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Let \mathrm{T} and \mathrm{C} respectively be the transverse and conjugate axes of the hyperbola 16 x^2-y^2+64 x+4 y+44=0. Then the area of the region above the parabola x^2=y+4, below the transverse axis \mathrm{T}and on the right of the conjugate axis \mathrm{C} is:

Option: 1

4 \sqrt{6}+\frac{28}{3}


Option: 2

4 \sqrt{6}-\frac{44}{3}


Option: 3

4 \sqrt{6}+\frac{44}{3}


Option: 4

4 \sqrt{6}-\frac{28}{3}


Answers (1)

best_answer

\begin{aligned} & 16\left(x^2+4 x\right)-\left(y^2-4 y\right)+44=0 \\ & 16\left\{(x+2)^2-4\right\}-(y-2)^2+4+44=0 \\ & 16(x+2)^2-(y-2)^2=16 \\ & \frac{(x+2)^2}{1}-\frac{(y-2)^2}{16} \end{aligned}

Area  =\int_{-2}^{\sqrt{6}}\left(y_2-y_1\right) d x

\begin{aligned} & =\int_{-2}^{\sqrt{6}}\left(2-\left(x^2-4\right)\right) d x \\ & =4 \sqrt{6}+\frac{28}{3} \end{aligned}

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Sayak

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