Find an equation of the tangent plane to the surface at the point . Question 14 options: a) b) c) d) e)

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.

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

ye

step-by-step explanation: