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

Guest May 12, 2019

#1**+3 **

**a)**

x = 2√2 cos θ

y = 2 sin θ

x = 2√2 cos θ

Square both sides of this equation

x^{2} = 8 cos^{2} θ

Divide both sides by 8.

x^{2} / 8 = cos^{2} θ

y = 2 sin θ

Square both sides of this equation.

y^{2} = 4 sin^{2} θ

By the Pythagorean identity, we can substitute (1 - cos^{2}θ) in for sin^{2}θ

y^{2} = 4(1 - cos^{2}θ)

Substitute x^{2} / 8 in for cos^{2}θ

y^{2} = 4(1 - x^{2} / 8)

Divide both sides of the equation by 4 .

y^{2} / 4 = 1 - x^{2} / 8

Add x^{2} / 8 to both sides of the equation.

x^{2} / 8 + y^{2} / 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

c^{2} = a^{2} - b^{2} = 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)

hectictar May 12, 2019

#2**+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

CPhill
May 12, 2019

#3**+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 ]

CPhill
May 12, 2019