Why is the Coordinate of Point P (3/2, 0) in this Differentiation Problem?

[david18]

"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

 

[sadhu]

your derivative is wrong
get it right , you will have your answer.

 

[EnumaElish]

If you rearrange the equation to y = (x - 6)/x^2, the derivative is (12-x)/x^3.