Find the equations of the hyperbola satisfying the given conditions Foci (0, ±13), the conjugate axis is of length 24.

Asked by Sakshi | 1 year ago |  72

##### Solution :-

Given:

Foci (0, ±13) and the conjugate axis is of length 24.

Here, the foci are on y-axis.

The equation of the hyperbola is of the form $$\dfrac{ x^2}{a^2} – \dfrac{y^2}{b^2} = 1$$

Since, the foci are (0, ±13), so, c = 13

Since, the length of the conjugate axis is 24,

2b = 24

b = $$\dfrac{24}{2}$$

= 12

It is known that, a2 + b2 = c2

a2 + 122 = 132

a2 = 169 – 144

= 25

The equation of the hyperbola is $$\dfrac{ y^2}{25} –\dfrac{ x^2}{144} = 1$$

Answered by Aaryan | 1 year ago

### Related Questions

#### An equilateral triangle is inscribed in the parabola y2 = 4ax,

An equilateral triangle is inscribed in the parabola y2 = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

#### A man running a racecourse notes that the sum of the distances from the two flag posts from him is always

A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

#### Find the area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends

Find the area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends of its latus rectum.