The equation of the hyperbola 4 x 2-32 x - y 2-4 y -24-0 in its standard form isA- - idfrac x-4- 216-ldfracy-2- 236-1 -B- - dfrac x-4- 29-dfracy-2- 236-1 -C- - Idfrac x-4- 29-dfracy-2- 236-1 -D- - dfrac x-4- 29- dfrac y-2- 216-1 -

The correct option is C (x 4)^29 (y+2)^236=14x^2 32x y^2 4y+24=0 ⇒4(x^2 8x) (y^2+4y)+24=0 ⇒4(x 4)^2 (y+2)^2 36=0 ⇒ (x 4)^29 (y+2)^236=1 ...

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Standard XII

Mathematics

Eccentricity of Hyperbola

The equation ...

Question

The equation of the hyperbola $4 x^{2} - 32 x - y^{2} - 4 y + 24 = 0$ in its standard form is

\frac{(x - 4)^{2}}{16}

- \dfrac{(y+2)^2}{36}=1$

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\frac{(x - 4)^{2}}{9}

- \dfrac{(y-2)^2}{36}=1$

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\frac{(x - 4)^{2}}{9}

- \dfrac{(y+2)^2}{36}=1$

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\frac{(x - 4)^{2}}{9}

- \dfrac{(y+2)^2}{16}=1$

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Solution

The correct option is C $\frac{(x - 4)^{2}}{9}$ - \dfrac{(y+2)^2}{36}=1$
$4 x^{2} - 32 x - y^{2} - 4 y + 24 = 0$
$\Rightarrow 4 (x^{2} - 8 x) - (y^{2} + 4 y) + 24 = 0$
$\Rightarrow 4 (x - 4)^{2} - (y + 2)^{2} - 36 = 0$
$\Rightarrow$ $\frac{(x - 4)^{2}}{9}$ $- \frac{(y + 2)^{2}}{36} = 1$

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The equation of the hyperbola 4x2−32x−y2−4y+24=0 in its standard form is

The correct option is C (x−4)29- \dfrac{(y+2)^2}{36}=1$4x2−32x−y2−4y+24=0 ⇒4(x2−8x)−(y2+4y)+24=0 ⇒4(x−4)2−(y+2)2−36=0 ⇒ (x−4)29−(y+2)236=1

The equation of the hyperbola $4 x^{2} - 32 x - y^{2} - 4 y + 24 = 0$ in its standard form is

The correct option is C $\frac{(x - 4)^{2}}{9}$ - \dfrac{(y+2)^2}{36}=1$
$4 x^{2} - 32 x - y^{2} - 4 y + 24 = 0$
$\Rightarrow 4 (x^{2} - 8 x) - (y^{2} + 4 y) + 24 = 0$
$\Rightarrow 4 (x - 4)^{2} - (y + 2)^{2} - 36 = 0$
$\Rightarrow$ $\frac{(x - 4)^{2}}{9}$ $- \frac{(y + 2)^{2}}{36} = 1$