sample 9 : Equation of Parabola

College algebra questions and problems on the equation of parabolas are presented. The solutions with explanations are located at the bottom of the page.

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\( 2(y^2 + 4y) + x + 1 = 0 \)

\( 2((y + 2)^2 - 4) + x + 1 = 0 \)

\( (y + 2)^2 = -\frac{1}{2}(x - 7) \)

Vertex at \( (7, - 2) \)

\( -\frac{1}{2} = 4a \), hence \( a = -\frac{1}{8} \)

Focus at \( (7 - \frac{1}{8}, -2) = (6.875, -2) \)

Axis of symmetry given by \( y = -2 \)

Directrix is a vertical line given by: \( x = h - a = 7 + \frac{1}{8} = 7.125 \)

\( x^2 - 8x - y + 2 = 0 \)

\( ((x - 4)^2 - 16) - y + 2 = 0 \)

\( (x - 4)^2 = y + 14 \)

Vertex at \( (4, -14) \)

\( 1 = 4a \), hence \( a = \frac{1}{4} \)

Focus at \( (4, -14 + \frac{1}{4}) = (4, -13.75) \)

Axis of symmetry given by \( x = 4 \)

Directrix is a horizontal line given by: \( y = k - a = -14 - \frac{1}{4} = -14.25 \)

\( (x - h)^2 = 4a(y - k) \), vertex at \( (h, k) \)

With the vertex at \( (h, k) = (-2, -2) \), the equation is given by

\( (x + 2)^2 = 4a(y + 2) \)

The distance from the vertex \( (-2, -2) \) to the focus \( (-2, -8) \) is \( |a| = 6 \). Since the parabola opens downward \( a = -6 \).

The equation of the parabola is given by: \( (x + 2)^2 = -24(y + 2) \)

From the given graph, the vertex is at \( (h, k) = (2, 4) \), therefore the equation of the parabola is given by

\( (y - 4)^2 = 4a(x - 2) \)

Use the point \( (-2, -2) \) to find \( a \) as follows

\( (-2 - 4)^2 = 4a(-2 - 2) \)

Solve the above for \( a \): \( a = -2.25 \)

Equation of the parabola: \( (y - 4)^2 = -9(x - 2) \)

The focus is at \( (h + a, k) = (2 - 2.25, 4) = (-0.25, 4) \)

\( (y - k)^2 = 4a(x - h) \), vertex at \( (h, k) \)

The vertex is the midpoint of the point \( (2, 0) \), which is the point of intersection of the directrix and the x-axis, and the focus \( (8, 0) \). Hence \( h = 5 \) and \( k = 0 \).

The distance between the directrix and the focus is twice \( |a| \). Hence \( 2|a| = 6 \), \( |a| = 3 \), and since the parabola opens to the right \( a = 3 \). The equation of the parabola is given by: \( y^2 = 12(x - 5) \)

\( 10 = \sqrt{(6^2 + (b + 8))^2} \)

Solve the above for \( b \): 2 solutions \( b = 0 \) and \( b = -16 \)

THERE ARE TWO SOLUTIONS TO THIS PROBLEM

First solution: the distance from the focus \( F(0, 0) \) to the directrix \( y = 2 \) is equal to \( 2|a| \), hence \( |a| = 1 \) and \( a = -1 \) since the parabola opens downward. The vertex is at \( (0, 1) \), at equal distances from the focus and the directrix.

Equation: \( x^2 = -4(y - 1) \)

Second solution: the distance from the focus \( F(0, -16) \) to the directrix \( y = 2 \) is equal to \( 2|a| \), hence \( |a| = 9 \) and \( a = -9 \) since the parabola opens downward. The vertex is at \( (0, -7) \), at equal distances from the focus and the directrix.

Equation: \( x^2 = -36(y + 7) \)

Point \( (100, 50) \) lies on the graph of the parabolic dish, hence \( 100^2 = 4a \times 50 \)

Solve to find: \( a = 50 \) which is also the distance from the vertex at \( (0, 0) \) to the focus. Hence the focus is at \( (0, 50 \, \text{cm}) \).

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