If $x^{2} + y^{2} = 9$ & $4 a^{2} + 9 b^{2} = 16$ , then maximum value of $4 a^{2} x^{2} + 9 b^{2} y^{2} - 12 a b x y$ is

81

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100

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121

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144

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Solution

The correct option is D $144$
Any parametric point on the given circle is $x = 3 cos θ, y = 3 sin θ$
Let $z = 4 a^{2} x^{2} + 9 b^{2} y^{2} - 12 a b x y = (2 a x - 3 b y)^{2} = 9 (2 a cos θ - 3 b sin θ)^{2}$
$∴ z_{m a x} = 9 (4 a^{2} + 9 b^{2}) = 144$
Since maximum value of $2 a cos θ - 3 b sin θ$ is $\sqrt{4 a^{2} + 9 b^{2}}$

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

\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}

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

\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}

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

\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}

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

\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}

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