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Question

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=5x=-5 so the hyperbola is vertical. So, we use the standard equation with center at (h,k)(h,k):

(yk)2a2(xh)2b2=1\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1

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

k+a=2k+a=2

0+a=20+a=2

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

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

k+c=3k+c=3

0+c=30+c=3

c=3c=3

Using the Pythagorean relationship, c2=a2+b2c^2=a^2+b^2, we can solve for bb:

b=c2a2=3222=94b=\sqrt{c^2-a^2}=\sqrt{3^2-2^2}=\sqrt{9-4}

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

So, the equation of the hyperbola is:

(y0)222(x(5))2(5)2=1\dfrac{(y-0)^2}{2^2}-\dfrac{(x-(-5))^2}{(\sqrt{5})^2}=1

y24(x+5)25=1\color{#c34632}\dfrac{y^2}{4}-\dfrac{(x+5)^2}{5}=1

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