Question

Find the equation to the hyperbola of given transverse axis whose vertex bisects the distance between the centre and the focus.

Answer 1

Standard equation of hyperbola is

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

x-axis is the transverse axis

So, centre $c=(0,0)$

Vertex $A=(a,0)$

Focus $S=(ae,0)$

Vertex bisects the distance between centre and focus

i.e, vertex is the mid point of line joining vertex and focus

Applying mid point formula

$a=\frac{0+ae}{2}$

$2a=ae$

$e=2$

We know that,

$e^2=1+\frac{b^2}{a^2}$

$3=\frac{b^2}{a^2}$

Hence the equation of parabola in terms of $a$ is

$\frac{x^2}{a^2}-\frac{y^2}{{3a}^2}=1$