Question

Show that for all real values of ${}^{\prime }{t}^{\prime }$ the line $2tx+y\sqrt{1-t^2}=1$ touches a fixed ellipse. Find the eccentricity of the ellipse.

Answer 1

$2+x+y\sqrt{1-t^2}=1$

$y=\sqrt{1-t^2}=1-2+x$

$y^2(1-t^2)=1+4t^2{x}^2-4+x$

$y^2-y^2{t}^2=1+4t^2{x}^2-4+4$

$t^2(4x^2+y^2)-y^2-4+x+1=0$

$t^2(4x^2+y^2)-4tx+x-y^2+1=0$

$D=0$

$16x^2-4(-y^2+1)(4x^2+y^2)=0$

$4x^2-(-4x^2{y}^2-y^2+4x^2+y^2)=0$

$4x^2+4x^2{y}^2+y^4-4x^2-y^2=0$

$4x^2{y}^2+y^4-y^2=0$

$y^2(4x^2+y^2-1)=0$

$4x^2+y^2=1$

$\frac{x^2}{44}+\frac{y^2}{1}=1$

which is a ellipse

$e=\sqrt{1-\frac{1}{4}}$

$e=\frac{\sqrt{3}}{2}$