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  • The mirror image of the parabola   in the tangent to the parabola to the point (1,2) is<div cl

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

Option: 1

(x-1)^2=4(y+1)


Option: 2

(x+1)^2=4(y+1)


Option: 3

(x+1)^2=4(y-1)


Option: 4

(x-1)^2=4(y-1)


Answers (1)

best_answer

Any point on the given parabola is \left(\mathrm{t}^2, 2 \mathrm{t}\right). The equation of the tangent at(1,2) \text { is } x-y+1=0 \text {. } The image (h,k) of the point\left(t^2, 2 t\right) \text { in } x-y+1=0 is 

given by \frac{\mathrm{h}-\mathrm{t}^2}{\mathrm{l}}=\frac{\mathrm{k}-2 \mathrm{t}}{-\mathrm{l}}=\frac{-2\left(\mathrm{t}^2-2 \mathrm{t}+\mathrm{l}\right)}{\mathrm{l}+\mathrm{l}}

\therefore \mathrm{h}=\mathrm{t}^2-\mathrm{t}^2+2 \mathrm{t}-1=2 \mathrm{t}-1

and  k=2 t+t^2-2 t+1=t^2+1

Eliminating t from \mathrm{h}=2 \mathrm{t}-1 \text { and } \mathrm{k}=\mathrm{t}^2+1

we get, (h+1)^2=4(k-1)

The required equation of reflection is (x+1)^2=4(y-1)

Posted by

Ritika Jonwal

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