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1. An ellipse has its center at the origin, its foci on the y-axis, and its major axis is three times as long as its minor axis. Given that the ellipse passes through the point (-4, 0), find its equation.

 

2. What is the minimum distance between a point on the circle x^2+y^2=16 and a point on the line x-y=8.

 

3. Find the foci of the ellipse whose major axis has endpoints (0,0) and (13,0) and whose minor axis has length 12.

 

4. A circle centered at the origin intersects the ellipse \(\frac{y^2}{16} + \frac{x^2}{9} = 1\) at the four vertices of a square. Find the area of that square.

 

5. Square ABCD has side length 60. An ellipse E is circumscribed about the square and there is a point P on the ellipse such that PC = PD =50. What is the area of E?

 

NOTE: It is not neccesary to solve all problems, any help would be appreciated. Thanks!

somebody  Nov 17, 2018
edited by somebody  Nov 17, 2018
 #1
avatar+9738 
+2

1. An ellipse has its center at the origin, its foci on the y-axis, and its major axis is three times as long as its minor axis. Given that the ellipse passes through the point (-4, 0), find its equation.


 

2. What is the minimum distance between a point on the circle x^2+y^2=16 and a point on the line x-y=8.

 

laugh

Omi67  Nov 17, 2018
edited by Omi67  Nov 17, 2018
 #2
avatar+92785 
+1

3. Find the foci of the ellipse whose major axis has endpoints (0,0) and (13,0) and whose minor axis has length 12.

 

We have this equation

 

(x - h)^2            (y - k)^2

_______    +   ________   =      1

   a^2                  b^2

 

The center of the ellipse is   at  (6.5, 0)     =  ( h, k)

The major axis is along the x axis.....its length is 13

"a" is 1/2 of this 6.5 

The minor axis is along y

And b is 1/2 of this  =  6

 

The y coordinate of both foci lies on the x axis

 

And the foci are given by   (  ±√ [a^2 - b^2 ] + h , 0 )    =   ( ±√ [6.5^2 - 6^2] + 6.5 , 0)   =

 

( 2.5 + 6.5, 0 )     and   (-2.5 + 6.5)    =

 

(9, 0)    and  ( 4, 0)

 

Here's the graph : https://www.desmos.com/calculator/hxa6eulntu

 

 

cool cool cool

CPhill  Nov 18, 2018
 #3
avatar+92785 
+1

4. A circle centered at the origin intersects the ellipse  x^2/9 + y^2/16 = 1   at the four vertices of a square. Find the area of that square.

 

Since the diagonals of a square are at right angles and the center of the elliipse is at (0,,0).....the lines y = x  and y = -x    will form diagonals that intersect at (0,0) and will be at right angles to each other

 

To find out where  y = x intersects the ellipse.....let us solve this

 

[ sub x for y  in the ellipse equation ]

 

x^2/9 + x^2/16  = 1       multiply through by 144

 

16x^2 + 9x^2  = 144

 

25x^2  = 144

 

x^2   =  144 / 25    take both roots

 

x = ±√ [144/25]   = ±12/5   =  ±2.4

 

And siince  y = x     and y = -x the coordinates  of the top  vertices of the square are

 

(2.4, 2.4) , ( -2.4, 2.4)

 

The distance between the points  is 4.8  =  the side of the square 

 

So....the area of the square  is  side^2   = 4.8^2   = 23.04 units^2

 

Here's a graph : https://www.desmos.com/calculator/uld9mzxaet

 

 

cool cool cool

CPhill  Nov 18, 2018
 #4
avatar+92785 
+2

5. Square ABCD has side length 60. An ellipse E is circumscribed about the square and there is a point P on the ellipse such that PC = PD =50. What is the area of E?

 

This one is a little difficult !!!!

 

Let the circle and the ellipse be centered at the origin

 

We can let the vertices of the square be  A = (-30,30) , B =(30,30), C = (30, -30)  and D = ( -30, -30)

 

The mid point of the bottom of the square is the point (-30, 0)

Call this point, M

And  PC = PD = 50

So....we can form right triangle MPC  such that PC forms the hypotenuse = 50

And MC = 30

So....MP  is the other leg  =  √ [ PC^2 - MC^2] = √[ 50^2 - 30^2]  = √ [ 2500 - 900 ] =

√1600 = 40

 

So......the distance from the origin to M  = 30

And the distance between M an P  =  40

So....using symmetry.....we can let the vertical axis of the ellipse  = 2* (30 + 40) = 140  = 2b

So......in the equation

 

x^2            y^2

___   +     ___    =  1

 a^2          b^2

 

We know that one point on the ellipse is (30, 30)  = (x , y)

And "b"  is (1/2) * 140  = 70

 

So.....we can find "a" as follows

 

30^2           30^2

_____  +    _____   =      1

  a^2            70^2

 

900            900

____ +      _____  =    1

a^2            4900

 

900         9

___  +   ____  =    1

a^2         49

 

900                   9

____  =   1  -    ___ 

 a^2                   49

 

900             [ 49 - 9]

___  =       _________

a^2                 49

 

900                 40

____   =        ____

 a^2                49

 

So

 

40 a^2 =  900*49

40a^2 = 44100

a^2  =  44100 / 40

a^2 = 4410/4

a = √ [ 4410 / 4 ]

 

And  the area of the ellipse   =

 

pi* a * b  =

 

pi * √[ 4410/ 4} * 70  ≈  7301.9   units^2

 

Here's the graph : https://www.desmos.com/calculator/r9mwhqlg2e

 

 

cool cool cool

CPhill  Nov 18, 2018

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