Question

use the binomial theorem to expand the binomial, ((x - 4y)^4)

select one:
○ a. (x^4 - 4x^3y + 6x^2y^3 - 4xy^3 + y^4)
○ b. (x^4 - 16x^3y + 96x^2y^2 - 256xy^3 + 256y^4)
○ c. (x^4 - 4x^3y + 16x^2y^2 - 64xy^3 + 256y^4)
○ d. (x^4 - 4(xy)^3 + 16(xy)^2 - 64(xy)^3 + 256y^4)

conic sections are created when a cuts through a at a specific angle. (dropdown options: vector, sphere, line, cylinder, double cone, plane)

find the eccentricity of the following conic section:
((x - 1)^2 + (y - 2)^2 = 30)

select one:
○ a. 0
○ b. 0.5
○ c. 30
○ d. 1

Explanation:

First Sub - Question (Binomial Expansion)

Step 1: Recall the Binomial Theorem

The Binomial Theorem states that \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\) and \(n!=n\times(n - 1)\times\cdots\times1\). For \((x-4y)^{4}\), we have \(a = x\), \(b=-4y\) and \(n = 4\).

Step 2: Calculate the binomial coefficients for \(k = 0,1,2,3,4\)

• When \(k = 0\): \(\binom{4}{0}=\frac{4!}{0!(4 - 0)!}=\frac{4!}{4!}=1\), and the term is \(\binom{4}{0}x^{4-0}(-4y)^{0}=1\times x^{4}\times1=x^{4}\)
• When \(k = 1\): \(\binom{4}{1}=\frac{4!}{1!(4 - 1)!}=\frac{4\times3!}{1\times3!}=4\), and the term is \(\binom{4}{1}x^{4 - 1}(-4y)^{1}=4\times x^{3}\times(-4y)=-16x^{3}y\)
• When \(k = 2\): \(\binom{4}{2}=\frac{4!}{2!(4 - 2)!}=\frac{4\times3\times2!}{2!\times2!}=\frac{12}{2}=6\), and the term is \(\binom{4}{2}x^{4-2}(-4y)^{2}=6\times x^{2}\times16y^{2}=96x^{2}y^{2}\)
• When \(k = 3\): \(\binom{4}{3}=\frac{4!}{3!(4 - 3)!}=\frac{4\times3!}{3!\times1!}=4\), and the term is \(\binom{4}{3}x^{4 - 3}(-4y)^{3}=4\times x\times(-64y^{3})=-256xy^{3}\)
• When \(k = 4\): \(\binom{4}{4}=\frac{4!}{4!(4 - 4)!}=\frac{4!}{4!}=1\), and the term is \(\binom{4}{4}x^{4-4}(-4y)^{4}=1\times1\times256y^{4}=256y^{4}\)

Step 3: Combine the terms

Combining all the terms, we get \((x - 4y)^{4}=x^{4}-16x^{3}y + 96x^{2}y^{2}-256xy^{3}+256y^{4}\)

Second Sub - Question (Conic Sections Formation)

Conic sections are defined as the curves formed when a plane cuts through a double cone at different angles. A plane intersecting a double cone can form circles, ellipses, parabolas, or hyperbolas depending on the angle of intersection.

Third Sub - Question (Eccentricity of Conic Section)

Step 1: Identify the type of conic section

The equation \((x - 1)^{2}+(y - 2)^{2}=30\) is in the standard form of a circle \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center of the circle and \(r\) is the radius.

Step 2: Recall the eccentricity of a circle

For a circle, the eccentricity \(e = 0\). This is because a circle is a special case of an ellipse where the two foci coincide, and the ratio of the distance from the center to a focus (which is \(0\) for a circle) to the distance from the center to a vertex (radius) is \(0\).

Answer:

b. $x^{4}-16x^{3}y + 96x^{2}y^{2}-256xy^{3}+256y^{4}$