Show that the gradient of the curve

[studentxlol]

Homework Statement

Show that the gradient of the curve $$\frac{a}{x}+\frac{b}{y}=1$$ is $$-\frac{ay^2}{bx^2}$$. The point (p,q) lies on both the straight line [itex]ax+by=1[/tex] and $$\frac{a}{x}+\frac{b}{y}=1$$ where ab =/= 0. Given that, at this point, the line and the curve have the same gradient, show that p=±q.<br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> $$\frac{a}{x}+\frac{b}{y}=1$$<br /> <br /> $$\frac{dy}{dx}=-ax^{-2}-by^{-2}$$<br /> <br /> $$-\frac{b}{y^2}\frac{dy}{dx}=\frac{a}{x^2}$$ <br /> <br /> $$\frac{dy}{dx}=-\frac{ay^2}{bx^2}$$<br /> <h2>Homework Statement </h2><br /> <br /> Not sure how to calculate the next part.. <br /> <br /> <br /> <br /> <h2>Homework Equations</h2><br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2>[/itex]

 

[grzz]

You already did the difficult part.

Now find the gradient of the straight line and equate the two gradients.

The put x = p and y = q.