Show that for all real values of 't' the line `2tx + ysqrt(1-t^2)=1` touches the ellipse.Find the eccentricity of the ellipse.

Options

A. `2//3`
B. `sqrt(3)//2`
C. `1//sqrt(3)`
D. `1//2`

Explanation

Let the ellipse be
`(x^(2))/(y^(2))+(y^(2))/(b^(2))=1`
The line `y==mx+-sqrt(a^(2)m^(2)+b^(2))` touches, the ellipse for all m.
Hence, it is identical with `y=-(2x)/(sqrt(1-p^(2)))+(1)/(sqrt(1-p^(2)))`
Hence, `m=-(2p)/(sqrt(1-p^(2)))`
and `a^(2)m^(2)+b^(2)=(1)/(1-p^(2))`
or `a^(2)(4p^(2))/(a-p^(2))+b^(2)(1)/(1-p^(2))`
or `p^(2)(4a^(2)-b^(2))+b^(2)-1=0`
This equation is true for all real p if
`b^(2)=1 and 4a^(2)=b^(2)`
`b^(2)=1 and a^(2)=(1)/(4)`
Therefore, the equation of the ellipse is
`(x^(2))/(1//4)+(y^(2))/(1)=1`
If e is its eccentricity, then
`(1)/(4)=1-e^(2) or e^(2)=(3)/(4) or e=(sqrt(3))/(2)`
be `=(sqrt(3))/(2)`
Hence, the foci are `(0,+-sqrt(3)//2)`
The equation of director circle is
`x^(2)+y^(2)=(5)/(4)`