# FIND AN EQUATION FOR THE CONIC SECTION WITH THE GIVEN PROPERTIES

Write the equation of the conic section using the information given. You may want to use the graph to help you.

Problem 1 :

Circle with center (4, -1) and point on the circle (2, -4).

Solution :

Equation of the circle :

(x - h)2 + (y - k)2 = r2

Center (h, k) ==> (4, -1)

Radius of circle = (x2 - x1)2 + (y2 - y1)2

Distance between center and point on the circle = radius

Center (4, -1) and (2, -4)

(2 - 4)2 + (-4 - (-1))2

(-2)2 + (-4 + 1)2

√4 + (-3)2

√(4 + 9)

√13 units

(x - 4)2 + (y - (-1))2 = √132

(x - 4)2 + (y + 1)2 = 13

Problem 2 :

Hyperbola with vertices (6, -4) and (6, 4) and foci (6, -6) and (6, 6).

Solution :

From the given information, the hyperbola is symmetric about y-axis.

Equation of the Hyperbola :

(y - k)2 / a2  - (x - h)2 / b2 = 1 ---(1)

Midpoint of vertices = Center

= (x1 + x2)/2, (y1 + y2)/2

vertices (6, -4) and (6, 4)

= (6 + 6)/2, (-4 + 4)/2

= (12/2, 0/2)

= (6, 0)

Distance between center and vertex = (x2 - x1)2 + (y2 - y1)2

Center (6, 0) vertex (6, -4)

(6 - 6)2 + (0 + 4)2

√(0 + 16)

a = 4

Distance between center and foci = c

Center (6, 0) and foci (6, -6)

(6 - 6)2 + (0 + 6)2

c = 6

For hyperbola, c2 = a2 + b2

62 = 42 + b2

b2 = 36 - 16

b2 = 20

Applying center, the values of a2 = 16 and b2 = 20

(y - 0)2 / 42  - (x - 6)2 / 20 = 1

[y2 / 16]  - (x - 6)2 / 20 = 1

Problem 3 :

Ellipse with vertices (-5, 1) and (-1, 1) and co-vertices (-3, 2) and (-3, 0)

Solution :

From the given information, the ellipse is symmetric about x-axis.

Equation of the Ellipse :

(x - h)2 / a2 (y - k)2 / b2 = 1 ---(1)

Midpoint of vertices = Center

= (x1 + x2)/2, (y1 + y2)/2

vertices (-5, 1) and (-1, 1)

= (-5 - 1)/2, (1 + 1)/2

= (-6/2, 2/2)

center = (-3, 1)

Distance between center and vertex = a

(x2 - x1)2 + (y2 - y1)2

Center (-3, 1) vertex (-5, 1)

(-3 + 5)2 + (1 - 1)2

√(4 + 0)

a = 2

Distance between center and covertex = b

Center (-3, 1) and co-vertex (-3, 2)

(-3 + 3)2 + (1 - 2)2

√(0 + 1)

b = 1

Applying center, the values of a2 = 4 and b2 = 1

(x + 3)2 / 4 (y - 1)2 / 1 = 1

(x + 3)2 / 4 (y - 1)2 = 1

Problem 4 :

Parabola with vertex (-1, -2) and focus (4, -2).

Solution :

From the given information, the parabola is symmetric about x-axis.

(x - h)2 = 4a(y- k)

Applying the vertex (-1, -2) in the above equation, we get

(x + 1)2 = 4a(y + 2)

Distance between vertex and focus :

(x2 - x1)2 + (y2 - y1)2

vertex (-1, -2) and focus (4, -2).

(-1 - 4)2 + (-2 + 2)2

(-5)2 + (0)2

√25

a = 5

(x + 1)2 = 4(5)(y + 2)

(x + 1)2 = 20(y + 2)

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