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

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

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

2. \(\cos \angle F_1 AF_2 = 5/7\)

(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

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:

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

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

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

Attn DragonLord,

NEXT time please post your questions seperately.

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

No problem, Melody!