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If line x-2y-1=0 intersects parabola y^(...

If line x-2y-1=0 intersects parabola `y^(2)=4x` at P and Q, then find the point of intersection of normals at P and Q.

Text Solution

Verified by Experts

The correct Answer is:
(19,4)
Any equation on the parabola `y^(2)=4x`, say `(t^(2),2t)`, lies on the line x-2y-1=0.
`:." "t^(2)-4t-1=0`
This equation has two roots `t_(1)andt_(2)`, which are parabola are parameters of the points of intersection of line and parabola.
`:." "t_(1)+t_(2)=4andt_(1)t_(2)=-1`
Now, point of intersection of normals at P and Q is
`N-=(2+(t_(1)^(2)+t_(2)^(2))+t_(1)t_(2)(t_(1)+t_(2)))`
`or" "N-=(2+(t_(1)+t_(2))^(2)-t_(1)t_(2),-t_(1)t_(2)(t_(1)+t_(2)))`
Putting the value of `t_(1)+t_(2)andt_(1)t_(2)`, we get
`N-=(2+16+1,1(4))-=(19,4)`
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