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The tangent at a point P(acosvarphi,bsin...

The tangent at a point `P(acosvarphi,bsinvarphi)` of the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` meets its auxiliary circle at two points, the chord joining which subtends a right angle at the center. Find the eccentricity of the ellipse.

Text Solution

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The equation of the auxliary circle is `x^(2)+y^(2)=a^(2) " "(1)`
Therefore, the equation of tangent to ellipse at a point `P(a cos phi, b sin phi)` is

`((x)/(a))cosphi+((y)/(b))sinphi=1(2)`
Which meets the auxiliary circle at points A and B
Therefore, the equation of the pair of lines OA and OB is obtained by making (1) homogeneous with the help of (2). Then,
`x^(2)+y^(2)=a^(2)((x)/(a)cosphi+(y)/(b)sinphi)^(2)`
But `angleAOB=90^(@)`. Therefore,
Cofficient of `x^(2)` + Coefficient of `y^(2)=0`
or `1-cos^(2)phi+1-(a^(2))/(b^(2))sin^(2)phi=0`
or `sin^(2)phi(1-(a^(2))/(b^(2)))+1=0`
`or (a^(-2)-b^(2))sin^(2)phi=b^(2`
`or a^(2)e^(2)sin^(2)phi=a^(2)(1-e^(2))`
or `(1+sin^(2)phi)e^(2)=1`
or `e=(1)/(sqrt(1+sin^(2)phi))`
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