Find the Foci 16x^2-64x+4y^2=0

Solution:

Step 1: Convert the given equation to the standard form of an ellipse.

Step 1.1: Complete the square for the $x$-terms, $16x^2 - 64x$.

Step 1.1.1: Identify the coefficients $a$, $b$, and $c$ from the quadratic expression $ax^2 + bx + c$.

• $a = 16$
• $b = -64$
• $c = 0$

Step 1.1.2: Recall the standard vertex form of a quadratic $a(x + d)^2 + e$.

Step 1.1.3: Calculate $d$ using $d = \frac{-b}{2a}$.

Step 1.1.3.1: Plug in the values for $a$ and $b$ into the formula.

• $d = \frac{-64}{2 \cdot 16}$

Step 1.1.3.2: Simplify the fraction.

Step 1.1.3.2.1: Factor out a common factor of 2 from the numerator.

• $d = \frac{2 \cdot -32}{2 \cdot 16}$

Step 1.1.3.2.1.2: Cancel out the common factors.

• $d = \frac{-32}{16}$

Step 1.1.3.2.2: Simplify the fraction by dividing.

• $d = -2$

Step 1.1.4: Determine $e$ using the formula $e = c - \frac{b^2}{4a}$.

Step 1.1.4.1: Substitute the known values into the formula.

• $e = 0 - \frac{(-64)^2}{4 \cdot 16}$

Step 1.1.4.2: Perform the arithmetic operations.

• $e = 0 - 64$
• $e = -64$

Step 1.1.5: Insert $d$ and $e$ into the vertex form to complete the square.

• $16(x - 2)^2 - 64$

Step 1.2: Replace $16x^2 - 64x$ with the completed square in the original equation.

• $16(x - 2)^2 - 64 + 4y^2 = 0$

Step 1.3: Add 64 to both sides to isolate the terms with $x$ and $y$ on one side.

• $16(x - 2)^2 + 4y^2 = 64$

Step 1.4: Divide the entire equation by 64 to normalize the right side to 1.

• $\frac{16(x - 2)^2}{64} + \frac{4y^2}{64} = 1$

Step 1.5: Simplify the fractions to find the standard form of the ellipse.

• $\frac{(x - 2)^2}{4} + \frac{y^2}{16} = 1$

Step 2: Identify the standard form of an ellipse $\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1$ to determine the center and axes lengths.

Step 3: Match the coefficients to find the values of $a$, $b$, $h$, and $k$.

• $a = 4$
• $b = 2$
• $h = 2$
• $k = 0$

Step 4: Calculate the distance $c$ from the center to a focus using the formula $c = \sqrt{a^2 - b^2}$.

Step 4.1: Substitute the values of $a$ and $b$ into the formula.

• $c = \sqrt{4^2 - 2^2}$

Step 4.2: Simplify the expression.

• $c = \sqrt{16 - 4}$
• $c = \sqrt{12}$
• $c = 2\sqrt{3}$

Step 5: Determine the coordinates of the foci.

Step 5.1: The foci are located at $(h, k \pm c)$.

Step 5.2: Substitute the known values to find the coordinates of the foci.

• First focus: $(2, 0 + 2\sqrt{3}) = (2, 2\sqrt{3})$
• Second focus: $(2, 0 - 2\sqrt{3}) = (2, -2\sqrt{3})$

Step 6: The foci of the ellipse are at $(2, 2\sqrt{3})$ and $(2, -2\sqrt{3})$.

Knowledge Notes:

To find the foci of an ellipse given by the equation $16x^2 - 64x + 4y^2 = 0$, we follow these steps:

1. Completing the Square: This technique is used to rewrite a quadratic expression in the form of a perfect square trinomial, which simplifies the process of converting the equation into the standard form of an ellipse.

2. Standard Form of an Ellipse: The standard form is $\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1$, where $(h, k)$ is the center of the ellipse, $a$ is the semi-major axis length, and $b$ is the semi-minor axis length.

3. Foci of an Ellipse: The foci are two fixed points inside the ellipse such that the sum of the distances from any point on the ellipse to the foci is constant. The distance $c$ from the center to each focus is found using the formula $c = \sqrt{a^2 - b^2}$.

4. Coordinates of the Foci: Once $c$ is determined, the coordinates of the foci can be found by adding and subtracting $c$ from the $y$-coordinate of the center (since the major axis is vertical in this case) to get $(h, k + c)$ and $(h, k - c)$.

5. Vertex Form of a Quadratic: The vertex form $a(x + d)^2 + e$ is used to complete the square, where $d$ and $e$ are determined using the formulas $d = \frac{-b}{2a}$ and $e = c - \frac{b^2}{4a}$.

Understanding these concepts is crucial for solving problems involving the geometry of ellipses and their foci.