Question 1179985: Find the x-coordinates of the points P and Q on y = (x − 7)^2 + 3 such that the tangents at P and Q
have gradients 1 and −1 respectively.
b: Show that the square formed by the tangents and normals at P and Q has area 1/2
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
I will describe the process for working the problem and let you do most of the actual work....
Draw a rough sketch of the graph -- it is a parabola that opens upward with vertex (7,3).
(1) Find the derivative of the function: dy/dx = 2(x-7)
(2) Set the derivative equal to 1 and solve the resulting equation to find the x value where the slope is 1; set it to -1 and solve to find the x value where the slope is -1
That answers the first question....
(3) Sketch the tangents at those two points
(4) Determine the y values at the two points of tangency (they should be the same)
(5) Determine the point of intersection of the two tangents (it should be on the axis of symmetry of the parabola, where x=7)
(6) Determine the side length of the square formed by the tangents and the normals at the points of tangency; that side length is the distance from each point of tangency to the intersection of the two tangents
(7) Square that side length of the square to verify that the area of the square is 1/2
If you need more help, re-post your question showing what work you have done and telling us what it is that is giving you difficulty
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