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If the ellipse x^2/a^2+y^2/b^2=1 (b > a)...

If the ellipse `x^2/a^2+y^2/b^2=1 (b > a)` and the parabola `y^2 = 4ax` cut at right angles, then eccentricity of the ellipse is (a) `(3)/(5)` (b) `(2)/(3)` (c) `(1)/(sqrt(2))` (d) `(1)/(2)`

A

`(3)/(5)`

B

`(2)/(3)`

C

`(1)/(sqrt(2))`

D

`(1)/(2)`

Text Solution

Verified by Experts

Given an ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1(bgta)` and a parabola `y^(2)=4ax`.
Differentating ellipse w.r.t.x, we get
`(2x)/(a^(2))+(2yy')/(b^(2))=0`
`:. y'=-(b^(2)x)/(a^(2)y)`
Dfferentiating parabola w.r.t.x, we get
`:.y'=(2a)/(y)`
Sicne curves intersects orthogonally, we have
`-(b^(2)x)/(a^(2)y*(2a)/(y)=-1`
`rArr(2b^(2)x)/(ay^(2))=1`
`rArr(2b^(2)x)/(a(4ax))=1`
`rArr (a^(2))/(b^(2))=(1)/(2)`
`rAr e^(2)=1-(a^(2))/(b^(2))=1-(1)/(2)=(1)/(2)`
`rArre=(1)/(sqrt(2))`
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