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Consider the point P(-1, 0) on the ellipse given by the equation 4x^2 + y^2 = 4. There are two points (a, b) and (a, c) on the ellipse whose distance from P is a maximum. What is the value of a? 

 Jan 11, 2020
 #1
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The value of a is 1/6.

 Jan 13, 2020
 #2
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Consider the point P(-1, 0) on the ellipse given by the equation 4x^2 + y^2 = 4. There are two points (a, b) and (a, c) on the ellipse whose distance (D) from P is a maximum. What is the value of a? 

 

Betrachten Sie den Punkt P (-1, 0) auf der Ellipse, die durch die Gleichung 4x ^ 2 + y ^ 2 = 4 gegeben ist. Es gibt zwei Punkte (a, b) und (a, c) auf der Ellipse, deren Distanz (D) von P ist maximal. Was ist der Wert von a?

 

Hello Guest!

 

Equation of the ellipse  Gleichung der Ellipse

\( 4x^2 + y^2 = 4\\ y=\sqrt{4-4x^2}\\ \color{blue}y=2 \sqrt{1-x^2}\)

 

Distance D to the ellipse                              Distanz D zur Ellipse                                  

Here the size a = x, the size b = y.               Hier ist die Größe a = x, die Größe b = y.

\(\color{blue}y=2\sqrt{1-x^2}\\ y^2=4(1-x^2)\)

 

\(D=\sqrt{y^2+x^2}\\D=\sqrt{4(1-x^2)+(x+1)^2}\\ D=\sqrt{4-4x^2+x^2+2x+1}\\ \color{blue}D=f(x)=\sqrt{-3x^2+2x+1}\)               Pythagoras

 

\({\large \frac{dD}{dx}}=D'=\frac{1}{2}(-3x^2+2x+5)^{-\frac{1}{2}}\cdot (-6x+2)=0\\ D'=\Large \frac{-3x+1}{\sqrt{-3x^2+2x+5}}=0\)   

                       \(\large -3x+1=0\)

\(\Large x=a=\frac{1}{3}\)    

 \(\)\(The\ value\ of\ a\ is\ \large\frac{1}{3}.\)

laugh  !

 Jan 13, 2020
edited by asinus  Jan 14, 2020

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