# 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

509 Points
13 years ago
x2/a2 + y2/b2 = 1

let 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 then

there is a quadrilateral OAPB in which AOB = 90 so

AOB + APB = 180o

90 + APB = 180

APB = 90o

APB is the angle bw tangents , since the angle is 90 so the product of slopes

of tangents will be -1 ...

m1m2 = -1

tangent of ellipse is : y = mx +(-) (a2m2 + b2)1/2

P will satisfy the eq of tangent

f = me +(-) (a2m2+b2)1/2

( f-me)2 = (a2m2+b2)

m2(e2-a2) - 2mef + f2+b2 = 0

m1m2 = c/a = f2+b2/e2-a2 = -1

e2+f2 = a2+b2

now replacing e,f by x,y we get

x2+y2 = a2 + b2

this is eq of circle & the required locus

approve if u like my ans
Anshul
23 Points
7 years ago
but the above answer is true if and only if OAPB is a cyclic quadrilateral ?? how do we know it is cyclic or not ??