The equations for the lines tangent to and normal to the curve at the point (2, 4) are y = 4 · x - 4 and y = (-1/4) · x + 9/2, respectively.
How to derive the equation of lines tangent and normal to a point of a curve
In this question we must use the definitions of linear function, tangent and normal lines and differential calculus to determine the equations needed. The slope of the tangent line (m) is found by implicit differentiation:
2 · x + 2 · y · y' - 3 · (y + x · y') = 0
2 · x + (2 · y - 3 · x) · y' - 3 · y = 0
(2 · y - 3 · x) · y' = 3 · y - 2 · x
y' = (3 · y - 2 · x)/(2 · y - 3 · x)
By (x, y) = (2, 4)
m = (3 · 4 - 2 · 2)/(2 · 4 - 3 · 2)
m = 4
And the slope of the normal line is:
m' = -1/m = -1/4
Lastly, we obtain the intercept for each line:
Tangent line
b = 4 - 4 · 2
b = 4 - 8
b = - 4
The equation of the line tangent to the curve at the point (2, 4) is y = 4 · x - 4.
Normal line
b = 4 - (-1/4) · 2
b = 4 + 1/2
b = 9/2
The equation of the line normal to the curve at the point (2, 4) is y = (-1/4) · x + 9/2.
To learn more on tangent lines: https://brainly.com/question/23265136
#SPJ1
