# Differential Calculus: center, vertices, foci of the ellipse: 16x² + 25y² - 160x - 200y + 400 = 0

Submitted by darbz11 on August 8, 2016 - 9:42pm

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Find the center, vertices, foci of the ellipse given the equation 16x² +25y² -160x - 200y + 400 = 0.

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## This is not really calculus.

This is not really calculus. The question might have been answered more promptly, if asked in the algebra section. You also had a fair chance of an answer in the algebra . com website.

16x² +25y² -160x - 200y + 400 = 0

16x² -160x+25y² - 200y = - 400

16(x² -10x)+25(y² - 8y) = - 400

16(x² -10x)+16(25) +25(y² - 8y) +25(16)= - 400+16(25)+25(16)

16(x² -10x+25)+25(y² - 8y+16) = 400

16(x-5)

^{2}+25(y-4)^{2}= 40016(x-5)

^{2}/400+25(y-4)^{2}/400 = 1(x-5)

^{2}/25+(y-4)^{2}/16 = 1(x-5)

^{2}/5^{2}+(y-4)^{2}/4^{2}= 1You can see that the last equation above represents an ellipse

centered at (5,4),

with horizontal major axis,

a semi-major axis a=5 ,

and a semi-minor axis b=4 .

That means that the vertices, at the ends of the major axis, are

(5-5,4) =(0,4) and (5+5,4) = (10,4) .

The co-vertices, at the end of the minor axis, are

(5,4-4) = (5,0) and (5,4+4) = (5,8) .

The foci are on the major axis, between the center and the vertices,

at (5-c,4) and (5+c,4) .

All that is left to do is find the focal distance, c .

For that you may remember that in an ellipse a

^{2}= b^{2}+ c^{2},or you may remember the definition of ellipse, and deduce the formula yourself.

In this case, substituting the values found for a and b,

5

^{2}= 4^{2}+ c^{2}, or25 = 16 + c

^{2}---> c^{2}= 25-16 = 9 ---> c = 3 .Then, the foci are at

(5-3,4) = (2,4) and (5+3,4) = (8,4) .

## Using Calculus for Vertices

Using Calculus for Vertices

$16x^2 + 25y^2 - 160x - 200y + 400 = 0$

For Upper and Lower Vertices

At the highest and lowest points, y' = 0

$32x - 160 = 0$

$x = 5$

$16(5^2) + 25y^2 - 160(5) - 200y + 400 = 0$

$25y^2 - 200y = 0$

$y = 8 ~ \text{and} ~ 0$

upper vertex = (5, 8)

lower vertex = (5, 0)

For Left and Right Vertices

At the extreme left and extreme right points, x' = 0

$50y - 200 = 0$

$y = 4$

$16x^2 + 25(4^2) - 160x - 200(4) + 400 = 0$

$16x^2 - 160x = 0$

$x = 10 ~ \text{and} ~ 0$

right vertex = (10, 4)

left vertex = (0, 4)

Below is the plot to locate the center and other points:

## Sir, pwede din bang gamitin

Sir, pwede din bang gamitin ang y' = infinity para kunin ang left and right vertices? In this way kasi, isa na lang ang kukunin, yung y' na lang, hindi na kailangan ang x'.

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