A hyperbola in the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant (less than to distance between th two fixed points)

The two fixed points are called the foci of the hyperbola.

The mid point of the line segment joining the foci is called the centre.

• The line through the foci is called, the transverse axis of the hyperbola

The line perpendicular to the transverse axis and passing through the centre is calls the conjungate axis of the hyperbola.

Let the distance between the two foci,

$F_1 F_2=2 c>0$

let centre be at the origin $(0,0)$ and the foci lie on the x axis.

Then $F_1 \equiv(-c, 0)$

$F_2 \equiv(c, 0)$

If $P$ is end point on the hyperbola, then

$\left|P F_1-P F_2\right|=$ constant

$=2 a \text { (say) }$

we have $a

Let $A=(x, 0)$ be a point on the hyperbola

If $0

$A F_1-A F_2=2 a$

but

$A F_1 =c+x$

$A F_2 =c-x$

$A F_1-A F_2 =(c+x)-(c-x)=2 x$

So, $2 x=2 a \Rightarrow x=a$.

So, $(a, 0)$ lies on the hyperbola.

similarly, $(-a, 0)$ also lies on the hyperbola

Also, if $|x| \geqslant c$, than $(x, 0)$ does not lie on the hyperbola.

Because if $x \geqslant c$ on $x \leqslant-c$, then the diff. of distances from

$F_1 to F_2$ is $F_1 F_2=2 c \neq 2 a$

$A F_1-A F_2=F_1 F_2$

$B F_2-8 F_1=F_1 F_2$.

So, there are exactly two points on the transverse axis which lie on the hyperbola

The coordinates of these two points are $(-a, 0) \text { and }(a, 0) \text {. }$

There two points are called the vertices of the hyperbola

1. Foci on the x-axis.

2. Foci on the y-axis.

(1) Let $P=(x,y)$ in any point on the hyperbola with $x>0$

then $PF_1 - PF_2 = 2a$

$PF_1 = \sqrt{(x+c)^2+y^2}$

$PF_2 = \sqrt{(x-c)^2+y^2}$

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

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

$\Rightarrow \quad(x+c)^2+y^2=4 a^2+(x-c)^2+y^2+4 a \sqrt{(x-c)^2+y^2}$

$\Rightarrow 4a \sqrt{(x-c)^2+y^2}=(x+c)^2-(x-c)^2-4 a^2$ $=4 c x-4 a^2$

$\Rightarrow a^2\left[(x-c)^2+y^2\right]=\left(c x-a^2\right)^2$

$\Rightarrow a^2 x^2+a^2 c^2-2 a^2 c x+a^2 y^2=c^2x^2+a^4-2 a^2 cx$

$\Rightarrow\left(c^2-a^2\right) x^2-a^2 y^2=a^2 c^2-a^4=a^2\left(c^2-a^2\right)$

$\text { Let’s put } c^2-a^2=b^2$

then $b^2 x^2-a^2 y^2=a^2 b^2$

Dividing by $a^2 b^2$,

we get $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \leftarrow {\text { symmetric about x-axis and y-axis }}$

$\frac{x^2}{a^2}=1+\frac{y^2}{b^2} \geqslant 1 \Rightarrow x^2 \geqslant a^2$

$\Rightarrow x \geqslant a \quad(for x>0)$

Similarly if $P \equiv(x, y)$ with $x<0$ on th Hyperbola, the

$P F_2-P F_1=2 a$

Proceeding like in previous case, we get the same equaion.

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

where $b^2=c^2-a^2$

So the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ looks like

• the hyperbola is symmetric about the transverse axis so will as the conjugate axis.

Focii on the y-axis.

The equation of the hyperbola with focii on the y-axis is given by

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

This intersect the y-axis at $(0,\pm a)$

Also, $|y| \geqslant a$ se. $y \geqslant a$ or $y \leqslant-a$

Line segment passing through on focus, perpendicular to the transverse axis, and having end points on the hyperbola

Let $A \equiv(c,-\beta) \quad and B \equiv(c, \beta)$

Then length of the latus rectum, $l=2 \beta$

since $(c, \beta)$ lies on th hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,

we get

$\frac{c^2}{a^2}-\frac{k^2}{b^2}=1 \Rightarrow \frac{p^2}{b^2}=\frac{c^2}{a^2}-1$

$=\frac{c^2-a^2}{a^2}=\frac{b^2}{a^2}$

$\Rightarrow \beta^2=\frac{b^4}{a^2} \Rightarrow \beta=\frac{b^2}{a}$

So, $l=\frac{2 b^2}{a}$

We define the eccentricity of the hyperbola as

$e = \frac{c}{a} >1$

Find the foci, vertices, eccentricity and length of latus rectum for

(i) $\frac{x^2}{16}-\frac{y^2}{9}=1$

(ii) $9 y^2-4 x^2=36$

(iii) $\quad \frac{x^2}{16}-\frac{y^2}{9}=1 \Rightarrow a=4, b=3$

Solution :

Foci lies on $x$-axis.

$c^2-a^2=b^2 \Rightarrow c^2=a^2+b^2=4^2+3^2=25$

$\therefore$ Foci at $( \pm 5,0)$

Vertices at $( \pm a, 0)=( \pm 4,0)$

$\begin{aligned} & \therefore c=\frac{c}{a}=\frac{5}{4} \ & =l=\frac{2 b^2}{a}=2 \times \frac{9}{4}=\frac{9}{2} . \end{aligned}$

(ii)

$9 y^2-4 x^2 =36$

$\quad \frac{y^2}{4}-\frac{x^2}{9} =1$

$\text { This is of the form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ }$
$\text { with}$ $a =2, b=3$

Focii lie on th $y$-axis $(0, \pm c)$

$c^2=a^2+b^2=2^2+3^2=13$
$c=\sqrt{17}$

$\therefore \text { Foci } \equiv(0, \pm \sqrt{13})$

$\text { Vertices } \equiv(0, \pm a)=(0, \pm 2)$

$e=\frac{c}{a}=\frac{\sqrt{12}}{2}$

$l=\frac{2 b^2}{a}=\frac{2 \times x^2}{2}=9$

Thank you