# 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 $(b,&space;f&space;(b))$ where $a\neq&space;b$.

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

At vertices, the tangents are parallel.
$f&space;'&space;(a)&space;=&space;f&space;'&space;(b).$
$-\frac{1}{{{(a+5)}^{2}}}=-\frac{1}{{{(b+5)}^{2}}}$
$(a&space;+&space;5)^2&space;=&space;(b&space;+&space;5)^2$
$a^2&space;+&space;10a&space;+&space;25&space;=&space;b^2&space;+&space;10b&space;+&space;25$
$a^2-b^2+10(a-b)&space;=&space;0$
$(a-b)&space;(a&space;+&space;b&space;+&space;10)&space;=&space;0$
Since $a\neq&space;b,&space;a-b\neq&space;0$
$\therefore&space;a&space;+&space;b&space;+&space;10&space;=&space;0$ and $b&space;=&space;-a-10$.
$\therefore&space;f&space;(b)&space;=&space;\frac{-2b-9}{b+5}=&space;\frac{2a+11}{-(a+5)}.$
Since the line passing through vertices is perpendicular to the respective tangents, its gradient is  $(a+5)^2$.
Hence $\frac{f(a)-f(b)}{a-b}=&space;(a&space;+&space;5)^2$
$\frac{\frac{-2a-9}{a+5}+\frac{2a+11}{a+5}}{2a+10}=&space;(a&space;+&space;5)^2$
$\frac{1}{{{(a+5)}^{2}}}=&space;(a&space;+&space;5)^2$
$(a&space;+&space;5)^4&space;=&space;1$
$a&space;+&space;5&space;=&space;\pm1$
$a&space;=-4$ (or) $a&space;=-6$
$\therefore&space;b=-6$ (or) $\therefore&space;b=-4$
$\therefore&space;f(a)=-1$ (or) $f(a)=-3$ and
$f(b)=-3$ (or) $f(b)=-1$
Therefore, the vertices are (– 4, – 1) and (– 6, – 3).
Hence the equation of required line is

$y-(-1)=\frac{-1-(-3)}{-4-(-6)}(x-(-4))$
$y&space;+&space;1&space;=&space;x&space;+&space;4$ (or) $y&space;=&space;x&space;+&space;3$