Question

how do you graph a conic sectionn thaat looks like this
(x-5)^2/4+(y+3)^2/16=1

Answer 1

Write it in the standard form:
\(\Large \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\)

Answer 2

\[\frac{ (x-5)^2 }{ 4 }+\frac{ (y+3)^2 }{ 16}=1\]

Answer 3

Here, the denominator under y^2 is greater than the denominator under the x^2 term and therefore, the 'a' should go with 'y'. The major axis is along the y-axis here.

\(\huge \frac{ (x-5)^2 }{ 2^2 }+\frac{ (y- (-3))^2 }{ 4^2}=1 \)

Answer 4

Compare it to the standard form (but with a and b switched for reasons explained above) and identify all the variables.
a = 4, b = 2, h = 5, k = -3

The center of the ellipse is at (h,k) which is (5, -3)
The length of the semi-major axis = a = 4 and it is along the y-axis
The length of the semi-minor axis = b = 2 and it is along the x-axis.

Time to draw the ellipse.

Answer 5

The center of the ellipse is at (5,-3)
The line segments marked 4 are the semi-major axis which is parallel to the y-axis.
The line segments marked 2 are the semi-minor axis which is parallel to the x-axis.