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 ]