Graph v=11i+11j

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Trigonometry

Graph v=11i+11j

$v = 11 i + 11 j$

Reorder $11 i$ and $11 x$ .

$y = 11 x + 11 i$

Find the asymptotes.

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Find the standard form of the hyperbola.

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Subtract $11 x$ from both sides of the equation.

$y - 11 x = 11 i$

Divide each term by $11 i$ to make the right side equal to one.

$\frac{y}{11 i} - \frac{11 x}{11 i} = \frac{11 i}{11 i}$

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 $1$ .

$\frac{y}{\frac{11}{i}} - x = 1$

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

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

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, $a$ .

$a = i \sqrt{11 i}$

$b = 1$

$k = 0$

$h = 0$

The asymptotes follow the form $y = \pm \frac{b (x - h)}{a} + k$ because this hyperbola opens left and right.

$y = \pm (- \frac{\sqrt{11 i}}{11} x) + 0$

Simplify $- \frac{\sqrt{11 i}}{11} x + 0$ .

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Add $- \frac{\sqrt{11 i}}{11} x$ and $0$ .

$y = - \frac{\sqrt{11 i}}{11} x$

Combine $x$ and $\frac{\sqrt{11 i}}{11}$ .

$y = - \frac{x \sqrt{11 i}}{11}$

This hyperbola has two asymptotes.

$y = - \frac{x \sqrt{11 i}}{11}, y = - \frac{x \sqrt{11 i}}{11}$

The asymptotes are $y = - \frac{x \sqrt{11 i}}{11}$ and $y = - \frac{x \sqrt{11 i}}{11}$ .

Asymptotes: $y = - \frac{x \sqrt{11 i}}{11}, y = - \frac{x \sqrt{11 i}}{11}$

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