Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XI  >>  Math  >>  Conic Sections
0 votes

Find the equations of the hyperbola satisfying the given conditions: Foci: $(\pm 3\sqrt 5, 0)$, Latus Rectum is of length $8$.

Can you answer this question?

1 Answer

0 votes
Given Foci: $(\pm 3\sqrt 5, 0) \rightarrow c = 3\sqrt5$.
Given Latus Rectum is of length $8 \rightarrow 8 = \large\frac{2b^2}{a}$$\rightarrow b^2 = 4a$
Since $a^2+b^2 = c^2 \rightarrow a^2+4a=3^2\sqrt5^2 \rightarrow a = 5$ (We discard $a=-9$ since $a$ is non-negative).
If $a = 5 \rightarrow b^2 = 4 \times 5 = 20$
Therefore, the equation of the hyperbola is as follows: $\;\large\frac{x^2}{25}$$-\large\frac{y^2}{20}$$=1$
answered Apr 8, 2014 by balaji.thirumalai

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App