Find the Foci 3x^2+y^2+3x-21/4=0

$3 x^{2} + y^{2} + 3 x - \frac{21}{4} = 0$

Step 1

Find the standard form of the ellipse.

Add $\frac{21}{4}$ to both sides of the equation.

$3 x^{2} + y^{2} + 3 x = \frac{21}{4}$

Complete the square for $3 x^{2} + 3 x$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 3$

$b = 3$

$c = 0$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{3}{2 \cdot 3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$d = \frac{3}{2 \cdot 3}$

Rewrite the expression.

$d = \frac{1}{2}$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{3^{2}}{4 \cdot 3}$

Simplify the right side.

Cancel the common factor of $3^{2}$ and $3$ .

Factor $3$ out of $3^{2}$ .

$e = 0 - \frac{3 \cdot 3}{4 \cdot 3}$

Cancel the common factors.

Factor $3$ out of $4 \cdot 3$ .

$e = 0 - \frac{3 \cdot 3}{3 \cdot 4}$

Cancel the common factor.

$e = 0 - \frac{3 \cdot 3}{3 \cdot 4}$

Rewrite the expression.

$e = 0 - \frac{3}{4}$

Subtract $\frac{3}{4}$ from $0$ .

$e = - \frac{3}{4}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $3 {(x + \frac{1}{2})}^{2} - \frac{3}{4}$ .

$3 {(x + \frac{1}{2})}^{2} - \frac{3}{4}$

Substitute $3 {(x + \frac{1}{2})}^{2} - \frac{3}{4}$ for $3 x^{2} + 3 x$ in the equation $3 x^{2} + y^{2} + 3 x = \frac{21}{4}$ .

$3 {(x + \frac{1}{2})}^{2} - \frac{3}{4} + y^{2} = \frac{21}{4}$

Move $- \frac{3}{4}$ to the right side of the equation by adding $\frac{3}{4}$ to both sides.

$3 {(x + \frac{1}{2})}^{2} + y^{2} = \frac{21}{4} + \frac{3}{4}$

Simplify $\frac{21}{4} + \frac{3}{4}$ .

Combine the numerators over the common denominator.

$3 {(x + \frac{1}{2})}^{2} + y^{2} = \frac{21 + 3}{4}$

Simplify the expression.

Add $21$ and $3$ .

$3 {(x + \frac{1}{2})}^{2} + y^{2} = \frac{24}{4}$

Divide $24$ by $4$ .

$3 {(x + \frac{1}{2})}^{2} + y^{2} = 6$

Divide each term by $6$ to make the right side equal to one.

$\frac{3 {(x + \frac{1}{2})}^{2}}{6} + \frac{y^{2}}{6} = \frac{6}{6}$

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 $1$ .

$\frac{{(x + \frac{1}{2})}^{2}}{2} + \frac{y^{2}}{6} = 1$

Step 2

This is the form of an ellipse. Use this form to determine the values used to find the center along with the major and minor axis of the ellipse.

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

Step 3

Match the values in this ellipse to those of the standard form. The variable $a$ represents the radius of the major axis of the ellipse, $b$ represents the radius of the minor axis of the ellipse, $h$ represents the x-offset from the origin, and $k$ represents the y-offset from the origin.

$a = \sqrt{6}$

$b = \sqrt{2}$

$k = 0$

$h = - \frac{1}{2}$

Step 4

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

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

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

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

$\sqrt{{(\sqrt{6})}^{2} - {(\sqrt{2})}^{2}}$

Simplify.

Rewrite ${\sqrt{6}}^{2}$ as $6$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{6}$ as $6^{\frac{1}{2}}$ .

$\sqrt{{(6^{\frac{1}{2}})}^{2} - {(\sqrt{2})}^{2}}$

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

$\sqrt{6^{\frac{1}{2} \cdot 2} - {(\sqrt{2})}^{2}}$

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

$\sqrt{6^{\frac{2}{2}} - {(\sqrt{2})}^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\sqrt{6^{\frac{2}{2}} - {(\sqrt{2})}^{2}}$

Rewrite the expression.

$\sqrt{6^{1} - {(\sqrt{2})}^{2}}$

Evaluate the exponent.

$\sqrt{6 - {(\sqrt{2})}^{2}}$

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

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

$\sqrt{6 - {(2^{\frac{1}{2}})}^{2}}$

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

$\sqrt{6 - 2^{\frac{1}{2} \cdot 2}}$

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

$\sqrt{6 - 2^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\sqrt{6 - 2^{\frac{2}{2}}}$

Rewrite the expression.

$\sqrt{6 - 2^{1}}$

Evaluate the exponent.

$\sqrt{6 - 1 \cdot 2}$

Simplify the expression.

Multiply $- 1$ by $2$ .

$\sqrt{6 - 2}$

Subtract $2$ from $6$ .

$\sqrt{4}$

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

$\sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$2$

Step 5

Find the foci.

The first focus of an ellipse can be found by adding $c$ to $k$ .

$(h, k + c)$

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

$(- \frac{1}{2}, 0 + 2)$

Simplify.

$(- \frac{1}{2}, 2)$

The first focus of an ellipse can be found by subtracting $c$ from $k$ .

$(h, k - c)$

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

$(- \frac{1}{2}, 0 - (2))$

Simplify.

$(- \frac{1}{2}, - 2)$

Ellipses have two foci.

${Focus}_{1}$ : $(- \frac{1}{2}, 2)$

${Focus}_{2}$ : $(- \frac{1}{2}, - 2)$

${Focus}_{1}$ : $(- \frac{1}{2}, 2)$

${Focus}_{2}$ : $(- \frac{1}{2}, - 2)$

Step 6

Find the Foci 3x^2+y^2+3x-21/4=0

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