Question

The coordinates of the vertices of a hyperbola are (9,2) and (1,2) and the distance between its two foci is 10. Find its equation and also the length of its latus rectum.

Answer 1

The ordinates of the vertices of the required hyperbola are equal. Therefore, the transverse axis of the hyperbola is parallel to axis and conjugate axis is parallel to y-axis.

The mid-point of the vertices $(\frac{9+1}{2},\frac{2+2}{2})=(5,2)$.

The mid-point of the vertices is the centre of the required hyperbola.

Therefore, the centre of the required hyperbola is (5,2).

Let the equation of the required hyperbola be $\frac{(x-\alpha {)}^2}{a^2}-\frac{(y-\beta {)}^2}{b^2}=1$

Now, length of transverse axis = the distance between the two vertices i.e., the distance between the points (9,2) and (1,2) = 8 unit

Thus, $2a=8$

$a=4$

Again, the distance between the two foci = $2ae=10$

$ae=5$

Now, $b^2=a^2(e^2-1)$

$b^2=5^2-4^2=25-16=9$

Now, from the equation of hyperbola, we get,

$9x^2-16y^2-90x+64y+17=0$

This is the equation of the hyperbola.

Length of latus rectum of the hyperbola = $\frac{2b^2}{a}=2\times \frac{9}{4}=\frac{9}{2}units$