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the locus of the foot of perpendicular d...

the locus of the foot of perpendicular drawn from the centre of the ellipse `x^2+3y^2=6` on any point:

A

`(x^(2)-y^(2))^(2)=6x^(2)+2y^(2)`

B

`(x^(2)-y^(2))^(2)=6x^(2)-2y^(2)`

C

`(x^(2)+y^(2))^(2)=6x^(2)+2y^(2)`

D

`(x^(2)+y^(2))^(2)=6x^(2)-2y^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
Equations of ellipse is `x^(2)+ 3y^(2)=6 or (x^(2))/(6)+(y^(2))/(2) =1.`
Equations of the tangent is `( x cos theta) /(a)+(y sin theta)/(b)=1`
Let (h,k) be point on the locus .
`therefore ( h) /(a) cos theta +(K) /(b) sin theta =1 .`
Slope of the tangent line is `(-b)/(a) cot theta.`
slope of perpendicular drawn from centre (0,0) to (h,k) is k/h
since , both the lines are perpendicular
` therefore ((K)/(h))xx(-(b)/(a) cot theta)=-1`
`implies ( cos theta)/( ha)= ( sin theta)/( kb ) =alpha`[ say]
`implies cos theta theta = aha`
`sin theta theta = alpha kb`
from Eq. `(i) (h) /(a) ( alpha h a) +(k)/(b)(alphakb)=1`
`implies h^(2) alpha + k^(2)alpha =1`
`implies alpha=(1)/(h^(2)+k^(2))`
Also ` sin ^(2) theta + cos ^(2) theta =1`
` implies ( akb)^(2) +(aha)^(2)=1`
`implies a^(2)k^(2)b^(2)+ a^(2)h^(2)a^(2)=1`
`implies (k^(2)b^(2))/((h^(2)+k^(2))^(2))+(h^(2)a^(2))/((h^(2)+k^(2))^(2))=1`
`implies (2k^(2))/((h^(2)+k^(2))^(2))+(6h^(2))/((h^(2)+k^(2))^(2))=1[ :' a^(2) = 6, b^(2)=2]`
` implies 6x^(2)2y^(2)=(x^(2)+y^(2))^(2)`
[ replacing K by and h by x]
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