College Algebra Questions and Problems With Solutions
Equation of Parabola

The distance from the vertex \((0, 0)\) to the focus \((0, 4)\) is \(|a| = 4\). Since the vertex is at \((0, 0)\) and the focus is at \((0, 4)\) on the y-axis, the parabola opens upward, which means \(a = 4\) and its equation is given by:

\[ x^2 = 4ay = 16y \]

Since the x-axis is the axis of symmetry of the parabola and its vertex is at the origin, the equation of the parabola has the form:

\[ y^2 = 4ax \]

Substitute the point \((-2, 4)\) into the equation:

\[ 4^2 = 4a(-2) \implies 16 = -8a \implies a = -2 \]

The equation is:

\[ y^2 = -8x \]

Since the vertex is at \((0, 2)\) and the focus is at \((0, 6)\), the parabola opens upward and the equation of a parabola with vertex at \((h, k)\) is given by:

\[ (x - h)^2 = 4a(y - k) \]

Here \(h = 0\) and \(k = 2\); the equation is given by:

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

\(|a|\) is the distance from vertex to focus = \(6 - 2 = 4\). Since the parabola opens upward, \(a = 4\).

The equation is given by:

\[ x^2 = 16(y - 2) \]

We first complete the square using the terms in \(y\) and \(y^2\) and write the given equation in the form \((y - k)^2 = 4a(x - h)\) where \((h, k)\) is the vertex, the focus is at \((h + a, k)\), the axis of symmetry is given by \(y = k\), and the directrix is given by \(x = h - a\).

\[ 2(y^2 + 4y) + x + 1 = 0 \] \[ 2((y + 2)^2 - 4) + x + 1 = 0 \] \[ 2(y + 2)^2 - 8 + x + 1 = 0 \implies 2(y + 2)^2 + x - 7 = 0 \implies (y + 2)^2 = -\frac{1}{2}(x - 7) \]

• Vertex: \((7, -2)\)
• Value of \(a\): \(4a = -\frac{1}{2} \implies a = -\frac{1}{8}\)
• Focus: \(\left(7 - \frac{1}{8}, -2\right) = (6.875, -2)\)
• Axis of symmetry: \(y = -2\)
• Directrix: \(x = h - a = 7 - \left(-\frac{1}{8}\right) = 7.125\)

We first complete the square using the terms in \(x\) and \(x^2\) and write the given equation in the form \((x - h)^2 = 4a(y - k)\) where \((h, k)\) is the vertex, the focus is at \((h, k + a)\), the axis of symmetry is given by \(x = h\), and the directrix is given by \(y = k - a\).

\[ (x^2 - 8x + 16 - 16) - y + 2 = 0 \] \[ (x - 4)^2 - 16 - y + 2 = 0 \implies (x - 4)^2 - y - 14 = 0 \implies (x - 4)^2 = y + 14 \]

• Vertex: \((4, -14)\)
• Value of \(a\): \(4a = 1 \implies a = \frac{1}{4}\)
• Focus: \(\left(4, -14 + \frac{1}{4}\right) = (4, -13.75)\)
• Axis of symmetry: \(x = 4\)
• Directrix: \(y = k - a = -14 - \frac{1}{4} = -14.25\)

The vertex and the focus are on the same vertical line \(x = -2\) with the focus below the vertex; therefore, the parabola opens downward and its equation has the form:

\[ (x - h)^2 = 4a(y - k) \]

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 \(|a| = |-8 - (-2)| = 6\). Since the parabola opens downward, \(a = -6\).

The equation of the parabola is given by:

\[ (x + 2)^2 = 4(-6)(y + 2) \implies (x + 2)^2 = -24(y + 2) \]

For the equation of the parabola of the form \((y - k)^2 = 4a(x - h)\), the vertex is at \((h, k)\).

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 \((-2, -2)\) from the graph to find \(a\):

\[ (-2 - 4)^2 = 4a(-2 - 2) \implies (-6)^2 = 4a(-4) \implies 36 = -16a \implies a = -\frac{36}{16} = -2.25 \]

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

Focus: \((h + a, k) = (2 + (-2.25), 4) = (-0.25, 4)\)

Since the directrix is the line \(x = 2\) and the focus is at \((8, 0)\), the parabola has the x-axis as an axis of symmetry and opens to the right. Its equation is of the form:

\[ (y - k)^2 = 4a(x - h) \]

The vertex is the midpoint between the point of intersection of the directrix and the x-axis \((2, 0)\) and the focus \((8, 0)\):

\[ h = \frac{2 + 8}{2} = 5, \quad k = 0 \]

The distance between the directrix and the focus is \(8 - 2 = 6\), which is equal to \(2|a|\). Hence \(|a| = 3\). Since the parabola opens to the right, \(a = 3\).

The equation of the parabola is given by:

\[ y^2 = 12(x - 5) \]

The focus is on the y-axis and therefore has the form \(F(0, b)\). By definition, any point on the parabola, particularly the point \((-6, -8)\), must be at an equal distance from the directrix \(y = 2\) and the focus \((0, b)\):

\[ \text{Distance to directrix} = |-8 - 2| = 10 \]

Setting this equal to the distance from \((-6, -8)\) to \((0, b)\):

\[ 10 = \sqrt{(-6 - 0)^2 + (-8 - b)^2} \implies 10 = \sqrt{36 + (b + 8)^2} \]

Square both sides:

\[ 100 = 36 + (b + 8)^2 \implies (b + 8)^2 = 64 \implies b + 8 = \pm 8 \]

This gives two solutions for \(b\): \(b = 0\) and \(b = -16\).

First Solution (\(b = 0\)): Focus is at \((0, 0)\) and directrix is \(y = 2\). The distance from the focus to the directrix is \(2|a| = 2\), so \(|a| = 1\). Since it opens downward, \(a = -1\). The vertex is at \((0, 1)\).

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

Second Solution (\(b = -16\)): Focus is at \((0, -16)\) and directrix is \(y = 2\). The distance from focus to directrix is \(|2 - (-16)| = 18 = 2|a|\), so \(|a| = 9\). Since it opens downward, \(a = -9\). The vertex is at \((0, -7)\).

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

Placing the vertex of the parabola at the origin \((0, 0)\) opening upward, the equation of the parabolic dish is of the form:

\[ x^2 = 4ay \]

Since the total diameter is 200 cm, the width from the center axis to the edge is \(100\) cm, and the maximum depth is \(50\) cm. Thus, the point \((100, 50)\) lies on the graph of the parabola.

Substitute this point into the equation to find \(a\):

\[ 100^2 = 4a(50) \implies 10000 = 200a \implies a = 50 \]

\(a\) represents the distance from the vertex to the focus. Therefore, the focus is located at \((0, 50 \text{ cm})\) from the vertex.