I am not too sure about this problem. Can you please help me?
Graph the ellipse and locate the foci.
9x^2=144-16y^2
I got (- squareroot 7, 0) and (squareroot 7,0), but I was looking at one problem in one book similar to it, and it had the answer as (0, - squareroot 7) and (0, squareroot 7).
You are correct, I got the same foci
The one you looked at in your text probably had the y-axis as its major axis, that is, the focal points were on the y axis.
Was the denominator of the x^2 smaller than the denominator of the y^2 term?
If so, that was the case.
The one in book, is 7x^2=35-5y^2
Rewrite as
x^2/(144/9) + y^2/(144/16) = 1
(x/4)^2 + (y/3)^2 = 1
The semimajor and semiminor axis distances are 4 and 3, respectively. The major axis lies along the x axis. The distance from the center to the foci is sqrt (4^2 - 3^2) = sqrt 7
7x^2=35-5y^2 rewritten in standard form is
x^2/5 + y^2/7 = 1
sure enough the y term denominator is 7 which is larger than the 5 under the x^2.
Which means the y axis is the major axis like I said before and the focal points are on the y-axis
Certainly! I can help you understand how to graph the given ellipse and locate its foci.
To begin with, let's rewrite the given equation in standard form for an ellipse:
9x^2 + 16y^2 = 144
To find the center of the ellipse, we need to isolate the x and y terms. Dividing both sides of the equation by 144, we get:
x^2/16 + y^2/9 = 1
From this form, we can determine that the center of the ellipse is at (0,0), as the x and y terms are both squared and have coefficients less than 1.
Next, let's look at the coefficients of the x and y terms. Since the coefficient of x^2 is greater than the coefficient of y^2, we know that the major axis of the ellipse is along the x-axis.
The distance from the center of the ellipse to the vertex on the major axis is determined by the square root of the coefficient of x^2. In this case, it is √16 = 4.
Therefore, the vertices on the major axis are located at (±4,0).
To find the foci of the ellipse, we need to calculate the value of c using the formula:
c = √(a^2 - b^2)
Where a is the distance from the center to the vertex on the major axis and b is the distance from the center to the vertex on the minor axis.
In this case, a = 4 and we need to find b.
To determine b, we need to rearrange the standard form of the equation to solve for y:
y^2/9 = 1 - x^2/16
y^2 = 9 - (9x^2)/16
y = ±√(9 - (9x^2)/16)
From this equation, we can see that when x = 0, y = ±3, which gives us the distance from the center to the vertex on the minor axis.
Thus, b = 3.
Now, we can calculate c:
c = √(4^2 - 3^2) = √(16 - 9) = √7
Therefore, the foci of the ellipse are located at (0, ±√7).
Hence, your initial answer of (±√7, 0) is correct, and the alternative answer given in the book as (0, ±√7) is also correct. Both pairs of coordinates represent the foci of the given ellipse.
I hope this explanation clarifies the problem for you.