Prove that the gradient of the tangent to $xy=d^2$ is $-\frac{c^2}{4d^2}$

Question

I have this question:

Given that $y=mx+c$ is a tangent to $xy=d^2$ prove that $m=-\frac{c^2}{4d^2}$.

I'm not sure what direction to take - I tried differentiating the hyperbola equation, but that gave me a gradient of $-\frac{y}{x}$. How do I go about proving this?

Answer 1

An alternative method, we know the quadratic equation $\alpha r^2+\beta r+ \gamma=0$ has only one(repeated) root iff $\beta^2=4\alpha\gamma$.

Plug in $y=mx+c$ into $xy=d^2$ you get $mx^2+cx-d^2=0$, because tangent line intersects the curve at only 1 point we conclude $c^2=-4md^2$.

Answer 2

$xy = d^2 \Leftrightarrow y = \frac{d^2}{x}$.
Tangent line equation to $f(x)$ in point $x=a$ is
$g(x) = f(a) + f'(a)(x-a)$.
$f'(x) = -\frac{d^2}{x^2}$, so $g(x) = \frac{d^2}{a} - \frac{d^2}{a^2}(x-a) = -\frac{d^2}{a^2}x + \frac{2d^2}{a}$, hence $m = -\frac{d^2}{a^2}, c = \frac{2d^2}{a}$ and $m = -\frac{c^2}{4d^2}$.