The area of figure enclosed by the curve $5 x^{2} + 6 x y + 2 y^{2} + 7 x + 6 y + 6 = 0$ is

$\frac{π}{4}$

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$\frac{π}{2}$

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$π$

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$2 π$

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Solution

The correct option is B
$\frac{π}{2}$

$y = \frac{- 3 (1 + x) \pm \sqrt{(3 - x) (x - 1)}}{2}, (\begin{matrix} (3 - x) (x - 1) \geq 0 1 \leq x \leq 3 \end{matrix})$
Required area $= \int_{1}^{3} | y_{1} - y_{2} | d x$
$= \int_{1}^{3} \sqrt{(3 - x) (x - 1)} d x$
put $x = 3 c o s^{2} θ + s i n^{2} θ$
$d x = (- 3 s i n 2 θ) d θ = - 2 s i n 2 θ d θ$
$A = \int_{0}^{\frac{π}{2}} s i n^{2} 2 θ d θ = \frac{π}{2}$

Suggest Corrections

Similar questions

\int_{0}^{π} \frac{s i n (n + \frac{1}{2}) x}{s i n (\frac{x}{2})} d x

f (x) = \int_{1}^{x} \frac{{tan}^{- 1} t}{t} d t

(x > 0)

f (e^{2}) - f (\frac{1}{e^{2}})

z

- \frac{π}{4} \leq arg (z) \leq \frac{π}{4}

| z | \leq 4

y = 5 sin x; (0 \leq x \leq \frac{π}{2})

y = sin 2 x, x = 0, x = π

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