Identify the type of conic section whose equation is given. x²= 4y - 2y². Find the vertices and foci.

Solution:

To identify the conic section from its general equation,

The graph of Ax² + Cy² + Dx + Ey + F = 0 is one of the following --------- (1)

1) Circle: A = C

2) Parabola: AC = 0 (either A or C = 0 but not both)

3) Ellipse: AC>0, (A and C have like signs)

4) Hyperbola: AC<0, (A and C have unlike signs)

Given, the equation is x²= 4y - 2y²

Rewriting the equation, x² + 2y² - 4y = 0 --------------- (2)

Comparing (1) and (2),

A = 1, C = 2

AC = 2

So, AC > 0

Therefore, the given equation represents an ellipse.

The standard form of equation of vertical ellipse is \(\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1\) ------------ (3)

Where a² > b²

a is the length of semi major axis

b is the length of semi minor axis

(h, k) is the center

(h±a, k) are the vertices

(h±c, k) are the foci, where c² = a² - b²

Rewritten equation (2) into standard form,

x² + 2(y² - 2y) = 0

x² + 2(y²- 2y + 1 - 1) = 0

x² + 2(y² - 2y + 1) - 2 = 0

x² + 2(y - 1)² = 2

Dividing by 2 on both sides,

\(frac{x^{2}}{2}+\frac{2(y-1)^{2}}{2}=\frac{2}{2}\)

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

Comparing (3) and (4),

a² = 2

So, a = √2

b² = 1

So, b = 1

center (h, k) = (0, 1)

c² = 2 - 1

c² = 1

So, c = 1

Vertices:

(h±a, k) = (0±√2, 1)

= (±√2, 1)

Foci:

(h±c, k) = (0±1, 1)

= (±1, 1)

Therefore, the vertices and foci are (±√2, 1) and (±1, 1).