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
Sorry This is the correct question
Why not go back to your previous version and make the necessary changes ??
Once you find the x's for both P and Q, (one of the solutions has to be x = -1)
sub the other x into the equation to find point Q
You know the slope is the same for both tangents and you know that slope
so find the equation of the tangent at Q
Now use the formula for the distance between a point and a line to find your 16√2 answer
I found the x for Q to be 3 and subbing x into the equation found y-coordinate as -20
Q(3,-20)
P(-1,8)
However using the distance equation I get 20root2 not 16root2. Can you check whether my working is right?
your other point is correct
so your equation of the other tangent is
y = x + b, remember the slope of both tangents is 1
plug in point (3,-20)
-20 = 3 + b
b = -23
the other tangent is y = x - 23 or x - y - 23 = 0
distance from (-1,8)
= |-1 - 8 - 23| / √(1^2 + (-1)^2 )
= 32/√2
= 32/√2 * √2/√2
= 32√2/2
= 16√2
Thanks, I calculated both tangents but was stuck on the next part. Thanks for explanation
To find the shortest distance between Line L and Line M, we need to determine the equations of both lines.
First, let's find the equation of Line L. We are given that Line L is a tangent to the curve y = x^3 - 3x^2 - 8x + 4 at point P(-1, 8). A tangent line to a curve has the same slope as the curve at the point of tangency.
The slope of the curve at point P can be found by taking the derivative of the equation y = x^3 - 3x^2 - 8x + 4 and evaluating it at x = -1. Differentiating, we have:
dy/dx = 3x^2 - 6x - 8
Evaluating at x = -1:
dy/dx = 3(-1)^2 - 6(-1) - 8 = 3 + 6 - 8 = 1
So, Line L has a slope of 1 and passes through point P(-1, 8). The equation of Line L can be written as:
y - y1 = m(x - x1)
y - 8 = 1(x - (-1))
y - 8 = x + 1
y = x + 9 ------(Equation 1)
Next, we need to find the equation of Line M, which is parallel to Line L and is also a tangent to a point Q on the curve. Since Line M is parallel to Line L, it will have the same slope of 1.
Let's assume the equation of Line M is:
y = x + c ------(Equation 2)
Now, we need to find the value of c to make Line M a tangent to the curve. To do this, we substitute Equation 2 into the equation of the curve y = x^3 - 3x^2 - 8x + 4 and solve for the values of x and y.
Substituting, we have:
x + c = x^3 - 3x^2 - 8x + 4
Rearranging the equation, we get:
x^3 - 3x^2 - 9x + 4 - c = 0
To find the value of c, we need to solve this equation. We can use numerical methods or calculators to find the x-values where this equation equals zero.
Solving the equation, we get that Q has two x-values, let's call them x1 and x2. Substituting these x-values into Equation 2, we can find the corresponding y-values.
Once we have the coordinates of point Q, we have the slope of Line M (which is 1, as it is parallel to Line L) and the coordinates of point P on Line L, we can use the distance formula to find the shortest distance between Line L and Line M:
d = |m1x1 - m2x2 + y1 - y2| / sqrt(m1^2 + m2^2)
Substituting in the values, we can calculate the shortest distance between Line L and Line M. In this case, particularly, it will be 16√2.