Question

If $2a$ and $2b$ are the lengths of transverse axis and conjugate axis respectively, then the equation of hyperbola with origin $(0,0)$ and transverse axis along $X$-axis, is

• A. $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
• B. $\frac{x^2}{b^2}-\frac{y^2}{a^2}=1$
• C. $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$
• D. $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

Answer 1

Answer: (A)

We will derive the equation for the hyperbola with foci on the $X$-axis.
Let $F_1$ and $F_2$ be the foci and $O$ be the mid-point of the line segment $F_1{F}_2$. Let $O$ be the origin and the line through $O$ through $F_2$ be the positive $X$ - axis and that through $F_1$ as the negative $X$-axis. The line through $O$ perpendicular to the $X$-axis be the $Y$-axis. Let the coordinates of $F_1$ be $(-c,0)$ and $F_2$ be $(c,0)$. Let $P(x,y)$ be any point on the hyperbola such that the difference of the distances from $P$ to the farther point minus the closer point be $2a$.
So given, $P{F}_1-P{F}_2=2a$

Using the distance formula, we have
$\sqrt{(x+c{)}^2+y^2}-\sqrt{(x-c{)}^2+y^2}=2a$
i.e.,$\sqrt{(x+c{)}^2+y^2}=2a+\sqrt{(x-c{)}^2+y^2}$
Squaring both sides, we get
$(x+c{)}^2+y^2=4a^2+4a\sqrt{(x-c{)}^2+y^2}+(x-c{)}^2+y^2$
and on simplifying, we get
$\frac{cx}{a}-a=\sqrt{(x-c{)}^2+y^2}$
On squaring again and further simplifying, we get
$\frac{x^2}{a^2}-\frac{y^2}{c^2-a^2}=1$
i.e.,$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$($ since, $c^2-a^2=b^2)$
Hence, any point on the hyperbola satisfies $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Conversely, let $P(x,y)$ satisfy the above equation with $0<a<c$.
Then, $y^2=b^2\left(\frac{x^2-a^2}{a^2}\right)$
Therefore, $P{F}_1=+\sqrt{(x+c{)}^2+y^2}$
$=+\sqrt{(x+c{)}^2+b^2\left(\frac{x^2-a^2}{a^2}\right)}=a+\frac{c}{a}x$
Similarly, $P{F}_2=a-\frac{a}{c}x$
In hyperbolac $>a$, and since $P$ is to the right of the line $x=a,x>a,\frac{c}{a}x>a$. Therefore, $a-\frac{c}{a}x$ becomes negative.
Thus, $P{F}_2=\frac{c}{a}x-a$
Therefore, $P{F}_1-P{F}_2=a+\frac{c}{a}x-\frac{cx}{a}+a=2a$
Also, note that if $P$ is to the left of the line $x=-a$, then
$P{F}_1=-\left(a+\frac{c}{a}x\right),P{F}_2=a-\frac{c}{a}x.$
In that case $P{F}_2-P{F}_1=2a$.
So, any point that satisfies $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, lies on the hyperbola.
Thus, we proved that the equation of hyperhola with origin $(0,0)$ and transverse axis along $X$-axis is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.