The curve $a^{2} y^{2} = x^{2} (a^{2} - x^{2})$ is defined for

x \leq a

and

x \geq - a

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x < a

and

x > - a

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x \leq - a

and

x \geq a

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x \leq a

and

x > - a

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Solution

The correct option is C $x \leq a$ and $x \geq - a$
Since we know that, square of any real number can never be negative,
And LHS of the given curve $a^{2} y^{2} = x^{2} (a^{2} - x^{2})$ is non negative

So RHS must also be non-negative

$\Rightarrow x^{2} (a^{2} - x^{2}) \geq 0$
$\Rightarrow a^{2} - x^{2} \geq 0$ , since $x^{2} \geq 0 \forall x \in R$

$\Rightarrow x^{2} - a^{2} \leq 0$

$\Rightarrow (x + a) (x - a) \leq 0$

$∴ x \leq a$ and $x \geq - a$

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The curve a2y2=x2(a2−x2) is defined for

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