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Question from DragonLord:

An ellipse has foci at \(F_1 = (0,2)\) and \(F_2 = (3,0)\).
The ellipse intersects the x-axis at the origin, and one other point.
What is the other point of intersection?

Hello DragonLord my attempt:

 

\(\text{Let $P(0,0) = $origin} \\ \text{Let $\angle F_1F_2P = \varphi $} \\ \text{Let $F_1F_2 = 2c $} \\ \text{Let $PF_1 = 2 $} \\ \text{Let $PF_2 = 3 $}\)

 

1. Parameter of the ellipse:

\(\begin{array}{rcll} PF_1 + PF_2 &=& 2a \\ 2+3 &=& 2a \\ \mathbf{a} &=& \mathbf{2.5} \qquad \mathbf{a^2} = \mathbf{6.25} \\ \end{array}\)

 

\(\begin{array}{rcll} PF_1^2 + PF_2^2 &=& (2c)^2 \\ 4 + 9 = 4c^2 \\ \mathbf{c^2} &=& \mathbf{\dfrac{13}{4} } \qquad \mathbf{c} = \mathbf{\dfrac{\sqrt{13}}{2}} \\ \end{array}\)

 

\(\begin{array}{rcll} c^2 &=& a^2 - b^2 \\ b^2 &=& a^2-c^2 \\ b^2 &=& 6.25-\dfrac{13}{4} \\ b^2 &=& \dfrac{4*6.25-13}{4} \\ \mathbf{b^2} &=& \mathbf{3} \qquad \mathbf{b} = \mathbf{\sqrt{3}} \\ \end{array}\)

 

Center M of the ellipse

\(\begin{array}{rcll} M &=& \left(\dfrac{PF_2}{2},~ \dfrac{PF_1}{2} \right) \\ M &=& \left(\dfrac{3}{2},~ \dfrac{2}{2} \right) \\ \mathbf{M} &=& \mathbf{(1.5,~ 1 )} \\ \end{array}\)

 

2. Ellipse in standard form, center in (0,0):

\(\begin{array}{rcll} \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} &=& 1 \quad | \quad a^2 = 6.26,~ b^2 = 3 \\\\ \dfrac{x^2}{6.25} + \dfrac{y^2}{3} &=& 1 \\ \end{array}\)

 

3. Translation to M and rotation with angle \(\varphi\):

\(\begin{array}{|lrcl|} \hline \text{move to }~M: \\ \quad \dfrac{(x-1.5)^2}{6.25} + \dfrac{(y-1)^2}{3} &=& 1 \\\\ \text{Rotation with $\varphi$}: \\ \quad \dfrac{ \Big( (x-1.5)*\cos(\varphi)-(y-1)*\sin(\varphi) \Big)^2}{6.25} + \dfrac{ \Big( (x-1.5)*\sin(\varphi)+(y-1)*\cos(\varphi) \Big)^2}{3} &=& 1 \\\\ \quad \boxed{ \sin(\varphi)=\dfrac{PF_1}{F_1F_2} \\ \sin(\varphi)=\dfrac{2}{2c} \\ \sin(\varphi)=\dfrac{1}{c} \\ \sin(\varphi)=\dfrac{2}{\sqrt{13}} \\ } \quad \boxed{ \cos(\varphi)=\dfrac{PF_2}{F_1F_2} \\ \cos(\varphi)=\dfrac{3}{2c} \\ \cos(\varphi)=\dfrac{3}{2}*\dfrac{2}{\sqrt{13}} \\ \cos(\varphi)=\dfrac{3}{\sqrt{13}} \\ } \\ \hline \end{array} \)

 

4. The ellipse intersects the x-axis. (We set y = 0):

\(\begin{array}{|lll|} \hline \dfrac{\Big( (x-1.5)*\dfrac{3}{\sqrt{13}}-(y-1)*\dfrac{2}{\sqrt{13}} \Big)^2}{6.25} + \dfrac{\Big( (x-1.5)*\dfrac{2}{\sqrt{13}}+(y-1)*\dfrac{3}{\sqrt{13}} \Big)^2}{3} = 1 \\\\ \boxed{y=0} \\ \dfrac{\Big( (x-1.5)*\dfrac{3}{\sqrt{13}}+\dfrac{2}{\sqrt{13}} \Big)^2}{6.25} + \dfrac{\Big( (x-1.5)*\dfrac{2}{\sqrt{13}}-\dfrac{3}{\sqrt{13}} \Big)^2}{3} = 1 \quad | \quad \times 3\times 6.25 \\\\ 3\Big[ \dfrac{1}{\sqrt{13}} \Big( 3(x-1.5)+2 \Big) \Big]^2 +6.25\Big[ \dfrac{1}{\sqrt{13}} \Big( 2(x-1.5)-3 \Big) \Big]^2 = 18.75 \\\\ \dfrac{3}{13}\Big( 3(x-1.5)+2 \Big)^2 +\dfrac{6.25}{13}\Big( 2(x-1.5)-3 \Big)^2 = 18.75 \quad | \quad \times 13 \\\\ 3\Big( 3(x-1.5)+2 \Big)^2 +6.25\Big( 2(x-1.5)-3 \Big)^2 = 243.75 \\\\ 3*\Big(9(x-1.5)^2+12(x-1.5)+4\Big) + 6.25*\Big(4(x-1.5)^2-12(x-1.5)+9\Big) = 243.75 \\\\ (27+4*6.25)(x-1.5)^2+(36-12*6.25)(x-1.5) +12+9*6.25 = 243.75 \\\\ \mathbf{52(x-1.5)^2-39(x-1.5)-175.5 = 0} \\\\ (x-1.5) = \dfrac{39\pm \sqrt{39^2-4*52*(-175.5)} }{2*52} \\\\ (x-1.5) = \dfrac{39\pm \sqrt{38025} }{104} \\\\ (x-1.5) = \dfrac{39\pm 195 }{104} \\\\ \boxed{ x_1-1.5 = \dfrac{39+195}{104} \\ x_1 -1.5 = 2.25 \\ x_1 = 2.25 + 1.5 \\ \mathbf{x_1=3.75} } \boxed{x_2-1.5 = \dfrac{39-195}{104} \\ x_2 -1.5 = -1.5 \\ x_2 = -1.5+1.5 \\ \mathbf{x_2=0} } \\ \hline \end{array}\)

 

The other point of intersection is \(x=3.75\)

 

laugh

 Jan 30, 2020
 #1
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Bravo on this solution! 

 Jan 30, 2020
 #3
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Thank you Guest,

 

laugh

heureka  Jan 30, 2020
 #2
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Thank you so so much for your hard work, heureka. I appreciate all of it. Thank you for your time and effort. <3

 Jan 30, 2020
 #4
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Thank you, DragonLord !

 

laugh

heureka  Jan 30, 2020

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