A maths

(1)

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(2)

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(3)

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1)
y=x^3-3x
y' = 3x^2 - 3
To make y' = 3x^2 - 3 = 9
x = -2 or x = 2
y = -2 or y = 2
There are 2 points on the curve (-2,-2) & (2,2) where the normals are parallel to the line x+9y-5=0.
At (-2,-2), the equation is
(y+2)/(x+2) = -1/9
9y+18 = -x-2
x+9y+20 = 0
At (2,2), the equation is
(y-2)/(x-2) = -1/9
9y-18 = -x+2
x+9y-20 = 0
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2)
C1 : 3y^2-8x-18y+48=0
6y(y')-8-18y' = 0
y' = 8/(6y-18) = 4/(3y-9)
At (3,4), y' = 4/3

C2 : x^2+y^2=25
2x+2y(y') = 0
y' = -2x/2y = -x/y
At (3,4), y' = -3/4

As the product of the y' = (4/3)(-3/4) = -1, the 2 tangents are perpendicular.
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3)
x=a cos^3 t
dx/dt = 3a cos^2 t (-sin t)
y = a sin^3 t
dy/dt = 3a sin^2 t (cos t)
So, dy/dx = -tan t = -1 at t = π/4
At t = π/4 , x = a(sqrt2)/4, y = a(sqrt2)/4
The tangent is
y - a(sqrt2)/4 = -1 (x - a(sqrt2)/4)
x+y - a(sqrt2)/2 = 0