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$xy(x - 6y) = 9a^3$ , a is not 0

The Question

Show that there is only one point on the curve at which the tangent is parallel to the x axis, and find the coordinate of this point.

How would I go about solving this question?

codez
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1 Answers1

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We have

$(*)$ $x^2y(x)-6xy(x)^2=9a^3$.

Implicit differentiation yields

$2xy(x)+x^2y'(x)-6y(x)^2-12xy'(x)y(x)=0$.

If $x_0$ is a point in the domain of $y$ with $y'(x_0)=0$, then

$2x_0y(x_0)-6y(x_0)^2=0$. Thus $y(x_0)=0$ or $y(x_0)=\frac{1}{3}x_0$

By $(*)$ we see that $y(x_0)=0$ is impossible, since $ a \ne 0$. Thus $y(x_0)=\frac{1}{3}x_0$.

Now it is your turn to show that $x_0=-3a$ and $y(x_0)=-a$.

Fred
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  • Look's like the question is out of scope for me. I am an A Level student. – codez Nov 29 '16 at 12:39
  • What are your problems ? – Fred Nov 29 '16 at 12:42
  • I sorta don't understand anything after the second line. – codez Nov 29 '16 at 13:43
  • @HassanAlthaf: just explaining a bit more. $y'(x_0)$ represent the inclination of the tangent at the point $x_0$, so once the tangent is parallel to the $x$ axis then its inclination must be zero, that's why $y'(x_0)=0$. The rest is calculation. – Arnaldo Nov 29 '16 at 14:35