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Thanks for answering my first question, hope you have time to help me with this one too. 

 

 May 12, 2019
 #1
avatar+9003 
+3

a)

x  =  2√2 cos θ

y  =  2 sin θ

 

x  =  2√2 cos θ

                            Square both sides of this equation

x2  =  8 cos2 θ

                            Divide both sides by  8.

x2 / 8  =  cos2 θ

 

y  =  2 sin θ

                                   Square both sides of this equation.

y2  =  4 sin2 θ

                                   By the Pythagorean identity, we can substitute  (1 - cos2θ)  in for  sin2θ

y2  =  4(1 - cos2θ)

                                   Substitute  x2 / 8  in for  cos2θ

y2  =  4(1 -  x2 / 8)

                                   Divide both sides of the equation by  4 .

y2 / 4  =  1 - x2 / 8

                                   Add  x2 / 8  to both sides of the equation.

x2 / 8  +  y2 / 4  =  1

 

Here is what the ellipse looks like:

https://www.desmos.com/calculator/o8fvqgkljx

 

the distance from the center of the ellipse to a focus  =  c

 

c2  =  a2 - b2  =  8 - 4  =  4

 

c  =  √4  =  2

 

the foci are located at  (0 + 2, 0)  and  (0 - 2, 0)

 

the foci are located at  (2, 0)  and  (-2, 0)

 May 12, 2019
 #2
avatar+111456 
+1

THX, hectictar  !!!!

 

Here's the "b"  part

 

Since the hperbola has the same foci its center is an equal distance from both foci = (0,0)....and one branch cuts the x axis at x  = 1 , then  a  =  1  and c = 2

Therefore....b  =   sqrt(c^2 - a^2)  =   sqrt (2^2 - 1^2)  = sqrt (4 - 1)  = sqrt (3)

So a^2  = 1  and b^2  = 3

 

So the equation of the hyperbola is

 

x^2          y^2

__    -     _____  =     1

 1              3

 

Here is the graph of the ellipse and hyperbola :  https://www.desmos.com/calculator/ur21asuwwk

 

 

 

cool cool cool

CPhill  May 12, 2019
 #3
avatar+111456 
+1

Here's the c part

 

Using implicit differentiation....the slope of a tangent line at any point on the ellipse is given by

 

(1/4)x + (1/2)y y'  = 0

(1/2)yy' = -(1/4)x

y' =  -(1/4)x / (1/2)y  =  -x / [ 2y]

 

And the slope of a tangent line to the hyperbola at any point is

2x - (2/3)yy' = 0

-(2/3)yy' = -2x

y' = 2x /[ (2/3)y ] =   3x / y

 

To find the itersection points.....we have that

 

(1/8)x^2 + (1/4)y^2  = 1    [multiply through by - 8 ] ⇒ - x^2 - 2y^2  = -8      (1)

x^2 - (1/3)y^2   = 1     (2)         add (1)  and (2) and we have that

 

-(7/3)y^2 = -7    [divide through by -7 ]

(1/3)y^2  = 1

y^2  = ±3

y = ±sqrt(3)

 

And using  (1)

 

-x^2 - 2(3) = -8

-x^2 - 6 = -8

x^2 + 6 = 8

x^2 = 2

x =±sqrt (2)

 

So.....the intersection points are

(sqrt(2), sqrt(3) )

(- sqrt(2),sqrt (3) )

(-sqrt(2), -sqrt(3) )

(sqrt (2), - sqrt(3))      

 

Testing the first point 

The slope of the tagent line to the ellipse at (sqrt(2), sqrt(3)  )   we have     -sqrt(2) / [2 sqrt(3)] = -1 / sqrt(6)

The slope of the tangent line to the hyperbola at this point is  3sqrt(2) / sqrt(3) = sqrt(6)

 

Since these are negative reciprocals.....the tangent lines are perpendicular

 

[I'll let you test the other three points ]

 

cool cool cool

CPhill  May 12, 2019
 #4
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0

Thank you so much for answering all parts of the question. 

 May 13, 2019

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