10.2
can someone show me this graph(x+2)^2/4+(y+2)^2/15=1
wolframalpha . com can. Just enter
plot (x+2)^2/4+(y+2)^2/15=1
Hyperbola with center at (-2,-2), eccentricity = √15/2
oops
ellipse, semi-major axes 2 and √15
major axis parallel to y-axis
thank you
To graph the equation `(x+2)^2/4+(y+2)^2/15=1`, we can follow these steps:
Step 1: Identify the type of conic section
The given equation is in the form `(x-h)^2/a^2 + (y-k)^2/b^2 = 1`, which represents an ellipse. The center of the ellipse is at the point (-2, -2).
Step 2: Determine the major and minor axes
The lengths of the major and minor axes can be found by taking the square root of the denominators of the x and y terms in the equation. In this case, `a = 2` and `b = sqrt(15)`.
Step 3: Plot the center and major and minor axes
Start by plotting the center of the ellipse at (-2, -2) on a coordinate plane. Then, draw the major and minor axes, which are horizontal and vertical lines passing through the center. The major axis runs horizontally with a length of 2 units on both sides of the center, and the minor axis runs vertically with a length of `sqrt(15)` units above and below the center.
Step 4: Draw the ellipse
Using the plotted center and the lengths of the major and minor axes, sketch an elliptical shape that fits within the major and minor axes.
The resulting graph will be an ellipse with center (-2, -2), major axis length of 4 units, and minor axis length of `2*sqrt(15)` units.