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).

right. 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 4 is equal to 1, 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 4, 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 the 16 is associated with a, and specifically 16 would be a squared, meaning that 4 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 4. 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 it'll get the square to cancel giving me that a is equal to the square root of 16 which is 4 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 4. 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 4, which is 2. So we have that a is equal to 4 and b is equal to 2. 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

19. Conic Sections

Ellipses: Standard Form

Precalculus

19. Conic Sections

Ellipses: Standard Form

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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

Start by identifying the standard form of the ellipse equation, which is $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $. This equation helps us determine the lengths of the semi-major and semi-minor axes.

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 $.

Calculate the semi-major axis $ a $ by taking the square root of $ a^2 $. Thus, $ a = \sqrt{16} = 4 $.

Calculate the semi-minor axis $ b $ by taking the square root of $ b^2 $. Thus, $ b = \sqrt{4} = 2 $.

Conclude that the semi-major axis is $ a = 4 $ and the semi-minor axis is $ b = 2 $. This matches the correct answer provided.

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