This question is incomplete, the complete question is;
Find an equation of the tangent plane to the surface z=1y²−1x² at the point (-4, 3, -7).
z = .....................
Note: Your answer should be an expression of x and y; e.g. 3x - 4y + 6.
Answer:
Z = 8x + 6y + 7
Step-by-step explanation:
Given that z = 1y² - 1x²
with points (-4, 3, -7)
so
Z = y² - x²
Equation of tangent plane to the surface Z = f(x,y) at the points ( x₀,y₀,z₀) is
Z - Z₀ = fx (x₀,y₀)(x-x₀) + fy (x₀y₀)(y-y₀)
therefore
Z = f(x,y) = y² - x²
fx = -2x, fx (-4, 3, -7) = -2(-4) = 8
fy = -2y, fy ( -4, 3, -7 ) = 6
Z + 7 = 8(x+4) + 6(y-3)
Z + 7 = 8x + 32 + 6y - 18
Z + 7 = 8x + 32 + 6y - 18
Z + 7 = 8x + 6y + 14
Z = 8x + 6y + 14 - 7
Z = 8x + 6y + 7
