Q11

Geometry Level 3

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


The answer is 7.

Sahil Bansal by Sahil Bansal , 21, India

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
Mar 3, 2019

The foci of the ellipse and the hyperbola are respectively:

F E = a 2 − b 2 = 16 − b 2 F_{E} = \sqrt{a^2 - b^2} = \sqrt{16-b^2}

F H = a 2 + b 2 = 144 25 + 81 25 = 9 = 3 F_{H} = \sqrt{a^2 + b^2} = \sqrt{\frac{144}{25} + \frac{81}{25}} = \sqrt{9} = 3

If these foci coincide, then:

F E = F H ⇒ 16 − b 2 = 3 ⇒ b 2 = 7 . F_{E} = F_{H} \Rightarrow \sqrt{16 - b^2} = 3 \Rightarrow \boxed{b^2 = 7}.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...