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# Find the equation for the conic in standard form. Hyperbola: center (-5,0), one focus is (-5,3), one vertex is (-5,2)

Solution

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The vertices lie on the focal axis which is the vertical line $x=-5$ so the hyperbola is vertical. So, we use the standard equation with center at $(h,k)$:

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

From the given, $(h,k)=(-5,0)$. The vertices are at $(h,k\pm a)$. Since $(-5,2)$ is above the center, we use $k+a$ to equate:

$k+a=2$

$0+a=2$

$\color{#4257b2}a=2$

The vertices are at $(h,k\pm c)$. Since $(-5,3)$ is above the center, we use $k+c$ to equate:

$k+c=3$

$0+c=3$

$c=3$

Using the Pythagorean relationship, $c^2=a^2+b^2$, we can solve for $b$:

$b=\sqrt{c^2-a^2}=\sqrt{3^2-2^2}=\sqrt{9-4}$

$\color{#4257b2}b=\sqrt{5}$

So, the equation of the hyperbola is:

$\dfrac{(y-0)^2}{2^2}-\dfrac{(x-(-5))^2}{(\sqrt{5})^2}=1$

$\color{#c34632}\dfrac{y^2}{4}-\dfrac{(x+5)^2}{5}=1$

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