How can I find a tangent having a slope of $-\frac{3}{4}$, to the curve $xy = 18$ and the co-ordinates of the point of tangency?

Question

Assuming the tangent to be $-y = -\frac{3}{4}x + c$, which is tangent to $xy = 18$, how do I find $c$?

Answer 1

First of all we should remember the equation of the line tangent at $x_0$ to the graph of a function $f$, that is $y=f'(x_0)(x-x_0)+f(x_0)$. So its slope is given by $f'(x_0)$.

Your function is $f(x)=18/x$, and therefore $f'(x)=-18/x^2$. Solve the equation $-18/x^2=-3/4$ and you will find two values, $x_1=2\sqrt{6}$ and $x_2=-2\sqrt{6}$. In fact, here there are two tangent lines with slope $-3/4$: for the first one the point of tangency is $(x_1,f(x_1))=(2\sqrt{6},3\sqrt{3/2})$ and for the second one its is $(x_2,f(x_2))=(-2\sqrt{6},-3\sqrt{3/2})$.

Answer 2

Tangent to $xy=k^2$ at $(kt, k/t)$ is $x/t + ty = 2k$. Slope is $-1/t^2 = -\frac{3}{4}$ and hence $t = \pm \frac{2}{\sqrt{3}}$. Hence $c = 2k/t = \pm 2\sqrt{18}\frac{\sqrt{3}}{2} = \pm \sqrt{54}$