The focii of the parabolas $y^{2} = 4 x, x^{2} = 4 y, y^{2} + 4 x = 0, x^{2} + 4 y = 0$ taken in order will form the vertices of a

rectangle

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parallelogram

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rhombus

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square

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Solution

The correct option is D square
The focii of the parabolas $y^{2} = 4 x, x^{2} = 4 y, y^{2} + 4 x = 0, x^{2} + 4 y = 0$ will be $A (1, 0), B (0, 1), C (- 1, 0), D (0, - 1)$ respectively.
On calculating, $A B, B C, C D, D A$ , we get
$A B = B C = C D = D A = \sqrt{2}$
Also $A C = B D$
$∴$ It forms a square.

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