IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry - Ellipse-9

kane

from any pt. on x²/a²+y²/b²=1,tangents are drawn to x²+y²=r².show that chord of contact touches a²x²+b²y²=r⁴

rhd92781

while solving this type of probl. u have to find the eqn of line (here chord of contact) first and then eliminate the parameter.
if u get a quadratic in the parameter, equate its discriminant to zero.

here, let point of ellipse be (acosw,bsinw)

eqn of chord of contact on circle is:
axcosw + bysinw = r^2
axcosw + byrt(1-cos^2w) = r^2

now make it a quadratic:
(a^2x^2 + b^2y^2)cos^2w - 2axr^2cosw + (r^4-b^2y^2)=0
equate its discriminant to zero and solve, u'll get

a^2x^2 + b^2y^2 = r^4

joyfrancis

General point on the ellipse = acost,bsint

Eqn of chord of contact to a circle is T=0 which in this case is

=> x(acost) + y(asint) - r² = 0

i.e y = (-acost/bsint)x + r²/bsint.........(1)

The curve given is

a²x²+b²y²=r⁴

= x² / (r²/a)² + y²/ (r²/b)² = 1........(2)

hence it is an ellipse

Condition for a line to be a tangent to the ellipse : c² = a²m²+b²

LHS = c² = r⁴/b²sin²t

RHS = a²m²+b² = r⁴(a²cos²t)/a²b²cos²t + r⁴/b² = r⁴/b²sin²t = LHS

Hence Proved

Ask & Discuss Questions with Community & Experts