Question

if the line x+y+k is a normal to the hyperbola xsquare/9-ysquare/4=1 then k=

Answer 1

I will assume you meant:
if the line x+y = k is a normal to the hyperbola x^2/9-y^2/4 = 1 then k=

your hyperbola is 4x^2 - 9y^2 = 36
8x - 18y dy/dx = 0
dy/dx = 8x/18y = 4x/9y
at any point on the hyperbola, the slope of a tangent is 4x/9y
so the slope of the corresponding normal would be -9y/4x

but the slope of x + y = k is -1
so -9y/4x = -1
9y = 4x
y = 4x/9
sub into 4x^2 - 9y^2 = 36
4x^2 - 9(16x^2/81) = 36
4x^2 - 16x^2/9 = 36
36x^2 - 16x^2 = 324
x^2 = 324/20 = 81/5
x = ± 9/√5
then y = ± 4x/9 = ±4(9/√5)/9 = ± 4/√5

but from my sketch , x+y = k can only be in quadrants I or III
using (9/√5 , 4/√5)
9/√5 + 4/√5 = k = 13/√5

x + y = 13/√5 in quad I
x + y = -13/√5 in quad III

checking with Wolfram:
www.wolframalpha.com/input/?i=solve+4x%5E2+-+9y%5E2+%3D+36+,+x+%2B+y+%3D+13%2F%E2%88%9A5

Wolfram has both graphed and solved my equations.