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The mirror image of the parabola y^2= 4x...

The mirror image of the parabola `y^2= 4x` in the tangent to the parabola at the point (1, 2) is:

Text Solution

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Equation of tangent at point `(1,2)` is,
`y-2 = 1(x-1)`
` =>x-y+1= 0->(1)`
Let `(h,k)` is the mirror image at point `(t^2,2t)` for the tangent `x-y+1 = 0`.
Then,
`(h-t^2)/1 = (k-2t)/-1 = (-t^2+2t-1)/1`
`=>h = 2t-1 => t = (h+1)/2`
`=>k = t^2+1`
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