Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

6. Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

$49y^2 - 16x^2 = 784$

Answers (1)

best_answer

Given a Hyperbola equation,

$49y^2 - 16x^2 = 784$

Can also be written as

$\frac{49y^2}{784} - \frac{16x^2}{784} = 1$

$\frac{y^2}{16} - \frac{x^2}{49} = 1$

$\frac{y^2}{4^2} - \frac{x^2}{7^2} = 1$

Comparing this equation with the standard equation of the hyperbola:

$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

We get,

$a=4$ and $b=7$

Now, As we know the relation in a hyperbola,

$c^2=a^2+b^2$

$c=\sqrt{a^2+b^2}$

$c=\sqrt{4^2+7^2}$

$c=\sqrt{65}$

Therefore,

Coordinates of the foci:

$(0,c) \:and\:(0,-c)=(0,\sqrt{65})\:and\:(0,-\sqrt{65})$

The Coordinates of vertices:

$(0,a) \:and\:(0,-a)=(0,4)\:and\:(0,-4)$

The Eccentricity:

$e=\frac{c}{a}=\frac{\sqrt{65}}{4}$

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(49)}{4}=\frac{98}{4}=\frac{49}{2}$

Posted by

Pankaj Sanodiya

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Board Exams 2025: Guidelines issued for Class 10, 12 practicals

anam-khan

Board Exam Date Sheet 2025 Live: ICSE, ISC, CBSE, MP time table out; BSEB routine soon

anam-khan

Board exam dates 2025: CBSE, CISCE, UP, MP, Maharashtra, TN time table out; state-wise updates

Latest Question