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A hard Algebra 2 Ellipse question. Thanks for the help.

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The figure below shows the ellipse $$\frac{(x-20)^2}{20}+\frac{(y-16)^2}{16}=2016$$.

Let $$R_1,R_2,R_3,$$ and $$R_4$$ denote those areas within the ellipse that are in the first, second, third, and fourth quadrants, respectively. Determine the value of $$R_1-R_2+R_3-R_4$$. Thanks!!!

Jan 30, 2020

#2
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The figure below shows the ellipse $$\dfrac{(x-20)^2}{20}+\dfrac{(y-16)^2}{16}=2016$$.

Let  $$R_1,~R_2,~R_3~$$and $$R_4$$ denote those areas within the ellipse that are in the first, second, third, and fourth quadrants, respectively.
Determine the value of $$R_1-R_2+R_3-R_4$$.

The center of the ellipse is $$(20,~16)$$

$$\text{Let the area of ellipse = A } \\ \text{Let the area of D = 20*16=320 }$$

$$\begin{array}{|rcll|} \hline R_1 &=& \dfrac{1}{4}A + B + D + E \\ R_2 &=& \dfrac{1}{4}A - B + C \\ R_3 &=& \dfrac{1}{4}A - C - D - F \\ R_4 &=& \dfrac{1}{4}A - E + F \\\\ R_1-R_2+R_3-R_4 &=& \dfrac{1}{4}A + B + D + E \\ && -\left(\dfrac{1}{4}A - B + C \right) \\ && + \dfrac{1}{4}A - C - D - F \\ && -\left(\dfrac{1}{4}A - E + F \right) \\\\ R_1-R_2+R_3-R_4 &=& \dfrac{1}{4}A + B + D + E \\ && -\dfrac{1}{4}A + B - C \\ && + \dfrac{1}{4}A - C - D - F \\ && -\dfrac{1}{4}A + E - F \\\\ R_1-R_2+R_3-R_4 &=& B + D + E \\ && B - C \\ && - C - D - F \\ && E - F \\\\ R_1-R_2+R_3-R_4 &=& 2(B - C + E - F ) \quad | \quad B = D+F,~ E=C+D \\ R_1-R_2+R_3-R_4 &=& 2(D+F - C + C+D - F ) \\ R_1-R_2+R_3-R_4 &=& 2(D+D) \\ R_1-R_2+R_3-R_4 &=& 2(2D) \\ R_1-R_2+R_3-R_4 &=& 4D \quad | \quad D =320 \\ R_1-R_2+R_3-R_4 &=& 4*320 \\ \mathbf{R_1-R_2+R_3-R_4} &=& \mathbf{1280} \\ \hline \end{array}$$

Jan 30, 2020
#3
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Oh, thats how you do it!!! Thank you!!! The person who posted this must be very good at math!!! Thank you for your time!

Jan 30, 2020
#4
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Thank you guys :D

Feb 4, 2020