Find the foci of the graph of x^2/4+y^2/9=1.?

Question

Can someone help me with this problem? Please explain your steps

Answer 1

x ²/4 + y ² /9 = 1

The equation is y ²/9 + x ²/4 = 1.

The standard form of the equation of an ellipse center (h, k) with major and minor axes of lengths 2a and 2b (where 0 < b < a) is (x - h )²/a ² + (y - k )²/b ² = 1 or (x - h )²/b ² + (y - k )²/a ² = 1.

The vertices and foci lie on the major axis, a and c units, respectively, from the center (h, k )  and the relation between a, b and c is c ² = a ² - b ².

Because the denominator of the y ² - term (9) is larger than the denominator of the x ² - term (4), the major axis is vertical.

Compare the equation (x - 0)²/9 + (y - 0)²/4 = 1 with (x - h )²/b ² + (y - k )²/a ² = 1.

b ² = 4, a ² = 9, h = 0 and k = 0.

b = 2 and a = 3.

To find the value of c, substitute the value of a² = 9 and b² = 4 in c² = a² - b².

c² = 9 - 4

c² = 5

c =√5

Center = (h, k ) = (0, 0).

Foci = (h, k ± c) = (0, +√5) = (0, +√5) and (0, -√5)

Foci = (0, +√5) and (0, -√5) or (0 , 2.23) and (0, -2.23)

Semi major axis a = 3 and minor axis b = 2

Vertices (h ,k ± a) = (0,3) and (0,-3).

1.Draw a coordinate plane.

2.Plot the coordinate points are vertices , foci and center.

3. use the lengths of minor axis(b ) and major axis(a ) lengths of the ellipse.

4.  connect the ellipse.