Thanks for answering my first question, hope you have time to help me with this one too.
+9468 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)
+128707 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
+128707 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 ]
