Question

If $2 a$ and $2 b$ 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 $2 a$.
So given, $P F_1-P F_2=2 a$

Using the distance formula, we have
$\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}=2 a$
i.e.,$\sqrt{(x+c)^2+y^2}=2 a+\sqrt{(x-c)^2+y^2}$
Squaring both sides, we get
$(x+c)^2+y^2=4 a^2+4 a \sqrt{(x-c)^2+y^2}+(x-c)^2+y^2$
and on simplifying, we get
$\frac{c x}{a}-a=\sqrt{(x-c)^2+y^2}$
On squaring again and further simplifying, we get
$\frac{x^2}{a^2}-\frac{y^2}{c^2-a^2}=1$
i.e.,$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\left(\right.$ since, $\left.c^2-a^2=b^2\right)$
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{c x}{a}+a=2 a$
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 $P F_2-P F_1=2 a$.
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 $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.