Question

Find an equation for the hyperbola that satisfies the given conditions.

Foci: (0,+-3), hyperbola passes through (4, 5)

Answer 1

Solution-

Since foci lies along y - axis.So, hyperbola must be open along y - axis.

Let the equation for the hyperbola that satisfies the given conditions be

$\frac{(y-k)^2}{a^2}+\frac{(x-h)^2}{b^2}=1$.....(1)

Here, (h, k) = centre of hyperbola

a = distance between centre and vertex and

b = distance between centre and end point of conjugate axis.

Foci = (h, k ±c)

On comparing foci coordinates with given coordinates we get

Foci: = (h, k ±c) (0,±3)

h = 0 , k = 0 and c =3

On putting (h, k) = (0,0) in equation (1) to get

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

Now, Since c² = a² + b² and c = 3

So, a² + b² = c²

Or a² + b² = 3²

Or a² + b² = 9

Or b² = 9 - a² ......(3)

Since hyperbola passes through (4, 5)

So, putting (x,y) = (4,5) in equation (2) we get

$\frac{5^2}{a^2}-\frac{4^2}{b^2}=1$

$\frac{25}{a^2}-\frac{16}{b^2}=1$

Or 25b² - 16a² = a²b²

Or 25b² - a²b² -16a² =0

Or b²( 25 - a²) - 16a² =0

Putting the value of b² from equation (3) to get

(9 - a²)(25 - a²) - 16a² =0

Or 225 -9a² -25a² + a⁴ -16a² =0

Or a⁴ -50a² +225 =0

Solving it by factorisation method to get

a⁴ -45a² -5a² +225 = 0

a²(a² -45) -5(a² - 45) =0

(a² - 45)(a² - 5) =0

This implies

a² = 45 , a² = 5

Or a = √(45) , a= √5

Or a = 3√5 , a √5

Putting the values of a² = 45, 5 in equation (3) to get

b² = 9 - 45 = -36 (not possible) ,

b² = 9 - 5 = 4 implies b = 2

So, a = √5 and b = 2

On putting a = √5 and b = 2 in equation (2) we get

The equation of hyperbola is

$\frac{y^2}{\sqrt{5}^2}-\frac{x^2}{2^2}=1$

$\frac{y^2}{5}-\frac{x^2}{4}=1$

Hence, the required equation of hyperbola is

$\frac{y^2}{5}-\frac{x^2}{4}=1$