Q. 7 B

# Find the equation

Given: Vertices are (-8, -1) and (16, -1) and focus is (17, -1)

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Center is the mid-point of two vertices

The distance between two vertices is 2a

The distance between the foci and vertex is ae – a and b2 = a2(e2 – 1)

The distance between two points (m, n) and (a, b) is given by

Mid-point theorem:

Mid-point of two points (m, n) and (a, b) is given by

Center of hyperbola having vertices (-8, -1) and (16, -1) is given by

= (4, -1)

The distance between two vertices is 2a and vertices are (-8, -1) and (16, -1)

The distance between the foci and vertex is ae – a, Foci is (17, -1) and the vertex is (16, -1)

b2 = a2(e2 – 1)

The equation of hyperbola:

25(x2 + 16 – 8x) – 144(y2 + 1 + 2y) = 3600

25x2 + 400 – 200x – 144y2 – 144 – 288y – 3600 = 0

25x2 – 144y2 – 200x – 288y – 3344 = 0

Hence, required equation of hyperbola is 25x2 – 144y2 – 200x – 288y – 3344 = 0

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

The eccentricity RD Sharma - Mathematics

The latus-rectum RD Sharma - Mathematics

The eccentricity RD Sharma - Mathematics

The foci of the hRD Sharma - Mathematics

The equation of tRD Sharma - Mathematics

The equation of tRD Sharma - Mathematics

The eccentricity RD Sharma - Mathematics

A point moves in RD Sharma - Mathematics

The equation of tRD Sharma - Mathematics

The length of theRD Sharma - Mathematics