4Thanks Anant Sir
Let PQ be a chord of the ellipse x2/a2 + y2/b2 = 1 , which subtends an angle of pi/2 radians at the centre . If L is the foot of perpendicular from (0,0) to PQ , then
(A) Locus of L is an ellipse
(B) locus of L is circle concentric with given ellipse
(C) locus of L is a hyperbola concentric with given ellipse
(D) a square concentric with given ellipse
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2 Answers
66Let the coordinates of L be (h,k). Then equation of PQ becomes
y-k=-\frac{h}{k}(x-h)\quad \Rightarrow \ hx+ky =h^2+k^2
Homogenize the equation of the ellipse using this equation to obtain
\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=\left(\dfrac{hx+ky}{h^2+k^2}\right)^2
This is the equation of the pair of straight lines joining the center to P and Q. Since PQ subtends a right angle at the center, the angle between the two straight lines is 90°. Hence sum of the coefficients of x2 and y2 must be zero. SO we get
\dfrac{h^2}{(h^2+k^2)^2}-\dfrac{1}{a^2}+\dfrac{k^2}{(h^2+k^2)^2}-\dfrac{1}{b^2}=0
That is
h^2+k^2 = \dfrac{a^2b^2}{a^2+b^2}
As such the locus of L is
x^2+y^2 = \dfrac{a^2b^2}{a^2+b^2}
which is a circle centered at the origin having a radius \dfrac{ab}{\sqrt{a^2+b^2}}