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# Algebra 2

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1. One focus of the ellipse $$\frac{x^2}{2} + y^2 = 1$$ 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 $$p.$$

2. The asymptotes of a hyperbola are $$y = x + 1$$ and $$y = 3 - x.$$ Also, the hyperbola passes through $$(3, 3).$$ Find the distance between the foci of the hyperbola.

Thank you guys so so much. I appreciate all the help!

Jan 26, 2020

#1
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1. p = 4/3

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

Jan 26, 2020
#2
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Thanks, but I could look in the back of the book for the bare answer.

The help that I really need is to know how the answer is calculated / derived.

Guest Jan 26, 2020
#3
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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.

Jan 26, 2020
#4
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Hi DragonLord,

Welcome to the Web2.0calc  forum

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.

Jan 26, 2020
#5
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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

Jan 27, 2020
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Jan 27, 2020
edited by heureka  Jan 27, 2020
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Heureka, thank you so much for your help! :)

Jan 27, 2020
#8
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I also contacted Geno for advice. :D

Jan 28, 2020
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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:

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

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

Jan 28, 2020
edited by heureka  Jan 28, 2020
edited by heureka  Jan 29, 2020
edited by heureka  Jan 29, 2020
edited by heureka  Jan 29, 2020
edited by heureka  Jan 29, 2020
edited by heureka  Jan 29, 2020
edited by heureka  Jan 29, 2020
edited by heureka  Jan 29, 2020
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Thanks Heureka :)

Jan 28, 2020
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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.

Jan 28, 2020