Problem Study (Analytic Geometry)

Find the equation of the line passing through the vertices of this curve $f(x)=\frac{-2x-9}{x+5}$ .

Solution
$f(x)=\frac{-2x-9}{x+5}$

Let the vertices be (a,f(a))

and

where $a\neq b$ .

∴ $f (a) = \frac{-2a-9}{a+5}$ and $f (b) = \frac{-2b-9}{b+5}$
The gradient of tangent to the curve is $f ' (x) =\frac{1}{{{(x+5)}^{2}}}.$

At vertices, the tangents are parallel.
∴ f ' (a) = f ' (b).

$-\frac{1}{{{(a+5)}^{2}}}=-\frac{1}{{{(b+5)}^{2}}}$
(a + 5)^2 = (b + 5)^2

Since $a\neq b, a-b\neq 0$ .
$\therefore a + b + 10 = 0$ and b = -a-10

.
$\therefore f (b) = \frac{-2b-9}{b+5}= \frac{2a+11}{-(a+5)}.$
Since the line passing through vertices is perpendicular to the respective tangents, its gradient is (a+5)^2

.
Hence $\frac{f(a)-f(b)}{a-b}= (a + 5)^2$
   $\frac{\frac{-2a-9}{a+5}+\frac{2a+11}{a+5}}{2a+10}= (a + 5)^2$
   $\frac{1}{{{(a+5)}^{2}}}= (a + 5)^2$
   (a + 5)^4 = 1