- #1
david18
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"the equation of a curve is x^2y=x-6 (x^2y is x squared times y)
The normal of the curve at the point where x=-2 meets the x-axis at p.
Find the coordinates of P"
The method I used was to find y by rearranging the given equation and differentiate it.
This gave me f ' (x)= 1+12x^-3
Then I subbed x=-2 in so that I could find the gradient of the tangent at that point. It worked out as -1/2, so then the gradient of the normal would be 2.
After finding the equation of the line and making y=0, I got the wrong answer: x=-1 and y=0
apparently the answer is (3/2, 0)
any help would be appreciated
The normal of the curve at the point where x=-2 meets the x-axis at p.
Find the coordinates of P"
The method I used was to find y by rearranging the given equation and differentiate it.
This gave me f ' (x)= 1+12x^-3
Then I subbed x=-2 in so that I could find the gradient of the tangent at that point. It worked out as -1/2, so then the gradient of the normal would be 2.
After finding the equation of the line and making y=0, I got the wrong answer: x=-1 and y=0
apparently the answer is (3/2, 0)
any help would be appreciated