• Parabola
• Ellipse $\rightarrow$ special case is circle
• Hyperbola

A circle is a set of points in a plane which are equidistant from a fixed point in the plane.

The fixed point is called the centre of the circle and the fixed distance of points on the circle from the center is called the radius of the circle.

Let the centre be $C(h, k)$ and radius be $r>0$

$P C=r$

$\Rightarrow \sqrt{(x-h)^2+(y-k)^2}=r$

$\Rightarrow (x-h)^2+(y-k)^2=r^2 \cdots(*)$

conversaly if (x,y) satisfies the above equation then the distance between (x,y) and (h,k) is $\sqrt{(x-h)^2+(y-h)^2}=\sqrt{r^2}=r$

Thus, the above equation (*) is the equation of the circle.

A parabola is the set of all points in a plane, that are equidistant from a fixed line ’ $l$ ’ and a fixed point (not on the line $l$ ) in the plane The line $l$ is called the directrix of the parabola. and the point $F$ is called the focus of the parabola

The line passing through the focus and perpendicular to the directrix is called the axis of the parabola. The point of intersection of the axis with the parabola is called the vertex of the parabola

We will discuss parabolas whose vertex is at the origin and the directrix is parallel to one of the coordinate axis.

Focus is $F(a, 0)$

Directrix is $x=-a$

vertex is $O(0,0)$

Let $P(x, y)$ be any point on the parabola. Then the distance of $P$ from the focus $F(a, 0)$ is equal to the perpendicular distance of $P$ from the directrix : $x=-{a}$

ie. $\quad P M=P F$

$\text { Now, } \quad P M=|x+a|$

$\quad P F=\sqrt{(x-a)^2+y^2}$

$\text { So, } \sqrt{(x-a)^2+y^2}=|x+a|$

$\Rightarrow(x-a)^2+y^2=(x+a)^2$

$$\Rightarrow \quad x^2-2 a x+a^2+y^2=x^2+2 a x+a^2$$

$\Rightarrow y^2=4 a x \quad(a>0)$

Here, vertex is at the origin and the focus lies on the positive $x$-axis.

• Symmetric about the $x$-axis, which is the axis of the parabola

Now, consider the parabola whose vertex is at the origin. and the focus lies on the negative $x$-axis, say $F(-a, 0) \quad(a>0)$ let $P(x, y)$ be any point on the parabola.

Then,

$\Leftrightarrow \quad P M=P F$

$\Leftrightarrow|x-a|=\sqrt{(x+a)^2+y^2}$

$\Leftrightarrow \quad(x-a)^2=(x+a)^2+y^2$

$y^2=-4 a x$

Focus on the $y$-axis vertex at the origin

$P F=\sqrt{x^2+(y-a)^2}$

$P M=|y+a|$

$P F=P M$

$\Leftrightarrow x^2+(y-a)^2=(y+a)^2$

$\Leftrightarrow x^2=(y+a)^2-a(y-a)^2$

$\Leftrightarrow x^2=4 a y$

Similarly, if $F(0,-a)$ then the equation is $x^2=-4 a y$

Latus rectum is the line segment AB which passes through the focus $F$ is perpendicular to the axis of the parabola and has end points on the parabola.

In the figure, $A B$ is the latus rectum of the parabola $y^2=4 a x$. length of latus rectum = ?

Equation of the parabola

$y^2=4 a x$

putting $x=a$,

gives $y^2=4 a^2$

$\Rightarrow y= \pm 2 a$

so, length of the latus rectum:

$l=4 a$

Find the equation of the parabola which is symmetric about the $y$-axis and passes through $(2,-3)$

Solution:

Since the parabola is symmetric about the $y$-axis, $y$-axis is the axis of the parabola and the vertex is at the origin.

$x^2=-4 a y$

since $(2,-3)$ lies on the parabola, $(2)^2=-4 a(-3)$

$\Rightarrow a=1 / 3$

Therefore, the equation of the required parabola is

$x^2=-4\left(\frac{1}{3}\right) y$

ie. $\quad x^2=-\frac{4}{3} y$

General form of the parabola given the focus and the directrix:

Let the focus be $F(\alpha, \beta)$ and the equation of the directrix be $a x+b y+c=0$

Let $P(x, y)$ be an arbitrary point on the parabola.

Then,

$P F=\sqrt{(x-\alpha)^2+(y-\beta)^2}$

and $P M=\frac{|a x+b y+c|}{\sqrt{a^2+b^2}}$

$P F=P M$

$\Leftrightarrow \quad(x-\alpha)^2+$ $(y-\beta)^2=\frac{(a x+b y+c)^2}{(a^2+b^2)}$

$\Leftrightarrow(a^2+b^2)[x^2+y^2-2\alpha x-2\beta y+\alpha^2+\beta^2]$

$=a^2 x^2+b^2 y^2+2 a b x y+2 a c x+2 b cy+c^2$

$\Leftrightarrow b^2 x^2+a^2 y^2-2 a b x yx$ $-2({\alpha(a^2+b^2)+ac}) -2({\beta(a^2+b^2)+b c})y$ $+(a^2+b^2)(\alpha^2+\beta^2)-c^2=0$

Thank you