Fun With Ellipse 12

Geometry Level 4

If 2 x 2 + x y − 3 y 2 = 0 2 x^2+x y-3y^2=0 is the equation of a pair of conjugate diameters of an ellipse x 2 a 2 + y 2 b 2 = 1 \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 , then find its eccentricity .

1 3 \frac{1}{3} 2 3 \frac{2}{\sqrt 3} 2 3 \frac{2}{3} 1 3 \frac{1}{\sqrt 3}

Akshay Sharma by Akshay Sharma , 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.

2 solutions

2 x 2 + x y − 3 y 2 = 0 ⟹ ( 2 x + 3 y ) ( x − y ) = 0. S o t h e c o n j u g a t e d i a m e t e r s a r e y = − 2 3 x , a n d y = x . S o t h e s l o p e s a r e − 2 3 , a n d 1. B u t ( − 2 3 ) ∗ ( 1 ) = p r o d u c t o f t h e i r s l o p e s = − b 2 a 2 . E c c e n t r i c i t y = e = 1 − b 2 a 2 = 1 − 2 3 = 1 3 . 2x^2+xy-3y^2=0~~\implies~(2x+3y)(x-y)=0.\\ So~the~conjugate~diameters~are~y= -\dfrac 2 3x,~~and~~y=x.\\ So~the~slopes~are~~~~ - \dfrac 2 3,~and~ 1.\\ But~\Big (- \dfrac 2 3 \Big )*(1)=product~of~their~slopes= - \dfrac {b^2}{a^2}.\\ Eccentricity=e=\sqrt{1-\dfrac{b^2}{a^2}}=\sqrt{1-\dfrac 2 3}=\dfrac 1{\sqrt 3}.

2 Helpful 1 Interesting 1 Brilliant 0 Confused
Aryan Goyat
May 27, 2016

Any line passing through center of ellipse is a diameter . two diameters are said to be conjugate of each other if each bisects the chord || to other.

let the two line be y=m1x and y=m2x

let a pt be l,m on y=m1x

now it is mid pt of chord

T=S1

find the slope of this and equate it to m2

we get m1*m2= -b^(2)/a^(2)

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...