Answer:
Firstly you have to find the intersections of the two curves via simultaneous equations:
x^2 - 4y^2 = 5
x^2 = 5 + 4y^2
sub this into the other equation
4(5 + 4y^2) + 9y^2 = 45
20 + 16y^2 +9y^2 = 45
y^2 = 1
y = +1, -1
sub this back to get x
x^2 = 5 + 4y^2
x^2 = 5 + 4(+-1)^2
x^2 = 9
x = +3, -3
So the points of intersection are (3, 1), (3, -1), (-3, 1), (-3, -1)
You will now need to find the slope of tangents at these points on the two curves. To do this, you need to differentiate the two.
4x^2 + 9y^2 = 45
Differentiating both sides (implicitly)
8x + 18y.dy/dx = 0
dy/dx = -8x/18y
dy/dx = -4x/9y
The other one:
x^2 -4y^2 = 5
2x - 8y.dy/dx = 0
dy/dx = 2x/8y
dy/dx = x/4y
So taking the point (3, 1)
The slope at this point on the 1st curve would be
dy/dx = -4(3)/(9.1) = -12/9 = -4/3
The 2nd curve:
dy/dx = 3/4.1 = 3/4
Multiply these two you get: -4/3 x 3/4 = -1
Hence the slopes are at right angles.
Do the same for the other 3 points and you will come to the same conclusion,
