4x^2+9y^2+8x-54y+49=0
We are asked to write this equation in sstandard form. Let's first remember how the standard form of an ellipse is written.
(x- h)^2/a^2+(y- k)^2/b^2=1
Now we will rewrite our equation by completing the perfect square trinomials.
Great! Let's now compare our obtained equation with the general form to determine the type of ellipse that our equation represents.
(x - h)^2/a^2+ & (y- k)^2/b^2 = 1
⇓
(x-( -1))^2/3^2+ & (y- 3)^2/2^2 = 1
As we can see, the binomials containing the variables are both raised to the power of 2 and are both positive. Moreover, the denominator of the binomial containing the x-variable is greater than the denominator of the binomial that contains the y-variable.
a> b>0 ⇒ 3> 2>0
Therefore, our equation matches the format of a horizontal ellipse.
b
We are asked to find the foci and vertices of the given ellipse. To do so, we will start by recalling the main characteristics of a horizontal ellipse.
Horizontal Ellipse
Standard-Form Equation
(x- h)^2/a^2+(y- k)^2/b^2=1
Center
( h, k)
Vertices
( h± a, k)
Co-vertices
( h, k± b)
Foci
( h± c, k)
a,b,c relationship, a>b>0
c^2= a^2- b^2
Now, let's consider our equation in standard form one more time.
(x-( -1))^2/3^2+ & (y- 3)^2/2^2 = 1
We see that a= 3, b= 2, h= - 1, and k= 3. The only value we do not know is c. Let's find it!
Note that we only took the principal root, because to find the foci we will add and subtract the value of c. Therefore, its sign is irrelevant. We can now write the desired information.
