Given the ellipse equation x216+y24=1\frac{x^2}{16}+\frac{y^2}{4}=116x2+4y2=1, determine the magnitude of the semi-major axis (a) and the semi-minor axis (b).

a = 4 a=4 a = 4 , b = 2 b=2 b = 2

Given the ellipse equation x 2 16 + y 2 4 = 1 \frac{x^2}{16}+\frac{y^2}{4}=1 16 x 2 + 4 y 2 = 1 , determine the magnitude of the semi-major axis (a) and the semi-minor axis (b).

Alright. So let's see how we can solve this problem. So in this problem, we are given the equation x squared over 16 plus y squared over four is equal to one, and we're asked to determine the magnitude of the semi major axis a and the semi minor axis b. Okay. So to solve this problem, what we should first do is take a look at our equation and figure out which one of these numbers, either 16 or four, is associated with the semi major axis. In this case, it's gonna go to the larger number, which is 16. So you can see that 16 is associated with a. And specifically 16 would be a squared meaning that four would have to be the semi minor axis squared which is b. So now that we have this we can recognize that a squared is equal to 16 and that b squared is equal to four. And all we really have to do is solve either one of these equations, or I should say both of them. So we'll start with this one, a squared is equal to 16. I'll take the square root on both sides. Then I'll get the square to cancel, giving me that a is equal to the square root of 16, which is four. We don't need to do a plus or minus because we only care about the magnitude in this problem. So a is equal to four. Now to solve for b, we'll take the square root on both sides of this equation, canceling the square, giving us that b is equal to the square root of four, which is two. So we have that a is equal to four and b is equal to two. And these are the semi major axis and the semi minor axis respectively. Hope you found this helpful.

Given the ellipse equation x 2 16 + y 2 4 = 1 \frac{x^2}{16}+\frac{y^2}{4}=1 16 x 2 + 4 y 2 = 1 , determine the magnitude of the semi-major axis (a) and the semi-minor axis (b).

a = 4 a=4 a = 4 , b = 2 b=2 b = 2

a = 16 a=16 a = 16 , b = 4 b=4 b = 4

a = 4 a=4 a = 4 , b = 16 b=16 b = 16

a = 2 a=2 a = 2 , b = 4 b=4 b = 4

16. Parametric Equations & Polar Coordinates

Conic Sections

Calculus

16. Parametric Equations & Polar Coordinates

Conic Sections

Struggling with Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Given the ellipse equation $\frac{x^2}{16}+\frac{y^2}{4}=1$ , determine the magnitude of the semi-major axis (a) and the semi-minor axis (b).

$a=16$ , $b=4$

$a=4$ , $b=16$

$a=4$ , $b=2$

$a=2$ , $b=4$

Verified step by step guidance

Step 1: Recall the standard form of an ellipse equation: $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, where $a$ represents the semi-major axis and $b$ represents the semi-minor axis.

Step 2: Compare the given equation $ \frac{x^2}{16} + \frac{y^2}{4} = 1 $ with the standard form. Here, $a^2 = 16$ and $b^2 = 4$.

Step 3: Solve for $a$ and $b$ by taking the square root of $a^2$ and $b^2$. Specifically, $a = \sqrt{16}$ and $b = \sqrt{4}$.

Step 4: Determine which axis corresponds to the semi-major axis and which corresponds to the semi-minor axis. The semi-major axis is the larger value, and the semi-minor axis is the smaller value.

Step 5: Conclude that the semi-major axis $a$ and semi-minor axis $b$ are determined based on the values calculated in Step 3.

3:8m

Watch next

Master Geometries from Conic Sections with a bite sized video explanation from Patrick

Start learning

Conic Sections practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice