Question

P is a variable point on the ellipse x^2/a^2+y^2/b^2= 2 whose foci are F1 and F2.The maximum area of PF1F2 is ?

Answer 1

hey first tell me how do we know is a greater or b ? :/ how do we decide will it be horizontal one or vertical..

Answer 2

We don't need to know which is greater, just assume any to be greater, form the required equation, everything will be adjusted when values will be given.

Answer 3

okay..

Answer 4

Now, F1 and F2 are fixed, you can find those co-ordinates right ?

Answer 5

\[\LARGE b^2=a^2(1-e^2)\]
\[\LARGE e=\sqrt{\frac{a^2-b^2}{a^2}}\]
and distance between focus=2ae
F1=(ae,0)
F=(-ae,0)

Answer 6

\[\LARGE AREA=\frac{1}{2} 2ae \times y\]

Answer 7

Yep, you're on the right track,

Answer 8

and y I suppose is
\[\LARGE y=b \sqrt{\frac{x^2}{a^2}-2}\]

Answer 9

y is the y co-ordinate of P right?

Answer 10

yes y also equals that, but no need of that.

Answer 11

yes

Answer 12

kaise nikalenge :o

Answer 13

So, area will be max when y is max.

Answer 14

y will be max when P will be on the y-axis.

Answer 15

As easy as that.

Answer 16

\[\LARGE A=a \times\sqrt{\frac{a^2-b^2}{a^2}} \times b \sqrt{\frac{x^2}{a^2}-2}\]

Answer 17

\[\LARGE A=\sqrt{2}b \sqrt{a^2-b^2}\]

Answer 18

Let \[\LARGE P=(\sqrt2acos \phi ,\sqrt2b \sin \phi)\]
F1 and F2=(+ - sqrt ae ,0)