FIND CENTER VERTICES AND CO-VERTICES OF AN ELLIPSE

Ellipse - Symmetric About x-Axis

where a² > b² and major axis is along x-axis.

Center : (0, 0).

Vertices : A(a, 0) and A'(-a, 0).

Co-vertices : B(0, b) and B(0, -b).

Foci : F(ae, 0) and F'(-ae, 0)

Ellipse - Symmetric About y-Axis

where a² > b² and major axis is along y-axis.

Center : (0, 0).

Vertices : A(0, a) and A'(0, -a).

Co-vertices : B(b, 0) and B(-b, 0).

Foci : F(o, ae) and F'(0, -ae)

Find the center, vertices and co-vertices of the following ellipses.

Example 1 :

Solution :

The above ellipse is symmetric about x-axis.

Center : (0, 0).

Vertices :

A(a, 0) and A'(-a, 0)

A(5, 0) and A'(-5, 0)

Co-vertices :

B(0, b) and B'(0, -b)

B(0, 3) and B'(0, -3)

Example 2 :

Let X = x - 1 and Y = y + 1.

The above ellipse is symmetric about Y-axis.

Center :

(0, 0)

X = 0  and  Y = 0

Substitute X = x - 1 and Y = y + 1.

x - 1 = 0  and  y + 1 = 0

x = 1  and  y = -1

The center is (1, -1)

Vertices :

A(0, a)  and  A'(0, -a)

A(0, 4)  and  A'(0, -4)

The vertices are (1, 3) and (1, -5).

Co-vertices :

B(b, 0)  and  B'(-b, 0)

B(3, 0)  and  B'(-3, 0)

The co-vertices are (4, -1) and (-2, -1).

Example 3 :

4x² + 32x + 36y² - 72y - 44 = 0

Solution :

The given equation of ellipse is not in standard form. Convert it to standard form.

4x² + 32x + 36y² - 72y - 44 = 0

4(x² + 8x) + 36(y² - 2y) - 44 = 0

4[x²+ 2x(4) + 4² - 4²] + 36[y² - 2y(1) + 1² - 1²] - 44 = 0

4[(x + 4)²- 16] + 36[(y - 1)²- 1] - 44 = 0

4(x + 4)²- 64 + 36(y - 1)²- 36 - 44 = 0

4(x + 4)² + 36(y - 1)² - 144 = 0

4(x + 4)² + 36(y - 1)² = 144

Divide both sides by 144.

The above ellipse is symmetric about X-axis.

Center :

(0, 0)

X = 0  and  Y = 0

Substitute X = x + 4 and Y = y - 1.

x + 4 = 0  and  y - 1 = 0

x = -4  and  y = 1

The center is (-4, 1)

Vertices :

A(a, 0)  and  A'(-a, 0)

A(6, 0)  and  A'(-6, 0)

The vertices are (2, 1) and (-10, 1).

Co-vertices :

B(0, b)  and  B'(0, -b)

B(0, 2)  and  B'(0, -2)

The co-vertices are (-4, 3) and (-4, -1).

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