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Write the equation in standard form for the hyperbola $-x^{2}+4y^{2}-100=0$ .

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Convert equations of conic sections from general to standard form

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Q. Write the equation in standard form for the hyperbola

-x^{2}+4y^{2}-100=0

.

Move constant to right: To write the equation of the hyperbola in standard form, we need to isolate the terms with variables on one side and the constant on the other side. $\newline$ The given equation is $-x^2 + 4y^2 - 100 = 0$ . $\newline$ We want to get it into the form $(x^2/a^2) - (y^2/b^2) = 1$ , where $a$ and $b$ are constants. $\newline$ First, we move the constant term to the right side of the equation by adding $100$ to both sides. $\newline$ $-x^2 + 4y^2 = 100$
Divide by $-100$ : Next, we need to divide the entire equation by $-100$ to get the $x^2$ term positive and to set the right side of the equation equal to $1$ . $(\frac{-x^2 + 4y^2}{-100} = \frac{100}{-100})$ This simplifies to: $(\frac{x^2}{100}) - (\frac{4y^2}{100}) = -1$
Divide $y^2$ by $4$ : Now, we need to divide the $y^2$ term by $4$ to get it in the form of $(y^2/b^2)$ . $\newline$ $\left(\frac{x^2}{100}\right) - \left(\frac{y^2}{25}\right) = -1$
Multiply by $-1$ : Finally, we multiply the entire equation by $-1$ to get the right side of the equation to be positive, which is the standard form for a hyperbola. $\newline$ $-\left(\frac{x^2}{100}\right) + \left(\frac{y^2}{25}\right) = 1$ $\newline$ This gives us the standard form of the hyperbola: $\newline$ $\left(\frac{y^2}{25}\right) - \left(\frac{x^2}{100}\right) = 1$

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