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If normal at any point P to ellipse `(x^2)/(a^2) + (y^2)/(b^2) = 1(a > b)` meet the x & y axes at A and B respectively. Such that `(PA)/(PB) = 3/4` , then eccentricity of the ellipse is:

A

`1/3`

B

`1/2`

C

`1/4`

D

`1/9`

Text Solution

Verified by Experts

The correct Answer is:
B

Equation of normal to ellipse
`(ax)/(cos theta)==-(by)/(sin theta)=a^(2)-b^(2)`
For point A
`y=0`
`x=cos theta((a^(2)-b^(2))/(a))`

For point B
`x=0`
`y=sin theta ((b^(2)-a^(2))/(3))`

`(a cos theta, b cos theta)(cos theta((a^(2)-b^(2))/(a)),0)(0, sin theta((b^(2)-a^(2))/(b)))`
`0=(b sin thetaxx1+3sin theta((b^(2)-a^(2))/(b)))/(4)`
`b^(2)+3(b^(2)-a^(2))=0`
`4b^(2)-3a^(2)=0 rarr (b^(2))/(a^(2))=(3)/(4)rarr" can be directly used"`
`e=sqrt(1-(3)/(4))=(1)/(2)`
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