Question
The equation of the hyperbola with vertices (3, 0) and (–3, 0) and semilatusrectum 4, is given by

None of these



easy
Solution
We have,
Hence, the equation of the hyperbola is
.
SIMILAR QUESTIONS
Find the asymptotes of the hyperbola xy – 3y – 2x = 0.
A ray emanating from the point (5, 0) is incident on the hyperbola 9x^{2} – 16y^{2} = 144 at the point P with abscissa 8. Find the equation of the reflected ray after first reflection and point P lies in first quadrant.
The equations of the transverse and conjugate axes of a hyperbola are respectively 3x + 4y – 7 = 0, 4x – 3y + 8 = 0 and their respective lengths are 4 and 6. Find the equation of the hyperbola.
A, B, C are three points on the rectangular hyperbola xy = c^{2}, find
1. The area of the triangle ABC
2. The area of the triangle formed by the tangents at A, B and C.
Find the coordinates of the foci and the equation of the directrices of the rectangular hyperbola xy = c^{2}.
Find the equation of the hyperbola whose asymptotes are x + 2y + 3 = 0 and 3x + 4y + 5 = 0 and which passes through the point
(1,–1 ). Find also the equation of the conjugate of the conjugate hyperbola.
The vertices of the hyperbola
The centre of the hyperbola
The eccentricity of the hyperbola with latusrectum 12 and semiconjugate axis , is
The equation of the tangent to the curve 4x^{2} – 9y^{2} = 1 which is parallel to 4y = 5x + 7, is