If a triangle is inscribed in an ellipse and two of its sides are parallel to the given straight lines, then prove that the third side touches the fixed ellipse.

Explanation

Let the eccntric angls of the vertices P,Q, R of `DeltaPQR "be" theta_(1),theta_(2),theta_(3)`, respectively.
Then the equations of PQ and PR are, respectively,
`(x)/(a)cos.(theta_(1)+theta_(2))/(2)+(y)/(b) sin.(theta_(1)+theta_(2))/(2)=cos.(theta_(1)-theta_(2))/(2)`
`and (x)/(a)cos.(theta_(2)+theta_(3))/(2)+(y)/(b) sin.(theta_(2)+theta_(3))/(2)=cos.(theta_(2)-theta_(3))/(2)`
If PQ and PR are parallel to the given straigth lines, then we have
`theta_(1)+theta_(2)` (Constatn = `2alpha` (say)
and `theta_(1)+theta_(3)` = Constant= `2beta" "(1)`
Hence, `theta_(1)-theta_(3)= 2 (alpha-beta)`
Now, the equation of QR is
`(x)/(a)cos.(theta_(2)+theta_(3))/(2)+(y)/(b)sin.(theta_(2)+theta_(3))/(2)=cos.(theta_(2)-theta_(3))/(2)" "(2)`
or `(x)/(a)cos.(theta_(2)+theta_(3))/(2)+(y)/(b)sin.(theta_(2)+theta_(3))/(2)=cos(alpha-beta)" "(3)`
Which shows that line (2), for different values of `theta_(2)+theta_(3)` , is tangent to the ellipse
`(x^(2))/(a^(2))+(y^(2))/(b^(2))=cos^(2)(alpha-beta)`