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  • The equation of parabola whose latus rectum is 2 units, axis of line is x+y-2=0 and tangent at the vertex is x-y+4=0 is given byOption: 1 <img alt="\

The equation of parabola whose latus rectum is 2 units, axis of line is x+y-2=0 and tangent at the vertex is x-y+4=0 is given by

Option: 1

\mathrm{(x+y-2)^2=4 \sqrt{2}(x-y+4)^2}


Option: 2

\mathrm{(x-y-4)^2=4 \sqrt{2}(x+y-2)}


Option: 3

\mathrm{(x+y-2)^2=2 \sqrt{2}(x-y+4)}


Option: 4

None of these


Answers (1)

best_answer

Let P(x, y) be any point on the parabola and say PM & PT are perpendiculars from P on the axis & tangent at the vertex.

\mathrm{\text { Now, }(P M)^2=\text { latus rectum } \times(P T)}

\mathrm{\begin{aligned} & \therefore \quad\left(\frac{x+y-2}{\sqrt{1^2+1^2}}\right)^2=2 \frac{(x-y+4)}{\sqrt{2}} \\ & \Rightarrow \quad(x+y-2)^2=2 \sqrt{2}(x-y+4) \end{aligned}}

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