Question

a parabola has a focus of (-3,4) and a directrix with equation: 3x-4y-25. Since the distance from the focus to a point (x,y) on the parabola is : squareroot (x+3)^2 +(7-4)^2 and is equal to the distance from the point (x.y) to the line which is 16x^2 + 9y^2 +300x - 400y +24xy=0. do the arithmetic to rotate this curve into recognizable form of a parabolawith axis of symmetry either veritical or horizontal.

Answer 1

First of all, I agree with your equation.

This is a pretty advanced topic for this forum, I will get you going, but will not be able to type the necessary matrices. It does not allow the proper formatting and doesn't line up.

Have you come across this:

The xy term can be eliminated from the equation
ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0
by a rotation through an angle ß, where
tan 2ß = 2h/(b-a)

so in your case of
16x^2 + 9y^2 + 300x - 400y + 24xy = 0
a=16
b=9
h=12

tan 2ß = 24/(9-16) = -24/7
(for reference angle:
2ß= 73.7º)
since the tangent is negative, 2ß lies in the second or fourth quadrants.
Let 2ß lie in the second quadrant so that 2ß = 180-73.7 = 106.3º
so ß = 53.15

So your required rotation matrix is
│cos 53.15 sin 53.15│
│-sin53.15 cos 53.15│

Does any of this make sense ?