Find the Foci 14y^2-28x^2=448

$14 y^{2} - 28 x^{2} = 448$

Step 1

Find the standard form of the hyperbola.

Divide each term by $448$ to make the right side equal to one.

$\frac{14 y^{2}}{448} - \frac{28 x^{2}}{448} = \frac{448}{448}$

Simplify each term in the equation in order to set the right side equal to $1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be $1$ .

$\frac{y^{2}}{32} - \frac{x^{2}}{16} = 1$

Step 2

This is the form of a hyperbola. Use this form to determine the values used to find vertices and asymptotes of the hyperbola.

$\frac{{(y - k)}^{2}}{a^{2}} - \frac{{(x - h)}^{2}}{b^{2}} = 1$

Step 3

Match the values in this hyperbola to those of the standard form. The variable $h$ represents the x-offset from the origin, $k$ represents the y-offset from origin, $a$ .

$a = 4 \sqrt{2}$

$b = 4$

$k = 0$

$h = 0$

Step 4

Find $c$ , the distance from the center to a focus.

Find the distance from the center to a focus of the hyperbola by using the following formula.

$\sqrt{a^{2} + b^{2}}$

Substitute the values of $a$ and $b$ in the formula.

$\sqrt{{(4 \sqrt{2})}^{2} + {(4)}^{2}}$

Simplify.

Simplify the expression.

Apply the product rule to $4 \sqrt{2}$ .

$\sqrt{4^{2} {\sqrt{2}}^{2} + {(4)}^{2}}$

Raise $4$ to the power of $2$ .

$\sqrt{16 {\sqrt{2}}^{2} + {(4)}^{2}}$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$\sqrt{16 {(2^{\frac{1}{2}})}^{2} + {(4)}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\sqrt{16 \cdot 2^{\frac{1}{2} \cdot 2} + {(4)}^{2}}$

Combine $\frac{1}{2}$ and $2$ .

$\sqrt{16 \cdot 2^{\frac{2}{2}} + {(4)}^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\sqrt{16 \cdot 2^{\frac{2}{2}} + {(4)}^{2}}$

Rewrite the expression.

$\sqrt{16 \cdot 2^{1} + {(4)}^{2}}$

Evaluate the exponent.

$\sqrt{16 \cdot 2 + {(4)}^{2}}$

Simplify the expression.

Multiply $16$ by $2$ .

$\sqrt{32 + {(4)}^{2}}$

Raise $4$ to the power of $2$ .

$\sqrt{32 + 16}$

Add $32$ and $16$ .

$\sqrt{48}$

Rewrite $48$ as $4^{2} \cdot 3$ .

Factor $16$ out of $48$ .

$\sqrt{16 (3)}$

Rewrite $16$ as $4^{2}$ .

$\sqrt{4^{2} \cdot 3}$

Pull terms out from under the radical.

$4 \sqrt{3}$

Step 5

Find the foci.

The first focus of a hyperbola can be found by adding $c$ to $k$ .

$(h, k + c)$

Substitute the known values of $h$ , $c$ , and $k$ into the formula and simplify.

$(0, 4 \sqrt{3})$

The second focus of a hyperbola can be found by subtracting $c$ from $k$ .

$(h, k - c)$

Substitute the known values of $h$ , $c$ , and $k$ into the formula and simplify.

$(0, - 4 \sqrt{3})$

The foci of a hyperbola follow the form of $(h, k \pm \sqrt{a^{2} + b^{2}})$ . Hyperbolas have two foci.

$(0, 4 \sqrt{3}), (0, - 4 \sqrt{3})$

Step 6

Find the Foci 14y^2-28x^2=448

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