find an equation of the tangent line to the graph of x^2/2 + y^2/3 = 5 at the point (8,1)

Answer 1

The equation of the curve is x²/2 + y²/3 = 5

Differentiating  on each side with respect of x .

d/dx(x²/2 + y²/3) =d/dx(5)

(2x/2) + (2y/3)dy/dx = 0

x + (2y/3)dy/dx = 0

dy/dx = -3x/2y

Substitute the values (x , y ) = (8,1) in above equation.

dy/dx = -3(8)/2(1) = -12

This is the slope of tangent line to the curve at (8, 1).

m = -12

To find the tangent line equation, substitute the values of m = -12 and (x, y ) = (8, 1).  in the slope intercept form of an equation.

1 = -12 (8) + b

b = 97

Substitute m = -12 and b = 97 in y = mx + b.

y = -12x + 97

Tangent line equation is  y = -12x + 97