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If the tangent to the ellipse x^2+2y^2=1...

If the tangent to the ellipse `x^2+2y^2=1` at point `P(1/(sqrt(2)),1/2)` meets the auxiliary circle at point `Ra n dQ` , then find the points of intersection of tangents to the circle at `Qa n dRdot`

Text Solution

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The correct Answer is:
`(1,sqrt(2),1)`
The equation of tangent to the ellipse at a given points is `x((1)/(sqrt(2)))+2y((1)/(2))=1`
or `x+sqrt(y)=sqrt(2)" "(1)`
Now, QR is the chord of contact of the circle `x^(2)_y^(2)=1` with respect to the point T(h,k). Then,
`QR-=hx+ky+ky=1 " "(2)`
Equations (1) and (2) represent the same straight line. Then comapring the ratio of coefficients, we have, `h//1=k//sqrt(2)=1sqrt(2)`.
Hence `(h,k)-=(1//sqrt(2))`
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