Algebra 2

1. p = 4/3

2. The distance between the foci is 4*sqrt(2).

Thank you so much for your time into doing these problems, but an explanation would be delightful. As much as I like answers, the explanations are more amusing. I would love one if someone doesn't mind.

Hi DragonLord,

Welcome to the Web2.0calc  forum

I would help you but I am not good with conics.

If you do not get a timely answer you could try asking Heureka, Omi, or Geno direct.

Others may also be able to help though. 

Hello Melody, thank you for the warm welcome to this site. Thank you for the advice. I am not very good at conics myself. Hehe

Question 2: answer see: https://web2.0calc.com/questions/help-thanks_15

Heureka, thank you so much for your help! :)

I also contacted Geno for advice. :D

One focus of the ellipse \(\dfrac{x^2}{2} + y^2 = 1 \)is at \(F = (1,0)\)
There exists a point \(P = (p,0)\), where p > 0 such that for any chord \(\overline{AB}\) that passes through \(F\)
angles \(\angle APF\) and \(\angle BPF\) are equal.
Find \(p\).

\(\text{Let $ \angle APF = \angle BPF = \varphi$ }\)

\(\begin{array}{|rcll|} \hline \tan{\varphi} = \dfrac{y_A}{x_P-x_A} &=& \dfrac{-y_B}{x_P-x_B} \\\\ \dfrac{y_A}{x_P-x_A} &=& \dfrac{-y_B}{x_P-x_B} \\\\ \mathbf{\dfrac{y_A}{y_B}} &=&\mathbf{ -\dfrac{(x_P-x_A)}{(x_P-x_B)}} \\ \hline \\ \tan{F} = \dfrac{y_A}{x_F-x_A} &=& \dfrac{-y_B}{x_B-x_F} \\\\ \dfrac{y_A}{x_F-x_A} &=& \dfrac{-y_B}{x_B-x_F} \\\\ \dfrac{y_A}{y_B} &=& -\dfrac{(x_F-x_A)}{(x_B-x_F)} \quad | \quad x_F = 1 \\ \\ \mathbf{\dfrac{y_A}{y_B}} &=&\mathbf{ -\dfrac{(1-x_A)}{(x_B-1)}} \\ \hline \dfrac{y_A}{y_B} = -\dfrac{(x_P-x_A)}{(x_P-x_B)} &=& -\dfrac{(1-x_A)}{(x_B-1)} \\\\ -\dfrac{(x_P-x_A)}{(x_P-x_B)} &=& -\dfrac{(1-x_A)}{(x_B-1)} \\\\ \dfrac{x_P-x_A}{x_P-x_B} &=& \dfrac{1-x_A}{x_B-1} \\\\ (x_B-1)(x_P-x_A) &=& (1-x_A)(x_P-x_B) \\ \cdots \\ \mathbf{x_P} &=& \mathbf{ \dfrac{2x_Ax_B-(x_A+x_B)} {x_A+x_B-2} } \\ \hline \end{array}\)

line \(y=mx+b\):

\(\begin{array}{|rcll|} \hline y &=& mx+b \quad | \quad F = (1,0) \text{ is on the line } \\ y_F &=& mx_F +b \quad | \quad y_F =0 \quad x_F = 1 \\ 0 &=& m*1 +b \\ b &=& -m \\\\ y &=& mx+b \quad | \quad b = -m \\ y &=& mx-m \\ y &=& m(x-1) \quad | \quad \text{equation of the line through F } \\ \mathbf{y^2} &=& \mathbf{m^2(x-1)^2} \\ \hline \end{array}\)

ellipse \(\dfrac{x^2}{2} + y^2 = 1\):

\(\begin{array}{|rcll|} \hline \dfrac{x^2}{2} + y^2 &=& 1 \\ \mathbf{ y^2} &=& \mathbf{1-\dfrac{x^2}{2}} \\ \hline \end{array}\)

intersection line - ellipse:

\(\begin{array}{|rcll|} \hline y^2 = m^2(x-1)^2 &=& 1-\dfrac{x^2}{2} \\\\ m^2(x-1)^2 &=& 1-\dfrac{x^2}{2} \\\\ m^2(x^2-2x+1) &=& 1-\dfrac{x^2}{2} \\\\ m^2 x^2-2m^2x+m^2 &=& 1-\dfrac{x^2}{2} \\\\ m^2 x^2-2m^2x+m^2 -1+\dfrac{x^2}{2} &=& 0 \\\\ x^2\left(m^2+\dfrac{1}{2}\right)-2m^2x+ (m^2 -1) &=& 0 \\\\ x &=& \dfrac{2m^2\pm \sqrt{4m^4 - 4 \left(m^2+\dfrac{1}{2}\right)(m^2 -1)} }{2\left(m^2+\dfrac{1}{2}\right)} \\\\ x &=& \dfrac{2m^2\pm 2\sqrt{m^4 - \left(m^2+\dfrac{1}{2}\right)(m^2 -1)} }{\left(2m^2+\dfrac{1}{2}\right)} \\\\ x &=& \dfrac{m^2\pm \sqrt{m^4 - \left(m^2+\dfrac{1}{2}\right)(m^2 -1)} }{\left(m^2+\dfrac{1}{2}\right)} \\\\ x &=& \dfrac{m^2\pm \sqrt{ \dfrac{1+m^2}{2} } }{\left(m^2+\dfrac{1}{2}\right)} \\\\ \mathbf{ x_A} &=& \mathbf{\dfrac{m^2+ \sqrt{ \dfrac{1+m^2}{2} } }{\left(m^2+\dfrac{1}{2}\right)}} \\\\ \mathbf{ x_B} &=& \mathbf{\dfrac{m^2- \sqrt{ \dfrac{1+m^2}{2} } }{\left(m^2+\dfrac{1}{2}\right)} }\\ \hline \end{array}\)

\(\mathbf{x_A =\ ?}\quad \mathbf{x_B =\ ?}\)
For example we set  \(m = 1 \text{$(~$for any chord $\overline{AB}$ that passes through $F$ angles $\angle APF$ and $\angle BPF$ are equal$~)$}\)

\(\begin{array}{|rcll|} \hline x_A &=& \dfrac{1^2+ \sqrt{ \dfrac{1+1^2}{2} } }{\left(1^2+\dfrac{1}{2}\right)} \\\\ x_A &=& \dfrac{1+ \sqrt{ \dfrac{2}{2} } }{\left(1+\dfrac{1}{2}\right)} \\\\ x_A &=& \dfrac{1+ 1}{\left(1+\dfrac{1}{2}\right)} \\\\ x_A &=& \dfrac{2}{ \dfrac{3}{2} } \\\\ \mathbf{x_A} &=& \mathbf{\dfrac{4}{3}} \\ \hline \\ x_B &=& \dfrac{1^2- \sqrt{ \dfrac{1+1^2}{2} } }{\left(1^2+\dfrac{1}{2}\right)} \\\\ x_B &=& \dfrac{1- \sqrt{ \dfrac{2}{2} } }{\left(1^2+\dfrac{1}{2}\right)} \\\\ x_B &=& \dfrac{1- 1}{\left(1^2+\dfrac{1}{2}\right)} \\\\ x_B &=& \dfrac{0}{\left(1^2+\dfrac{1}{2}\right)} \\\\ \mathbf{x_B} &=& \mathbf{0} \\ \hline \end{array}\)

\(\mathbf{x_P =\ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{x_P} &=& \mathbf{ \dfrac{2x_Ax_B-(x_A+x_B)} {x_A+x_B-2} } \\\\ x_P &=& \dfrac{2*\dfrac{4}{3}*0-\left(\dfrac{4}{3}+0\right)} {\dfrac{4}{3}+0-2} \\\\ x_P &=& \dfrac{ -\dfrac{4}{3}} {\dfrac{4}{3}-2} \\\\ x_P &=& \dfrac{ -\dfrac{4}{3}} {\dfrac{4-6}{3}} \\\\ x_P &=& \dfrac{ -4} {-2} \\\\ \mathbf{ x_P} &=& \mathbf{2} \\ \hline \end{array}\)

\(P = (p,~0)\\ p=x_P=2 \\ P = (2,~0)\)

In general:

\(\begin{array}{|rcll|} \hline x^2\left(m^2+\dfrac{1}{2}\right)-2m^2x+ (m^2 -1) &=& 0 \quad | \quad :\left(m^2+\dfrac{1}{2}\right) \\\\ x^2-\dfrac{2m^2}{\left(m^2+\dfrac{1}{2}\right)}x+ \dfrac{(m^2 -1)}{\left(m^2+\dfrac{1}{2}\right)} &=& 0 \\\\ x^2\underbrace{-\dfrac{2m^2}{\left(m^2+\dfrac{1}{2}\right)}}_{=-(x_A+x_B)}x+ \underbrace{\dfrac{(m^2 -1)}{\left(m^2+\dfrac{1}{2}\right)}}_{=x_A*x_B} &=& 0 \\ \boxed{\text{Vieta's formulas:}\\ \text{The roots } r_1 ,\ r_2\text{ of the quadratic polynomial } \\ P(x)=a x^2 + b x + c \text{ satisfy}\\ r_1 + r_2 = -\dfrac{b}{a},\ r_1 r_2 = \dfrac{c}{a}. } \\ \hline \mathbf{x_A+x_B} &=& \mathbf{\dfrac{2m^2}{\left(m^2+\dfrac{1}{2}\right)}} \\\\ \mathbf{x_A*x_B} &=& \mathbf{\dfrac{(m^2 -1)}{\left(m^2+\dfrac{1}{2}\right)}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x_P} &=& \mathbf{ \dfrac{2x_Ax_B-(x_A+x_B)} {x_A+x_B-2} } \\\\ && \mathbf{x_A+x_B} = \mathbf{\dfrac{2m^2}{\left(m^2+\dfrac{1}{2}\right)}} \\\\ && \mathbf{x_A*x_B} = \mathbf{\dfrac{(m^2 -1)}{\left(m^2+\dfrac{1}{2}\right)}} \\\\ x_P &=& \dfrac{2\dfrac{(m^2 -1)}{\left(m^2+\dfrac{1}{2}\right) } -\dfrac{2m^2}{\left(m^2+\dfrac{1}{2}\right)}} {\dfrac{2m^2}{\left(m^2+\dfrac{1}{2}\right)}-2} \\\\ x_P &=& \dfrac{\dfrac{2(m^2 -1)-2m^2}{\left(m^2+\dfrac{1}{2}\right) } } {\dfrac{2m^2}{\left(m^2+\dfrac{1}{2}\right)} -2\dfrac{\left(m^2+\dfrac{1}{2}\right)}{\left(m^2+\dfrac{1}{2}\right)}} \\\\ x_P &=& \dfrac{\dfrac{2(m^2 -1)-2m^2}{\left(m^2+\dfrac{1}{2}\right) } } {\dfrac{2m^2-2\left(m^2+\dfrac{1}{2}\right)}{\left(m^2+\dfrac{1}{2}\right)} } \\\\ x_P &=& \dfrac{2(m^2 -1)-2m^2} {2m^2-2\left(m^2+\dfrac{1}{2}\right) } \\\\ x_P &=& \dfrac{2m^2 -2-2m^2}{2m^2-2m^2-1} \\\\ x_P &=& \dfrac{-2}{-1} \\\\ \mathbf{x_P} &=& \mathbf{2} \quad (p=2) \\ \hline \end{array}\)

Thanks Heureka :)

Hello, Hereuka. Thank you so much for your time and effort to respond to it. I will thouroughly revise your response slowly. Since I don’t know much trigonometry. Once again, thank you.