Standard equation of a hyperbola is the form

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

Find the equation of the hyperbola whose vertices are at $( \pm 2,0)$ and foci at $( \pm 3,0)$.

Solution: Since the vertices are on $x$-axis, the equation of the hyperbola is of $u_0$ form

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

Vertices $\equiv( \pm a, 0)$, foci $\equiv( \pm c, 0)$

So, $a=2$ and $c=3$

$a^2+b^2=c^2 \Rightarrow b^2=c^2-a^2=5$

$\therefore ~$ Equation is $\frac{x^2}{4}-\frac{y^2}{5}=1$

Find the equation of the vertices are at $(0, \pm 5)$ and foci at $( 0, \pm 8)$.

Solution: Equation: $\quad \frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

where $a=5$ and $c=8$

$\therefore \quad b^2=c^2-a^2=8^2-5^2=39$

$\therefore$ The equation is $\frac{y^2}{25}-\frac{x^2}{39}=1$

Find the equation of the hyperbola whose foci are at $( \pm 4,0)$ and the length of the latex rectum is 12 .

Solution :

Since foci are on the $x$-axis, the Equation is

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

Foci $=( \pm 4,0) \Rightarrow c=4 \Rightarrow a^2+b^2=4^2$ (i)

Length of latex rectum, $l=\frac{2 b^2}{a}=12$

$\Rightarrow b^2=6 a \text {-(ii) }$

(i) $\Rightarrow b^2=6 a \quad \text { (i) } and (ii) \Rightarrow a^2+6 a=16\Rightarrow a^2+6 a-16=0$

$\Rightarrow(a-2)(a+8)=0$

$\Rightarrow \quad a=2 \quad[\because a>0]$

$\therefore \quad b^2=6 a=12$

$\therefore$ The equation is $\frac{x^2}{4}-\frac{y^2}{12}=1$

Find the equation of the hyperbola whose foci are at $(0, \pm \sqrt{10})$ and wide passes through $(2,3)$.

Solution :

Foci $\equiv(0, \pm \sqrt{10}) \Rightarrow c=\sqrt{10}$ $\operatorname{con} . \quad \frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

since it passes through $(2,1)$, we get $\frac{9}{a^2}-\frac{4}{b^2}=1-(i)$

Also, $\quad a^2+b^2=c^2=10$

$\Rightarrow b^2=10-a^2$

putting this in (i), gives

$\frac{9}{a^2}-\frac{4}{10-a^2}=1$

$\Rightarrow 9\left(10-a^2\right)-4 a^2=a^2\left(10-a^2\right)$

$\Rightarrow 90-9 a^2-4 a^2=10 a^2-a^4$

$\Rightarrow a^4-23 a^2+90=0$

$\Rightarrow \left(a^2-5\right)\left(a^2-18\right)=0$

$\Rightarrow a^2=5 \quad \frac{y^2}{5}-\frac{x^2}{5}=1$

$\text { But } a^2+b^2=10 \Rightarrow a^2=18$

But, $a^2+b^2=10 \Rightarrow a^2=18$ $\frac{y^2}{5}-\frac{x^2}{5}=1$

Hence $a^2+b^2=10 \Rightarrow a^2 \leq 10$

i.e. $y^2-x^2=5$

Consider the hyperbola:

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

The straight lines $y= \pm \frac{b}{a} x$ do not intersect the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ because

$y= \pm \frac{b}{a} x \Rightarrow \frac{y^2}{b^2}=\frac{x^2}{a^2} \Rightarrow \frac{x^2}{a^2}-\frac{y^2}{b^2}=0$

$\Rightarrow(x, y)$ does not lie on the hyperbola.

However, if $(x, y)$ on the hyperbola and $(x, y$) lies on the line $y=\frac{b}{a} x$, then

$\begin{aligned} \left|y-y_1\right| & =\left|b \sqrt{\frac{x^2}{a^2}-1}-\frac{b}{a} x\right| \ & =\frac{b}{a}\left(x-\sqrt{x^2-a^2}\right) \end{aligned}$

As $x \rightarrow+\infty$,

$\begin{aligned} & 1 \lim _{x \rightarrow+\infty}\left[x-\sqrt{x^2-a^2}\right] \\ & =\lim _{x \rightarrow+\infty} \frac{\left(x+\sqrt{x^2-a^2}\right)\left(x-\sqrt{x^2-a-1}\right)}{\left(x+\sqrt{x^2-a}\right)} \\ & =\lim _{x \rightarrow \infty} \frac{x^2-\left(x^2-a^2\right)}{x+\sqrt{x^2-a^2}} \\ & =\lim _{x \rightarrow+\infty} \frac{a^2}{x+\sqrt{x^2-a^2}}=0 \end{aligned}$

$\therefore$ The line $y=\frac{b}{a} x$ and hyperbola

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ approaches each other as $x \rightarrow+\infty$.

These lines are therefore the asymptotes of the hyperbola.

Define: If $a=b$ than the hyperbola is called a rectangular (or equilateral) hyperbola.

This is because in this case the asymptotes are $y= \pm x$, which are perpendicular to each other.

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

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

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

Putting $x^{\prime}=x-y \quad \& \quad y^{\prime}=x+y$, we get $x^{\prime} y^{\prime}=a^2$

This gives another form of rectangular hyperbola, i.e.

$x y=a^2$

Asymptotes are $x$-axis and $y$-axis.

If $a=1, x y=1$

$y=\frac{1}{2}, x+0$

Let’s find the vertices and foci of the rectangular hyperbola: $x y=a^2$.

Vertices are $(a, a), (-a,-a)$

let foci be $(c, c)$ , $(-c,-c)$

since $a=6$ for rectangular hyperbola, $c^2=2 a^2$

$\Rightarrow c=\sqrt{2} a$

Foci : $(\sqrt{2} a, \sqrt{2} a)$ and $(-\sqrt{2} a,-\sqrt{2} a)$.

Let us find the foci of $x y=a^2$ using the definition of hyperbola.

If $P$ is any point on the hyperbola lying in the 1st quadrant, then

$P F_2-P F_2=A B$

$\text { Let } P \equiv\left(c, \frac{a^2}{c}\right)$

$P F_2=\sqrt{(c+c)^2+\left(\frac{a^2}{2}+c\right)^2}$

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

So, $\sqrt{4 c^2+\left(\frac{a^2}{c}+c\right)^2}=\left(c-\frac{a^2}{c}\right)+2 \sqrt{2} a$

$\Rightarrow 4 c^2+\left(\frac{a^2}{c}+c\right)^2=\left(c-\frac{a^2}{c}\right)^2+8 a^2+4 \sqrt{2 a}\left(c-\frac{a^2}{c}\right)$

$\Rightarrow 4 c^2+4 a^2=3 a^2+4 \sqrt{2} a c-4 \sqrt{2} \frac{a^3}{c}$

Solving this, we will get $c=\sqrt{2} a$

$\therefore$ Foci $\equiv(\sqrt{2} a, \sqrt{2} a) R(-\sqrt{2} a,-\sqrt{2} a) \text {. }$

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