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If the normals from any point to the parabola `y^2=4x` cut the line `x=2` at points whose ordinates are in AP, then prove that the slopes of tangents at the co-normal points are in GP.

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The equation of the normal to the parabola `y^(2)=4` is given by
`y=-tx+2t+t^(3)` (1)
Since it intersects x=2, we get ` y=t^(3)`
Let the three ordinates `t_(1)^(3),t_(2)^(3), andt_(3)^(3)` be in AP. Then,
`2_(2)^(3)=t_(1)^(3)+t_(3)^(3)`
`=(t_(1)+t_(3))^(3)-3t_(1)t_(3)(t_(1)+t_(3))` (2)
Now, `t_(1)+t_(2)+t_(3)=0`
`or" "t_(1)+t_(3)=-t_(2)`
Hence, (2) reduces to
`2t_(2)^(3)=-(-t_(2))^(3)-3t_(1)t_(3)(-t_(2))`
`=-(t_(2)^(3)+3t_(1)t_(2)t)(3)`
`3t_(2)^(3)+3t_(1)t_(2)t_(3)ort_(2)^(2)=t_(1)t_(3)`
Therefore, `t_(1),t_(2)andt_(3)` are tangents `1//t_(1),1//t_(2),and1//t_(3)` are in GP.
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