Let PQ be a chord of the ellipse x2a2+y2b2=1 which subtends right angle at the centre (0,0). Then its distance from the centre is equal to
A
ab√a2+b2
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B
a√a2+b2
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C
b√a2+b2
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D
2ab√a2+b2
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Solution
The correct option is Aab√a2+b2 Let xcosα+ysinα=p be the chord PQ.
Then p is the desired distance.
Homogenizing the equation of the ellipse with the help of this equation, we get the combined equation of OP and OQ.
x2a2+y2b2=1 ⇒x2a2+y2b2=(xcosα+ysinαp)2 ⇒(1a2−cos2αp2)x2+(1b2−sin2αp2)y2−2xysinαcosαp2=0
As OP⊥OQ,
coefficient of x2+ coefficient of y2=0 (condition of perpendicular lines) 1a2−cos2αp2+1b2−sin2αp2=0 ⇒1p2=1a2+1b2 ⇒p=ab√a2+b2