Identifying Conics. Put equation in standard form and graph.
9x^2-4y^2-90x+189=0
9(x^2-10x+25-25)-4y^2+189=0
9(x-5)^2-225-4y^2+189=0
9(x-5)^2-4y^2=36
(x-5)^2/2^2 - y^2/3^2=1 hyperbola
To identify and graph the conic given by the equation 9x^2 - 4y^2 - 90x + 189 = 0, we will need to put it in standard form. The standard form for each type of conic is as follows:
1. For a circle: (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r is the radius.
2. For an ellipse: (x - h)^2 / a^2 + (y - k)^2 / b^2 = 1, where (h, k) represents the center of the ellipse and a and b are the semi-major and semi-minor axes, respectively.
3. For a hyperbola: (x - h)^2 / a^2 - (y - k)^2 / b^2 = 1 or (y - k)^2 / b^2 - (x - h)^2 / a^2 = 1, where (h, k) represents the center of the hyperbola, a and b are the semi-transverse and semi-conjugate axes, respectively.
4. For a parabola: y = a(x - h)^2 + k or x = a(y - k)^2 + h, where (h, k) represents the vertex of the parabola and a is a constant.
Let's now put the given equation into standard form:
9x^2 - 4y^2 - 90x + 189 = 0
First, let's group the x-terms and the y-terms separately:
(9x^2 - 90x) - 4y^2 + 189 = 0
Next, we can factor out the common coefficients:
9(x^2 - 10x) - 4y^2 + 189 = 0
Now, we complete the square for both the x-terms and the y-term:
To complete the square for the x-terms:
1. Divide the coefficient of x (which is -10) by 2 and square it: (-10/2)^2 = (-5)^2 = 25
2. Add the obtained value (25) inside the parentheses and subtract it from the equation:
9(x^2 - 10x + 25) - 4y^2 + 189 - (9 * 25) = 0
Simplifying further:
9(x - 5)^2 - 4y^2 + 189 - 225 = 0
9(x - 5)^2 - 4y^2 - 36 = 0
Now, we divide the entire equation by the constant term to simplify:
9(x - 5)^2 / 36 - 4y^2 / 36 - 1 = 0
(x - 5)^2 / 4 - y^2 / 9 = 1
Now that we have put the equation in standard form, we can determine that it represents a hyperbola. By comparing the equation to the standard form equations given above, we see that the terms involving x have a positive coefficient and the terms involving y have a negative coefficient which is characteristic of a hyperbola.
To graph the hyperbola, we can use the information from the standard form equation. The center of the hyperbola is at the point (5, 0) (opposite sign of the constant terms inside the parentheses). The semi-transverse axis is equal to 2 (the square root of the coefficient of (x - 5)^2) and the semi-conjugate axis is equal to 3 (the square root of the coefficient of y^2).
With these values, you can sketch the hyperbola on a graph by plotting the center, then marking points on either side along the x-axis (±2 units) and along the y-axis (±3 units), and then drawing the curve through those points.
To identify the conic and put the equation in standard form, we will need to complete the square for both the x and y variables. Let's start with the x terms:
9x^2 - 90x - 4y^2 + 189 = 0
First, let's move the constant term to the right side:
9x^2 - 90x - 4y^2 = -189
Next, let's divide each term by the coefficient of x^2 and y^2 to make the leading coefficients 1:
x^2 - 10x/9 - (4y^2)/9 = -189/9
Now let's complete the square for the x terms:
(x^2 - 10x/9 + (10/18)^2) - (10/18)^2 - (4y^2)/9 = -189/9
(x^2 - 10x/9 + 25/81) - (100/81) - (4y^2)/9 = -189/9
(x - 5/9)^2 - 100/81 - (4y^2)/9 = -21
Now let's complete the square for the y terms:
(x - 5/9)^2 - (4y^2)/9 = -21 + 100/81
(x - 5/9)^2 - (4y^2)/9 = 899/81
To put the equation in standard form, we will divide each term by the constant term on the right side:
[(x - 5/9)^2/(899/81)] - [(4y^2)/9]/(899/81) = 1
Simplifying further:
[(x - 5/9)^2/(899/81)] - [(4y^2)/9]/(899/81) = 1
Now that the equation is in standard form, we can identify the conic. Since the coefficient of the x term squared is positive and the coefficient of the y term squared is negative, the graph represents a hyperbola.
To graph the hyperbola, we can determine the center, vertices, foci, asymptotes, and the shape.
The center of the hyperbola is at the point (h, k), which in our case is (5/9, 0).
To find the vertices, we can solve for y = 0:
[(x - 5/9)^2/(899/81)] - [(4*0^2)/9]/(899/81) = 1
(x - 5/9)^2/(899/81) = 1
(x - 5/9)^2 = (899/81)
Taking the square root of both sides:
x - 5/9 = ± √(899/81)
x = 5/9 ± (√(899)/9)
So the vertices are V1: (5/9 + (√(899)/9), 0) and V2: (5/9 - (√(899)/9), 0).
To find the foci, we can use the formula:
c = √(a^2 + b^2)
Where c is the distance from the center to the foci, and a and b are the semi-major and semi-minor axes.
In our case, a^2 = 899/81 and b^2 = (899/81) + (4/9), so we have:
c = √(899/81 + (899/81) + (4/9))
c = √(899/81 + 899/81 + 8/9)
c = √((899 + 899 + 8) / 81)
c = √(1806/81)
c = √(22)
So the foci are F1: (5/9 + √(22)/9, 0) and F2: (5/9 - √(22)/9, 0).
To find the asymptotes, we can use the formula:
y = ± (b/a) * (x - h) + k
Where (h, k) is the center of the hyperbola, and a and b are the semi-major and semi-minor axes.
In our case, the equation becomes:
y = ± (√(899)/√(899/81)) * (x - 5/9)
Simplifying:
y = ± (√(899)/9) * (x - 5/9)
So the asymptotes are the lines with equations:
y = (√(899)/9) * (x - 5/9) and y = - (√(899)/9) * (x - 5/9).
Now we have all the information needed to graph the hyperbola.