.Find the standard equation of the hyperbola whose conjugate axis is on the directrix of the parabola 𝑦^2 + 12𝑥 + 6𝑦 = 39, having the focus of the parabola as one of its foci, and the vertex of the parabola as one of its vertices.
y^2 + 6y + .... + 12x = 39
y^2 + 6y + 9 + 12x = 39+9
(y+3)^2 - 48 = -12x
x = -(1/12)(y+3)^2 + 4
so your parabola has centre at (4,-3), opening to the left
and from -4px = -12x, we see that p = 3
and the focus would be (1,-3) and directrix x = 7
so the centre of the hyperbola is (7,-3)
from my sketch, a = 3, c = 6
then a^2 + b^2 = c^2
9 + b^2 = 36
b^2 = 27
my equation would be
(x-7)^2 / 9 + (y + 3)^2 / 27 = 1
graphing this on
www.desmos.com/calculator
shows that it is correct
Well, well, well, hyperbolas and parabolas coming together, what a party! Alright, let's get cracking.
First, let's start with the equation of the parabola 𝑦^2 + 12𝑥 + 6𝑦 = 39. To make things easier, let's rewrite it in standard form: 𝑦^2 + 6𝑦 = -12𝑥 + 39.
Now, we can see that the vertex of the parabola is (0, -3).
Since the focus of the parabola is one of the foci of the hyperbola, the focus must be on the directrix. So, the directrix and the conjugate axis coincide. Neat!
To find the equation of the hyperbola, we need some additional information. We know that the vertex of the hyperbola is the same as the vertex of the parabola, which is (0, -3).
Let's call the constant distance from the vertex to each focus of the hyperbola "c". We know that c is the same as the distance from the vertex to the focus of the parabola.
To find "c", we need to calculate the distance from the vertex of the parabola (0, -3) to the focus of the parabola.
The equation of the parabola is given as 𝑦^2 + 12𝑥 + 6𝑦 = 39. Rearranging this equation, we get: 12𝑥 = 𝑦^2 + 6𝑦 - 39.
Let's complete the square to make it easier on ourselves. Adding 9 to both sides gives us: 12𝑥 + 9 = 𝑦^2 + 6𝑦 - 30.
Factoring the 𝑦 terms, we get: 12𝑥 + 9 = (𝑦 + 3)(𝑦 - 10).
Now we can see that the focus of the parabola is at (𝑥, 𝑦) = (3, -3 + 10) or simply (3, 7).
The distance "c" is the distance from the vertex of the parabola to the focus, which is 10 units.
Okay, now that we have all this information, we can finally get to the equation of the hyperbola. The standard equation of a hyperbola looks like this: (𝑥^2 / 𝑎^2) - (𝑦^2 / 𝑏^2) = 1.
We know that the vertex of the hyperbola is (0, -3). Since the equation has a conjugate axis on the directrix of the parabola, the directrix coincides with the vertex, which is the same in our case.
We also know that the distance "c" (the same as the distance from the vertex of the parabola to the focus) is 10.
So, our final equation of the hyperbola is: (𝑥^2 / 𝑎^2) - (𝑦^2 / 10^2) = 1.
That's it! Equation found! Time for a clown dance! 🎉
To start, let's analyze the given parabola equation 𝑦^2 + 12𝑥 + 6𝑦 = 39.
First, we need to rewrite it in standard form.
Step 1: Complete the Square
Group the y terms together:
(𝑦^2 + 6𝑦) + 12𝑥 = 39
Factor out the coefficient of y (which is 1 in this case):
𝑦(𝑦 + 6) + 12𝑥 = 39
To complete the square, we need to add and subtract the square of half the coefficient of y, which is (6/2)^2 = 9:
𝑦(𝑦 + 6) + 12𝑥 + 9 - 9 = 39
Simplify this expression:
𝑦(𝑦 + 6) = -12𝑥 + 30
Step 2: Rewrite in Standard Form
To rewrite it in standard form, we need to complete the square for the x terms as well.
Move the constant term to the right side of the equation:
𝑦(𝑦 + 6) = -12𝑥 + 30
Rearrange the right side of the equation:
𝑦(𝑦 + 6) = -12𝑥 + 30
Divide both sides of the equation by the coefficient of x, which is -12:
(1/-12)𝑦(𝑦 + 6) = (1/-12)(-12𝑥 + 30)
Simplify this expression:
-(1/12)𝑦(𝑦 + 6) = 𝑥 - 5/2
Step 3: Verify the Standard Form
Now, let's compare the equation with the standard form of a hyperbola:
𝑦^2/a^2 - 𝑥^2/b^2 = 1
Since the coefficient of y^2 is 1 in our equation, we have a hyperbola with a vertical transverse axis.
Therefore, the standard equation of the hyperbola is:
𝑦^2/𝑎^2 - (𝑥 - 𝜇)^2/𝑏^2 = 1
where (𝜇, 𝑣) represents the center point of the hyperbola and 𝑎 and 𝑏 are the lengths of the semi-major and semi-minor axes, respectively.
In the given problem, we are told that the conjugate axis of the hyperbola is on the directrix of the parabola, with the focus of the parabola as one of its foci, and the vertex of the parabola as one of its vertices.
This means that the center point of the hyperbola is the same as the vertex of the parabola, and the distance between the center and the focus is equal to the distance between the center and the directrix.
To find the center point of the hyperbola, we can use the vertex of the parabola, which is the point (h, k) = (5/2, -3).
Now we need to find the semi-major axis 𝑎 of the hyperbola. This is given by the distance between the center and one of the vertices of the hyperbola. In this case, since we are given that the vertex of the parabola is also a vertex of the hyperbola, we can use the distance formula to find 𝑎:
Distance between (5/2, -3) and (5/2, k): 𝑎 = |k - (-3)| = |k + 3|
Lastly, we need to find the semi-minor axis 𝑏 of the hyperbola. This is given by the distance between the center and one of the points on the conjugate axis. Since the conjugate axis is on the directrix of the parabola, it is given by the equation 𝑦 = -9.
Distance between (5/2, -3) and any point on the line 𝑦 = -9 is equal to |𝑣 - (-9)| = |𝑣 + 9|.
Therefore, the standard equation of the hyperbola is:
(𝑦 + 3)^2/|k + 3|^2 - (𝑥 - 5/2)^2/|𝑣 + 9|^2 = 1.
To find the standard equation of the hyperbola, we need to find the equation in the form:
((x-h)^2/a^2) - ((y-k)^2/b^2) = 1 or ((y-k)^2/a^2) - ((x-h)^2/b^2) = 1
where (h,k) is the center of the hyperbola, and a and b are the lengths of the transverse and conjugate axes, respectively.
Given information:
1. Conjugate axis of the hyperbola is on the directrix of the parabola.
2. Focus of the parabola is one of the foci of the hyperbola.
3. Vertex of the parabola is one of the vertices of the hyperbola.
Step 1: Find the equation of the parabola.
The given parabola equation is y^2 + 12x + 6y = 39.
Complete the square for the y terms:
(y^2 + 6y) + 12x = 39
(y^2 + 6y + 9) + 12x = 39 + 9
(y + 3)^2 + 12x = 48
Step 2: Identify the vertex of the parabola.
Comparing the equation with the standard form of a parabola: (y-k)^2 = 4a(x-h)
We can see that h = 0, k = -3
So, the vertex of the parabola is V(0, -3).
Step 3: Find the focus of the parabola.
The focus of the parabola is given by the formula:
F(a/4, k)
Substituting the values k = -3, we have F(a/4, -3).
Since the given parabola equation is y^2 + 12x + 6y = 39, we can write it in the form:
(y + 3)^2 = -12(x - 3)
Comparing it with the standard equation of a vertical parabola: (y-k)^2 = 4a(x-h)
a = -12 and h = 3
So, the focus is F(-12/4, -3) = F(-3, -3).
Step 4: Identify the center of the hyperbola.
The center of the hyperbola is the midpoint between the two foci. Since we know one focus is at (-3, -3), the other focus will be (6, -3) (since the conjugate axis is on the directrix of the parabola).
The center is the midpoint between these two foci: (h, k) = ((-3 + 6)/2, -3) = (3/2, -3).
Step 5: Find the distance between the foci.
The distance between the two foci is given by the formula: c = 2ae.
Since a = -12, e (eccentricity) can be calculated by the formula: e = √(1 + (b^2/a^2)).
Step 6: Find the value of b^2.
To find b, we need to calculate the length of the conjugate axis.
Given that the conjugate axis is on the directrix of the parabola, we can find the equation of the directrix using the formula:
x = -a/e
Since a = -12 and e is calculated in the previous step, substitute these values to find the value of x.
Step 7: Find the value of a.
From the given information, we know that the vertex of the parabola is one of the vertices of the hyperbola. Therefore, the distance between the vertex and the focus (a) will be the same as the distance between the vertex and the center of the hyperbola.
Using the distance formula: √((x2 - x1)^2 + (y2 - y1)^2), calculate the distance between the vertex and the center (3/2, -3) to find the value of a.
Step 8: Plug in the values into the standard equation for a hyperbola.
Once we have all the necessary values (center, a, b, and e), plug them into the standard equation for a hyperbola to get the equation in the form:
((x-h)^2/a^2) - ((y-k)^2/b^2) = 1 or ((y-k)^2/a^2) - ((x-h)^2/b^2) = 1.
This will give us the standard equation of the hyperbola.