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QuestionConvert the hyperbola equation 16x2−9y2+32x−18y−137=016x^{2}-9y^{2}+32x-18y-137=0 to standard form and graph it.

Studdy Solution

STEP 1

Assumptions1. The given equation is of a hyperbola. . The general form of a hyperbola equation is Ax+By+Cx+Dy+=0Ax^{} + By^{} + Cx + Dy + =0.
3. The standard form of a hyperbola equation is (x−h)a−(y−k)b=1\frac{(x-h)^{}}{a^{}} - \frac{(y-k)^{}}{b^{}} =1 if the hyperbola opens horizontally or (y−k)a−(x−h)b=1\frac{(y-k)^{}}{a^{}} - \frac{(x-h)^{}}{b^{}} =1 if the hyperbola opens vertically, where (h,k) is the center of the hyperbola.

STEP 2

First, we need to group the x-terms and y-terms together in the given equation.
16x2+32x−9y2−18y−137=016x^{2}+32x-9y^{2}-18y-137=0

STEP 3

Next, we complete the square for the x-terms and y-terms. To do this, we take the coefficient of the x-term (or y-term), divide by2, and square it. We then add and subtract this value to the equation.
16(x2+2x+1)−1−9(y2−2y+1)+1−137=016(x^{2}+2x+1)-1-9(y^{2}-2y+1)+1-137=0

STEP 4

We simplify the equation by factoring the perfect square trinomials and combining like terms.
16(x+1)2−9(y−1)2−137=016(x+1)^{2}-9(y-1)^{2}-137=0

STEP 5

We isolate the hyperbola equation by moving the constant term to the other side of the equation.
16(x+1)2−9(y−1)2=13716(x+1)^{2}-9(y-1)^{2}=137

STEP 6

We divide through by the constant on the right side of the equation to get the equation in standard form.
(x+1)213716−(y−1)21379=1\frac{(x+1)^{2}}{\frac{137}{16}}-\frac{(y-1)^{2}}{\frac{137}{9}}=1

STEP 7

We simplify the fractions in the equation.
(x+1)2.5625−(y−1)215.2222=1\frac{(x+1)^{2}}{.5625}-\frac{(y-1)^{2}}{15.2222}=1

STEP 8

Now that we have the equation in standard form, we can identify the center, a, and b of the hyperbola. The center is (-1,1), a is 8.5625\sqrt{8.5625}, and b is 15.2222\sqrt{15.2222}.

STEP 9

To graph the hyperbola, we plot the center at (-,). Then, we move right and left from the center by a units and up and down from the center by b units to get the vertices and co-vertices of the hyperbola. We can then sketch the hyperbola by drawing curves that pass through these points.

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