The axis of a parabola is along the line and the distances of its vertex and focus from origin are and respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is [2006]

Question

The axis of a parabola is along the line $y = x$ and the distances of its vertex and focus from origin are $\sqrt{2}$ and $2 \sqrt{2}$ respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is [2006]

Smart Achievers · Accepted Answer

(4)

Since distance of vertex and focus of the parabola from origin is $\sqrt{2}$ and $2 \sqrt{2}$ ,

$∴$ Vertex is (1, 1) and focus is (2, 2), directrix $x + y = 0$

$∴$ Equation of parabola is

${(x - 2)}^{2} + {(y - 2)}^{2} = {(\frac{x + y}{\sqrt{2}})}^{2}$

$\Rightarrow 2 (x^{2} - 4 x + 4) + 2 (y^{2} - 4 y + 4) = x^{2} + y^{2} + 2 x y$

$\Rightarrow x^{2} + y^{2} - 2 x y = 8 (x + y - 2) \Rightarrow {(x - y)}^{2} = 8 (x + y - 2)$

Solution

Q.
The axis of a parabola is along the line $y = x$ and the distances of its vertex and focus from origin are $\sqrt{2}$ and $2 \sqrt{2}$ respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is [2006]

1 ${(x + y)}^{2} = (x - y - 2)$

2 ${(x - y)}^{2} = (x + y - 2)$

3 ${(x - y)}^{2} = 4 (x + y - 2)$

4 ${(x - y)}^{2} = 8 (x + y - 2)$

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Solution

Q. The axis of a parabola is along the line y=x and the distances of its vertex and focus from origin are 2 and 22 respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is [2006]

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

2 (x-y)2=(x+y-2)

3 (x-y)2=4(x+y-2)

4 (x-y)2=8(x+y-2)