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If alpha-beta= constant, then the locus ...

If `alpha-beta=` constant, then the locus of the point of intersection of tangents at `P(acosalpha,bsinalpha)` and `Q(acosbeta,bsinbeta)` to the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` is a circle (b) a straight line an ellipse (d) a parabola

A

a cricle

B

a straigth line

C

an ellipse

D

a parabola

Text Solution

Verified by Experts

Tangents to the ellipse at P and Q are, respectively,
`(x)/(a) cos alpha+(y)/(b) sin alpha=`
`and (x)/(a) cos beta+(y)/(b) sin beta=1`
Solving, (1) and (2) , we get

or `x=(a(sinalpha-sinbeta))/(sin(beta-alpha))`
and `y=(-b(cosalpha-cosbeta))/(sin(beta-alpha))`
`and =(-xsin(beta-alpha))/(a)=sinalpha-sinbeta`
`y=(-ysin(beta-alpha))/(a)=(cosalpha-cosbeta)`
Saqring and adding. we get
`sin^(2)(beta-alpha)((x^(2))/(a^(2))+(y^(2))/(b^(2)))=2`
`(x^(2))/(a^(2))+(y^(2))/(b^(2))=(2)/(sin^(2)c) " "` [ where `beta-alpha=c` (constant) given]
which is an ellipse
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