If normal at point P on parabola `y^(2)=4ax,(agt0)`, meets it again at Q in such a way that OQ is of minimum length, where O is the vertex of parabola, then `DeltaOPQ` is

Options

A. a right-angled triangle
B. an obtuse-angled triangle
C. an acute-angled triangle
D. none of these

Answer

A

Explanation

(1) Normal at `P(t_(1))` meets at `Q(t_(2))`. So
`t_(2)=-(2)/(t_(1))-t_(1)`

`|t_(2)|ge2sqrt(2)`
For minimum length of `Q Q,|t_(2)|` should be minimum. Therefore,
`|t_(2)|=2sqrt(2)ort_(2)=pm2sqrt(2)`
If `t_(2)=-2sqrt(2)`, then `t_(1)=sqrt(2)`.
Slope of `OQ=(2)/(t_(2))=m_(1)`
Slope of `OP=(2)/(t_(1))=m_(2)`
`:.m_(1)m_(2)=-1`
So, `DeltaOPQ` is a right-angled triangle.