The curve C has equation: (x-y)^2 = 6x +5y -4. Use Implicit differentiation to find dy/dx in terms of x and y. The point B with coordinates (4, 2) lies on C. The normal to C at B meets the x-axis at point A. Find the x-coordinate of A.
We start off by differentiating the equation implicitly which will give us:2(x-y) -2(x-y)dy/dx = 6 + 5dy/dxThen we rearrange to get dy/dx on it's own:dy/dx = (2x-2y-6)/(2x-2y+5)
For the second part of the question we must find the gradient of the tangent at point B, so M = dy/dx at (4,2). Thus M = -2/9. We then find the gradient of the normal which is the negative inverse of M = 9/2. Then we use the equation y-2 = 9/2 *(x-4) but let y=0 since we are finding the x-coordinate of A. Doing this, x must equal 32/9 in order for the equation to be satisfied.
JM
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