How do I write the equation of the conic section given this info?: An ellipse with the vertices #(0,-5)# and #(0,5)# and a minor axis of length 8

Answer 1

#x^2/16+y^2/25=1#

As the vertices are #(0,-5)# and #(0,5)# and length of its minor axis is #8#, the vertices must represent ends of major axis and hence length of its major axis is #10#.

Further center of ellipse is #(0,0)# and vertices are on #y#-axis and therefore major axis is parallel to #y#-axis and minor axis is parallel to #x#-axis.

Hence the equation of such an ellipse is

#x^2/(8/2)^2+y^2/(10/2)^2=1#

or #x^2/16+y^2/25=1#

or #25x^2+16y^2-400=0#

graph{(25x^2+16y^2-400)(x^2+y^2-10y+24.95)(x^2+y^2+10y+24.95)=0 [-12, 12, -6, 6]}

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email

Answer 2

Sadie Ackerman

To write the equation of the ellipse, you need to use the standard form of the equation for an ellipse centered at the origin:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis. Since the center of the ellipse is at the origin and the minor axis has a length of 8, $b = 4$ . The length of the semi-major axis is the distance from the origin to one of the vertices, which is 5 units. So, $a = 5$ .

Therefore, the equation of the ellipse is:

$\frac{x^2}{5^2} + \frac{y^2}{4^2} = 1$

$\frac{x^2}{25} + \frac{y^2}{16} = 1$

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.