If the vertices of a quadrilateral is given by $(x^{2} - 4)^{2} + (y^{2} - 9)^{2} = 0$ then area of quadrilateral is

36

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24

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16

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81

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Solution

The correct option is A $36$
Vertices are given by $x^{2} - 4 = 0$ and $y^{2} - 9 = 0$
So points will be $(2, 3), (2, - 3), (- 2, - 3), (- 2, 3)$
Area will be $= 4 \times 6 = 24 s q^{2} u n i t$

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Similar questions

E_{1} : \frac{x^{2}}{16} + \frac{y^{2}}{9} = 1

E_{2} : \frac{x^{2}}{9} + \frac{y^{2}}{16} = 1

\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1

(- 16)

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A (p, 0), B (4, 0), C (5, 6)

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\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1

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