I am having trouble with several problems:
1) Find the coordinates of the center, foci, and the vertices, and the equations of the asymptotes of the graph of the equation (x+1)^2/2 - y^2/8 =1
2) Write the standard form of the equation of the circle that is tangent to x= -2 and has its center at (2,-4).
Any help with these problems would be great!!
Sure, I can help you with those problems! Let's start with Problem 1.
1) To find the coordinates of the center, foci, and the vertices, and the equations of the asymptotes of the graph of the equation (x+1)^2/2 - y^2/8 = 1:
Step 1: Rewrite the given equation in standard form:
(x+1)^2/2 - y^2/8 = 1
Multiply both sides of the equation by 8 to clear the denominators:
4(x+1)^2 - y^2 = 8
Step 2: Group the x-terms and the y-terms together:
4(x+1)^2 = y^2 + 8
Step 3: Divide both sides of the equation by 4:
(x+1)^2 = (y^2 + 8)/4
Step 4: Rearrange the equation to get the standard form:
(x+1)^2/4 - y^2/8 = 1
Now we can identify the center, foci, vertices, and asymptotes.
Center: The given equation is in the form (x-h)^2/a^2 - (y-k)^2/b^2 = 1, where (h,k) represents the center of the hyperbola. Comparing this with our equation, we can see that the center is (-1, 0).
Vertices: The vertices of a hyperbola are located at a distance of a units from the center along the transverse axis. In this case, since a^2 = 4, a = 2. Therefore, the vertices are (-1+2, 0) = (1, 0) and (-1-2, 0) = (-3, 0).
Foci: The foci of a hyperbola are located at a distance of c units from the center along the transverse axis, where c is defined as c^2 = a^2 + b^2. In our case, since a^2 = 4, we have c^2 = 4 + 8 = 12. Taking the square root of 12, we get c = √(12) = 2√3. Therefore, the foci are (-1+2√3, 0) and (-1-2√3, 0).
Asymptotes: The equations of the asymptotes of a hyperbola can be found using the formula y = ±(b/a)(x - h) + k, where (h,k) represents the center of the hyperbola. In our case, the equation becomes y = ±(√2/2)(x + 1) + 0 = ±√2/2(x+1). Therefore, the equations of the asymptotes are y = √2/2(x+1) and y = -√2/2(x+1).
Now, let's move on to Problem 2.
2) To write the standard form of the equation of the circle that is tangent to x = -2 and has its center at (2,-4):
A circle is defined by the equation (x-h)^2 + (y-k)^2 = r^2, where (h,k) represents the center of the circle, and r represents the radius.
Step 1: Locate the center of the circle. In this case, we are given that the center is (2,-4).
Step 2: Determine the radius of the circle. Since the circle is tangent to the line x = -2, the distance between the center of the circle and the tangent line will be equal to the radius of the circle. In this case, the distance from the center (2,-4) to the line x = -2 is 2 units. Therefore, the radius of the circle is 2.
Step 3: Substitute the values of the center and radius into the general equation of a circle:
(x-h)^2 + (y-k)^2 = r^2
(x-2)^2 + (y+4)^2 = 2^2
Thus, the standard form of the equation of the circle that is tangent to x = -2 and has its center at (2,-4) is (x-2)^2 + (y+4)^2 = 4.
I hope this helps! Let me know if you have any further questions.