If the tangent drawn at point `(t^2,2t)` on the parabola `y^2=4x` is the same as the normal drawn at point `(sqrt(5)costheta,2sintheta)` on the ellipse `4x^2+5y^2=20,` then `theta=cos^(-1)(-1/(sqrt(5)))` (b) `theta=cos^(-1)(1/(sqrt(5)))` `t=-2/(sqrt(5))` (d) `t=-1/(sqrt(5))`

Options

A. `theta=cos^(-1)(-(1)/(sqrt(5)))`
B. `theta=cos^(-1)((1)/(sqrt(5)))`
C. `t=-(2)/(sqrt(5))`
D. `t=-(1)/(sqrt(5))`

Explanation

The eqaution of the tagent at `(t^(2),2t)` to the parabola `y^(2)=4x` is
`2ty=2(x+r^(2))`
or`ty=x+r^(2)`
or `x-ty+t^(2)=0" "(1)`
The equation of the normal at `(sqrt(5)cos theta, 2 sin theta)` on the ellipse `4x^(2)+5y^(2)=20`is
` (sqrt(5)sectheta)x-(2"cosec" theta)y=5-4`
` (sqrt(5)sectheta)x-(2"cosec" theta)y=1" "(2)`
Given that (1) and (2) represent the same line then.
`(sqrt(5)sectheta)/(1)=(-2"cosec" theta)/(-t)=(-1)/(r^(2))`
or `t=(2)/(sqrt(5))cot theta and t=-(1)/(2)sintheta`
or `(2)/(sqrt(5))cot theta =-(1)/(2)sintheta`
or `4 cos theta=-sqrt(5)(1-cos^(2)theta)`
on `sqrt(5)cos^(2) theta-4 costheta-sqrt(5)=0`
or `(cos theta-sqrt(5))(sqrt(5)cos theta+1)=0`
`or cos theta=-(1)/(sqrt(3))`
`or theta=cos^(-1)(-(1)/(sqrt(5)))`
Putting `cos theta-ssqrt(5) "in" t=-(1//2) sin theta,` we get
`t=-(1)/(2)sqrt(1-(1)/(5))=-(1)/(sqrt(5))`
Hence , `theta = cos^(-1)(-(1)/(sqrt(5))) and t=-(1)/(sqrt(5))`