 x^2+2 e f x+f^2$$

$$\Rightarrow\left(1-e^2\right) x^2-2 f(1+e) x+y^2=0$$

Putting $p=f(1+e)$, we get

$\left(1-e^2\right) x^2-2 p x+y^2=0$

If $e=1$, we get $y^2=2 p x$, which is a parabola.

If $e<1, a^2 x^2-2 p x+y^2=0$

$\Rightarrow \quad x^2-\frac{2 p}{a^2} x+\frac{y^2}{a^2}=0$

$\Rightarrow\left(x-\frac{p}{a^2}\right)^2-\frac{p^2}{a^4}+\frac{y^2}{a^2}=0$

$\Rightarrow\left(x-\frac{p}{a^2}\right)^2+\frac{y^2}{a^2}=\frac{p^2}{a^4}$

which is equation of the ellipse

Special case : $e=0$ gives

$x^2-2 p x+y^2=0$

which is equation of a circle.

If $e>1$,

${(e^2-1)} x^2+2 p x-y^2=0$

which is equation of a hyperbola.

$$\text{Consider the hyperbola:} \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$

$\text { Line : } x=\frac{a^2}{c}$

$c^2=a^2+b^2>a^2$

$\frac{a}{c}<1$

$\therefore \frac{a^2}{c}

$P F_1^2=(x-c)^2+y^2=x^2-2 c x+c^2+y^2$

$\therefore P F_1^2 =x^2-2 c x+c^2+\frac{b^2}{a^2} x^2-b^2$

$=\left(1+\frac{b^2}{a^2}\right) x^2-2 c x+\left(c^2-b^2\right)$

$=\frac{c^2}{a^2} x^2-2 c x+a^2$

$=\left(\frac{c}{a} x-a\right)^2=\left(\frac{c}{a}\right)^2\left(x-\frac{a^2}{c}\right)^2$

$=e^2 P l^2$

$\therefore \quad \frac{P F_1}{P l} =e>1$

So, th line $x= \pm \frac{a^2}{c}$ are the directrix of the hyperbola

$\frac{x^2}{a^2}-\frac{y^2}{x^2}=1$.

Thank you