InterviewSolution
Saved Bookmarks
| 1. |
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))`A. `theta=cos^(-1)(-(1)/(sqrt(5)))`B. `theta=cos^(-1)((1)/(sqrt(5)))`C. `t=-(2)/(sqrt(5))`D. `t=-(1)/(sqrt(5))` |
|
Answer» 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))` |
|