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P(t(1))andQ(t(2)) are points t(1)andt(2)...

`P(t_(1))andQ(t_(2))` are points `t_(1)andt_(2)` on the parabola `y^(2)=4ax`.
The normals at P and Q meet on the parabola. Show that the middle point of PQ lies on the parabola `y^(2)=2a(x+2a)`.

Text Solution

Verified by Experts

The equation of normal to `y^(2)=4ax` is
`y=-tx+2at+at^(3)`
It passes through the (h,k). Then
`k=-tx+2at+at^(3)`
`or" "at^(3)+(2a-h)t-k=0`
Since normal at `P(at_(1)^(2),2at_(1))andQ(at_(2)^(2),2at_(2))` passes through the point `R(at_(3)^(2),2at_(3))` on the parabola,
`t_(1)t_(2)t_(3)=(k)/(a)=(2at_(3))/(a)`
`:." "t_(1)t_(2)=2`
Also , if `(x_(1),y_(1))` is the midpoint of PQ, then
`x_(1)=(1)/(2)(at_(1)^(2)+at_(2)^(2))andy_(1)=(1)/(2)(2at_(1)+2at_(2))` (2)
`:." "(t_(1)+t_(2))^(2)=((y_(1))/(a))^(2)`
`or" "((y_(1))/(a))^(2)=t_(1)^(2)+t_(2)^(2)+2t_(1)t_(2)=(2x_(1))/(a)+4`
`or" "y_(1)^(2)=2a(x_(1)+2a)`
Hence, the locus of `(x_(1),y_(1))` is `y^(2)=2a(x+2a)`.
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