A geometry problem by A Former Brilliant Member

Geometry Level 3

If the foci of the ellipse x 2 16 + y 2 b 2 = 1 \frac{x^2}{16} + \frac{y^2}{b^2} = 1 and the hyperbola, x 2 144 − y 2 81 = 1 25 \frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25} coincide then find the value of b 2 b^2 .


The answer is 7.

A Former Brilliant Member by A Former Brilliant Member

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Tom Engelsman
Sep 25, 2020

For the ellipse, its focal length from the origin is: a 2 − b 2 = c \sqrt{a^2 - b^2} = c . For the hyperbola, its a 2 + b 2 = c \sqrt{a^2 + b^2} = c . If the foci are coincident, then we compute:

16 − b 2 = 144 / 25 + 81 / 25 \sqrt{16 - b^2} = \sqrt{144/25 + 81/25} ;

or 16 − b 2 = 225 / 25 ; 16 - b^2 = 225/25;

or 16 − 9 = b 2 16 - 9 = b^2 ;

or b 2 = 7 . \boxed{b^2 = 7}.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...