Foci (± 4, 0) and the latus rectum is of length 12

Here, the foci are on x-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 (± 4, 0), so, c = 4

Length of latus rectum is 12

\(\dfrac{2b^2}{a}\) = 12

2b^{2} = 12a

b^{2} = \( \dfrac{12a}{2}\)

= 6a

It is known that, a^{2} + b^{2} = c^{2}

a^{2} + 6a = 16

a^{2} + 6a – 16 = 0

a^{2} + 8a – 2a – 16 = 0

(a + 8) (a – 2) = 0

a = -8 or 2

Since, a is non – negative, a = 2

So, b^{2} = 6a

= 6 × 2

= 12

The equation of the hyperbola is

\( \dfrac{x^2}{4}-\dfrac{y^2}{12}=1\)

Answered by Abhisek | 11 months agoAn equilateral triangle is inscribed in the parabola y^{2} = 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 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 x^{2} = 12y to the ends of its latus rectum.

A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.