+0

# Algebra 2 Polynomials and Ellipses. Thank you all!!!

+1
50
11
+177

1. A polynomial of degree $$13$$ is divided by $$d(x)$$ to give a quotient of degree $$7$$ and a remainder of $$3x^3+4x^2-x+12$$. What is $$\deg d$$?

2. The ellipse $$\frac{x^2}{25} + \frac{y^2}{16} = 1$$ has foci at $$F_1$$ and $$F_2,$$ as shown below. The circle with diameter $$\overline{F_1 F_2}$$ is drawn.

Let $$B$$ be a point on the circle such that $$BF_2 = 5.$$ When $$\overline{BF_2}$$ is extended past $$B$$ it intersects the ellipse at $$A.$$ Compute $$\cos \angle F_1 AF_2.$$

3. (Bonus question) Recently, I've encountered a polynomial question $$\frac{9x^{4}+x^{3}-12x+21}{x+4}$$. I found the quotient is $$9x^{3}-35x^{2}-140x-572$$ and the remainder is $$-63,$$ but apparently it is wrong. I am confused and would like anyone to give me an explanation on this question!!

Thank you guys SO SO much for your time!!!

Feb 4, 2020
edited by DragonLord  Feb 4, 2020

#1
+177
0

Nvm, I found out my mistake on the bonus question!!! So, you guys don't need to help on that one anymore. Thank you once again :D

Feb 4, 2020
#2
0

1. Since the degree of the remainder is 3, the degree of d(x) is 4.

2. $$\cos \angle F_1 AF_2 = 5/7$$

.
Feb 4, 2020
#3
+107011
+2

(2)

Note that F1  and F2  are endpoints of the diameter of the circle....so triangle  F1BF2  is a right triangle inscribed in the circle....

And note that sqrt (25)   =  5   =  a

And 2a   = 10

And the focal length = the radius of the  circle  =  sqrt (a^2 - b^2) =  sqrt (25  -16)  = sqrt (9)   = 3

So  F1F2   = 2(3)    =  6

And  since triangle F1BF2   is right with F1F2  the hypotenuse and BF2  a leg = 5 then BF1  =

sqrt (6^2 - 5^2)   =sqrt (36 -25) = sqrt (11)

Call  F1A =  y

And AB  =x

Since  triangle F1BF2  is right, then so is triangle  F1AB  with ABF1  being a right angle

And the  sum of the distances F1A  and F2A  =  2a  = 10

So we have the following system

(F1A)^2 - (AB)^2   = (BF1)^2

F1A + F2A   = 6

So we have that

y^2  - x^2  =11         (1)

y + (5 + x)  =  10  ⇒  x + y  = 5  ⇒  y = 5 - x     (2)

Subbing (2)  into (1)  we have that

(5 - x)^2 - x^2  =  11

25 - 10x  = 11

14  = 10x

7 = 5x

x  =7/5

And  y  =  5 - (7/5)  =  25/5  -7/5  =18 / 5

So....the cosine  of  F1AF2  =

AB   / AF1  =

x  / y  =

(7/5) / (18/5)  =

7 / 18

Feb 4, 2020
edited by CPhill  Feb 4, 2020
edited by CPhill  Feb 4, 2020
edited by CPhill  Feb 4, 2020
#4
+24031
+3

2.
The ellipse $$\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$$
has foci at $$F_1$$ and $$F_2$$ as shown below. The circle with diameter $$\overline{F_1 F_2}$$ is drawn.

Let B be a point on the circle such that $$\overline{BF_2} = 5$$.
When $$\overline{BF_2}$$ is extended past B it intersects the ellipse at A.
Compute $$\cos (\angle F_1 AF_2)$$.

$$\text{Let \angle F_1 AF_2 = A } \\ \text{Ellipse: \dfrac{x^2}{25} + \dfrac{y^2}{16} = 1 \quad or \quad y^2=\dfrac{16}{25}(25-x_A^2)  }$$

$$\text{1. Parameter of the ellipse:}\\ \begin{array}{rclrcl} a^2 &=& 25 \qquad \text{or} \qquad \mathbf{a=5} \\\\ \mathbf{b^2} &=& \mathbf{16} \\\\ c^2 &=& a^2-b^2 \\ c^2 &=& 25-16\\ c^2 &=& 9 \\ \mathbf{c} &=& \pm3 \\\\ F_1 &=& (-c,0) = (-3,0) \\ F_2 &=& (c,0) = (3,0) \\ \end{array} \begin{array}{|rcll|} \hline \overline{AF_1} +\overline{AF_2} &=& 2a \\ \overline{AF_1} +\overline{AF_2} &=& 2*5 \\ \mathbf{\overline{AF_1} +\overline{AF_2}} &=& \mathbf{10} \\ \hline \end{array} \\ \text{2. Circle:}\\ \begin{array}{|rcll|} \hline x^2+y^2 &=& r^2 \quad | \quad r=c,\ c= 3 \\ \mathbf{x^2+y^2} &=& \mathbf{3^2} \\ \hline \end{array}$$

$$\text{Point B:}\\ \begin{array}{|rcll|} \hline B &=& 3\dbinom{\cos(2\varphi)}{\sin(2\varphi)} \\ &&& \boxed{\sin(\varphi)= \dfrac{2.5}{3}\\ \mathbf{\sin(\varphi)= \dfrac{5}{6}} \\ } &&& \boxed{ \cos(\varphi)= \dfrac{d}{3} \\ \cos(\varphi)= \dfrac{\sqrt{3^2-2.5^2} } {3} \\ \cos(\varphi)= \dfrac{\sqrt{2.75 }}{3} \\ \cos(\varphi)= \dfrac{ \sqrt{ \dfrac{11}{4} } } {3} \\ \mathbf{\cos(\varphi)}= \mathbf{\dfrac{ \sqrt{11} }{6}} \\ } \\ B &=& 3\dbinom{\cos^2( \varphi)-\sin^2(\varphi)}{2\sin(\varphi)\cos(\varphi)} \\ B &=& 3\dbinom{\frac{11}{36}-\frac{25}{36}}{2*\frac{5}{6} *\frac{11}{4}} \\ \ldots \\ \mathbf{B} &=& \mathbf{ \dbinom{-\frac{7}{6}} {\frac{5}{6}\sqrt{11}} } \\ \hline \end{array} \\ \text{Line y=mx+b:}\\ \begin{array}{|rcll|} \hline y &=& mx+b \quad | \quad x_{F_2} = 3, y_{F_2} = 0 \\ 0 &=& 3m+b \\ \mathbf{b} &=&\mathbf{ -3m } \\\\ y &=& mx+b \quad | \quad b=-3m \\ y &=& mx-3m \\ y &=& m(x-3) \quad | \quad x_B = -\dfrac{7}{6}, \ y_B = \dfrac{5}{6}\sqrt{11} \\ \dfrac{5}{6}\sqrt{11} &=& m\left(-\dfrac{7}{6}-3 \right) \\ \dfrac{5}{6}\sqrt{11} &=& m\left(-\dfrac{25}{6} \right) \\ 5\sqrt{11} &=& -25m \\ \mathbf{m} &=& \mathbf{-\dfrac{\sqrt{11}}{5} } \\\\ y &=& m(x-3) \quad | \quad m=-\dfrac{\sqrt{11}}{5} \\ \mathbf{y} &=& \mathbf{-\dfrac{\sqrt{11}}{5}(x-3)} \quad \text{or} \quad \mathbf{y^2} = \mathbf{ \dfrac{11}{25}(x-3)^2} \\ \hline \end{array}$$

Point A: Intersection line and ellipse:

$$\begin{array}{|rcll|} \hline y_A^2 = \dfrac{11}{25}(x_A-3)^2 &=& \dfrac{16}{25}\left(\dfrac{25-x_A^2}{25}\right) \\ \dfrac{11}{25}(x_A-3)^2 &=& \dfrac{16}{25}(25-x_A^2) \\ 11(x_A-3)^2 &=& 16(25-x_A^2) \\ \ldots \\ 27x_A^2 -66x_A -301 &=& 0 \\ \ldots \\ \mathbf{x_A} &=& \mathbf{-\dfrac {7}{3}} \\\\ y_A &=& -\dfrac{\sqrt{11}}{5}(x_A-3) \\ \ldots \\ \mathbf{y_A} &=& \mathbf{\dfrac {16}{15}\sqrt{11}} \\ \hline \end{array}$$

$$\overline{AB}:\\ \begin{array}{|rcll|} \hline \mathbf{A} &=& \mathbf{ \dbinom{-\frac {7}{3}} {\frac {16}{15}\sqrt{11}}} \\ \mathbf{B} &=& \mathbf{ \dbinom{-\frac{7}{6}} {\frac{5}{6}\sqrt{11}}} \\ \overline{AB} &=& \sqrt{(-\frac {7}{3}-(-\frac{7}{6}))^2+ (\frac {16}{15}\sqrt{11}-\frac{5}{6}\sqrt{11})^2 } \\ \ldots \\ \mathbf{\overline{AB}} &=& \mathbf{1.4} \\ \hline \end{array} \\ \overline{AF_1}:\\ \begin{array}{|rcll|} \hline \overline{AF_1} + \overline{AF_2} &=& 10 \quad | \quad \overline{AF_2} =\overline{AB} + \overline{AF_2} \quad \overline{AF_2} = 5 \\ \overline{AF_1} + \overline{AF_2} &=& 10 \quad | \quad \overline{AF_2} =\overline{AB} + 5 \quad \overline{AB}=1.4\\ \overline{AF_1} + \overline{AF_2} &=& 10 \quad | \quad \overline{AF_2} =1.4 + 5 \\ \overline{AF_1} + \overline{AF_2} &=& 10 \quad | \quad \mathbf{\overline{AF_2}} = \mathbf{6.4} \\ \overline{AF_1} + 6.4 &=& 10 \\ \overline{AF_1} &=& 10-6.4 \\ \mathbf{\overline{AF_1}} &=& \mathbf{3.6} \\ \hline \end{array}$$

cos-rule:

$$\begin{array}{|rcll|} \hline \mathbf{\overline{F_1F_2}^2} &=& \mathbf{\overline{AF_1}^2+\overline{AF_2}^2-2*\overline{AF_1}*\overline{AF_2}*\cos (\angle F_1 AF_2)} \\\\ 6^2 &=& 3.6^2+6.4^2-2*3.6*6.4*\cos (\angle F_1 AF_2) \\\\ \cos (\angle F_1 AF_2) &=& \dfrac{3.6^2+6.4^2-6^2 }{2*3.6*6.4} \\\\ \cos (\angle F_1 AF_2) &=& \dfrac{17.92}{46.08} \\\\ \cos (\angle F_1 AF_2) &=& \dfrac{1792}{4608} \\\\ \mathbf{\cos (\angle F_1 AF_2)} &=& \mathbf{\dfrac{7}{18}} \\ \hline \end{array}$$

Feb 4, 2020
#5
+177
+2

Thank you all so so much for your help. Thank you for returning as well, CPhill! We all missed you! <3

Feb 4, 2020
#6
+177
+1

Hello Guest, would you please explain why the degree is 4?

Feb 4, 2020
#7
+107414
+2

I suspect guest two is the one that is positing many one line incorrect answers.

Maybe he/she thinks it is funny.

$$\frac{ax^{13}+......}{d(x)}=(bx^{7}+.....)+\frac{cx^4+....}{d(x)}\\ so\\ ax^{13}+...... =[(bx^{7}+.....)*d(x)]\;\;+\;\;cx^4+...$$

For this to work, d(x) would have to have a  degree of 6

because x^7*x^6 = x^13  and that will be the highest power on both sides.

So the remainder has no relevance.

coding:

\frac{ax^{13}+......}{d(x)}=(bx^{7}+.....)+\frac{cx^4+....}{d(x)}\\
so\\
ax^{13}+...... =[(bx^{7}+.....)*d(x)]\;\;+\;\;cx^4+...

Melody  Feb 5, 2020
#8
+107414
0

Attn DragonLord,

Feb 5, 2020
#9
+177
0

Melody, ok will do. I think he keeps posting on my questions for some reasons. And the answers are always wrong.

Feb 7, 2020
#10
+107414
0

Thanks DL

Melody  Feb 7, 2020
#11
+177
0

No problem, Melody!

Feb 8, 2020