Answer to Question #104883 in Geometry for Joseph Ocran

Question

The tangent at any point on the eclipse x2/a2 +y2/b2 meets the tangents at the ends of the major axis Q and Q'. Show that the circle on QQ' has diameter passes through the foci.

Answer 1

Let ellipse be x²/a² +y²/b²=1

Let P(acosø,bsinø) be any point on this ellipse.

Equation of tangent at P(acosø,bsinø) is

(x/a)cosø+(y/b)sinø=1 (1)

The two tangents drawn at the ends of major axis are x=a and x=-a

Solving (1) and x=a we get

T={a,b(1-cosø)/sinø}={a,btan(ø/2)}

Solving (1) and x=-a we get

T¹ ={-a,b(1+cosø)/sinø}={-a,b cot(ø/2)}

Equation of circle on TT¹ as a diameter is (x-a)(x+a)+(y-b tan(ø/2))(y-b cot(ø/2))=0

x² +y² -by(tan(ø/2))+cot(ø/2))-a² +b² =0

Now put x=+/-as and y=0 in above equation, we get

a² e² +0-0-a² +b²=a²-b²-a²+b²=0

Thus foci lies in 0

In conics only circle has foci in zero,hence it's a circle