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Find the locus of the point of intersection of the normals at the end of the focal chord of the parabola `y^2=4a xdot`

Text Solution

Verified by Experts

The correct Answer is:
`y^(2)=a(x-3a)`
Point of intersection of normals to the parabola at points
`P(at_(1)^(2),2at_(1))andQ(at_(2)^(2)+t_(2)^(2)+t_(1)t_(2)),-at_(1)t_(2)(t_(1)+t_(2)))-=(h,k)`
Also, PQ is focal chord, then `t_(1)t_(2)=-1`.
`:." "k=a(t_(1)+t_(2))` (1)
`and" "h=2a+a[(t_(1)+t_(2))^(2)+t_(2)t_(2)]`
`or" "h=2a+a[(t_(1)+t_(2))^(2)+1]` (2)
Eliminating `(t_(1)+t_(2))` from (1) and (2), we get
`h=2a+a((k^(2))/(a^(2))+1)`
`:." "y^(2)=a(x-3a)`,
which is the required locus.
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