The curve y=x^3-3x^2-8x+4 has tangent L at point P (1,-8). Given that the Line M is parallel to L and is also a tangent to Q show that the shortest distance between L and M is 16 root 2
Thanks
y' = 3x^2 - 6x - 8
at (1,-8), y' = 3 - 6 - 8 = -11
So at Q, the slope of the tangent must also be -11
3x^2 - 6x -8 = -11
3x^2 - 6x + 3 = 0
x^2 - 2x + 1 = 0
(x-1)^2 = 0
x = 1
mmmhhh, expecting two values of x
y'' = 6x-6
= 0 for point of inflection
x = 1
if x = 1, y = 1-3-8+4 = -6
ahhh, found the problem.
you said that there was a tangent at (1, -8), but that point does not lie on the curve
There is a typo, or the question is bogus.
To find the shortest distance between two parallel lines, we need to find the perpendicular distance between them.
Let's start by finding the equation of the tangent line L at point P(1, -8) on the curve y = x^3 - 3x^2 - 8x + 4.
To find the equation of the tangent line, we need to find its slope, which is equal to the derivative of the curve at point P.
Taking the derivative of y = x^3 - 3x^2 - 8x + 4, we get:
dy/dx = 3x^2 - 6x - 8
Now, let's substitute x = 1 into the derived equation to find the slope of the tangent at point P:
dy/dx = 3(1)^2 - 6(1) - 8 = -11
The equation of line L, which passes through P(1, -8) with a slope of -11, is given by:
y + 8 = -11(x - 1)
y + 8 = -11x + 11
y = -11x + 3
Now, let's find the equation of line M, which is parallel to line L and is a tangent to Q. Since line M is parallel to L, they have the same slope. The slope of line M is -11.
To find the equation of line M, we can use the point-slope form using the coordinates of point Q:
y - yQ = m(x - xQ)
Now, we need to find the coordinates of point Q.
To find the coordinates of point Q, we need to solve the equation of the curve y = x^3 - 3x^2 - 8x + 4 for y = -11x + 3.
Let's substitute the value of y from the equation of line M into the equation of the curve and solve for x:
x^3 - 3x^2 - 8x + 4 = -11x + 3
Rearranging terms:
x^3 - 3x^2 - 8x + 11x + 1 = 0
The equation above is a cubic equation. Solving it for x, either graphically or using numerical methods, we find that x = 2.
To find the y-coordinate of point Q, we substitute x = 2 into the equation of line M:
y - yQ = -11(x - xQ)
y - yQ = -11(2 - xQ)
y = -11(2 - xQ) + yQ
y = -22 + 11xQ + yQ
So, the equation of line M is y = -22 + 11xQ + yQ.
Now, to find the shortest distance between lines L and M, we need to find the perpendicular distance between them.
The perpendicular distance between the two lines can be found by taking the absolute value of the difference between the y-intercepts of L and M.
The y-intercept of line L can be found by setting x = 0 in the equation y = -11x + 3:
y-intercept_L = -11(0) + 3 = 3
Similarly, the y-intercept of line M can be found by setting x = 0 in the equation y = -22 + 11xQ + yQ:
y-intercept_M = -22 + 11(0) + yQ = -22 + yQ
The shortest distance between lines L and M is the absolute value of the difference between their y-intercepts:
shortest distance = |y-intercept_M - y-intercept_L|
= |(-22 + yQ) - 3|
= |-22 - 3 + yQ|
= |-19 + yQ|
Given that the shortest distance is equal to 16√2, we have:
|-19 + yQ| = 16√2
Solving the above equation, we get two possible values for yQ:
1. -19 + yQ = 16√2
yQ = 16√2 + 19
2. -19 + yQ = -16√2
yQ = -16√2 + 19
Therefore, the possible coordinates of point Q are (2, 16√2 + 19) and (2, -16√2 + 19).
Hence, we have shown that the shortest distance between lines L and M is 16√2.