Math  /  Geometry

QuestionFind an equation of the hyperbola having foci at (2,152)(2,-1-\sqrt{52}) and (2,1+52)(2,-1+\sqrt{52}) and asymptotes at y=23x73y=\frac{2}{3} x-\frac{7}{3} and y=23x+13y=-\frac{2}{3} x+\frac{1}{3}.

Studdy Solution

STEP 1

What is this asking? We need to find the equation of a hyperbola, given its foci and asymptotes. Watch out! Don't mix up the formulas for ellipses and hyperbolas!
Also, remember the center isn't always at (0,0)(0,0).

STEP 2

1. Find the center.
2. Determine aa and bb.
3. Write the equation.

STEP 3

The center of the hyperbola is the midpoint of the foci.
Let's **calculate** it!
The x-coordinate is 2+22=42=2\frac{2+2}{2} = \frac{4}{2} = 2.
The y-coordinate is (152)+(1+52)2=22=1\frac{(-1-\sqrt{52})+(-1+\sqrt{52})}{2} = \frac{-2}{2} = -1.
So, the **center** is at (2,1)(2,-1).

STEP 4

The slopes of the asymptotes are 23\frac{2}{3} and 23-\frac{2}{3}.
Since the foci have the same x-coordinate, the hyperbola opens vertically.
For a vertical hyperbola, the slopes of the asymptotes are given by ±ab\pm\frac{a}{b}.
This tells us that ab=23\frac{a}{b} = \frac{2}{3}.

STEP 5

The distance from the center (2,1)(2,-1) to a focus (2,1+52)(2,-1+\sqrt{52}) is c=(22)2+(1+52(1))2=02+(52)2=52c = \sqrt{(2-2)^2 + (-1+\sqrt{52}-(-1))^2} = \sqrt{0^2 + (\sqrt{52})^2} = \sqrt{52}.
Remember, for a hyperbola, we have c2=a2+b2c^2 = a^2 + b^2.

STEP 6

We know ab=23\frac{a}{b} = \frac{2}{3}, so b=3a2b = \frac{3a}{2}.
Substituting this into c2=a2+b2c^2 = a^2 + b^2, we get 52=a2+(3a2)2=a2+9a24=13a2452 = a^2 + (\frac{3a}{2})^2 = a^2 + \frac{9a^2}{4} = \frac{13a^2}{4}.

STEP 7

Multiplying both sides by 413\frac{4}{13}, we get a2=45213=413413=16a^2 = \frac{4 \cdot 52}{13} = \frac{4 \cdot 13 \cdot 4}{13} = 16, so a=4a=4.
Then, since b=3a2b = \frac{3a}{2}, we have b=342=6b = \frac{3 \cdot 4}{2} = 6.

STEP 8

Since the hyperbola is centered at (2,1)(2,-1) and opens vertically, its equation is (y(1))2a2(x2)2b2=1\frac{(y-(-1))^2}{a^2} - \frac{(x-2)^2}{b^2} = 1.
Plugging in a=4a=4 and b=6b=6, we get (y+1)216(x2)236=1\frac{(y+1)^2}{16} - \frac{(x-2)^2}{36} = 1.

STEP 9

The equation of the hyperbola is (y+1)216(x2)236=1\frac{(y+1)^2}{16} - \frac{(x-2)^2}{36} = 1.

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