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`        locus of point of intersection of tangents whose chord of condact subtend right angle at the centre of the ellipse x^2/a^2+y^2/b^2 = 1 `
6 years ago

510 Points
```										x2/a2 + y2/b2 = 1let chord of contact intersect the ellipse at A,B ..if O is origin then AO , BO are perpendicular ...let point of intersection is P(e,f) , join A,B with P thenthere is a quadrilateral OAPB in which AOB = 90 so   AOB + APB = 180o   90 + APB = 180          APB = 90oAPB is the angle bw tangents , since the angle is 90 so the product of slopesof tangents will be -1 ...m1m2 = -1tangent of ellipse is : y = mx +(-) (a2m2 + b2)1/2P will satisfy the eq of tangent  f = me +(-) (a2m2+b2)1/2 ( f-me)2 = (a2m2+b2)m2(e2-a2) - 2mef + f2+b2 = 0m1m2 = c/a = f2+b2/e2-a2 = -1e2+f2 = a2+b2now replacing e,f by x,y we getx2+y2 = a2 + b2this is eq of circle & the required locusapprove if u like my ans
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6 years ago
Anshul
23 Points
```										but the above answer is true if and only if OAPB is a cyclic quadrilateral ?? how do we know it is cyclic or not ??
```
one year ago
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