Graph each ellipse and find its foci :
a) 3x^2+y^2=12
b) 5x^2+9y^2=45
We cannot graph on here posts.
Write the equation of the circle in standard form. Find the center, radius, intercepts, and graph the circle.
x2+y2+16x−18y+145=25
Standard form
To graph each ellipse and find its foci, we need to rewrite the equations of the ellipses in standard form. The standard form of an ellipse with its major axis aligned with the x-axis is given by:
(x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h,k) is the center of the ellipse, a is the length of the semi-major axis, and b is the length of the semi-minor axis.
a) To rewrite the equation 3x^2 + y^2 = 12 in standard form, we divide each term by 12 to get:
x^2/4 + y^2/12 = 1.
Comparing the equation with the standard form, we can see that the center of the ellipse is (0, 0), the semi-major axis a is 2, and the semi-minor axis b is 2√3. Therefore, the length of the major axis is 2a = 4, and the length of the minor axis is 2b = 4√3.
To find the foci, we can use the formula c = √(a^2 - b^2), where c is the distance from the center to each focus.
For this ellipse, since a^2 = 4^2 = 16 and b^2 = (2√3)^2 = 12, we have:
c = √(16 - 12) = √4 = 2.
The foci are located at (±c, 0), so the foci for this ellipse are (2, 0) and (-2, 0).
b) Similarly, let's rewrite the equation 5x^2 + 9y^2 = 45 in standard form. Dividing by 45, we have:
x^2/9 + y^2/5 = 1.
Comparing the equation with the standard form, we see that the center of the ellipse is (0,0), the semi-major axis a is 3, and the semi-minor axis b is √5. Therefore, the length of the major axis is 2a = 6, and the length of the minor axis is 2b = 2√5.
To find the foci, we use the formula c = √(a^2 - b^2). With a^2 = 3^2 = 9 and b^2 = (√5)^2 = 5, we have:
c = √(9 - 5) = √4 = 2.
The foci of this ellipse are located at (±c, 0), so the foci for this ellipse are (2, 0) and (-2, 0).
To graph the ellipses, we plot the center, major and minor axes, and then sketch the ellipse using these measurements.