Question

A tangent to the hyperbola $y=\frac{x+9}{x+5}$ passing though the origin is

• A. $x+25y=0$
• B. $5x+y=0$
• C. $5x-y=0$
• D. $x-25y=0$

Answer 1

Answer: (A)

$x+25y=0$

$y=\frac{x+9}{x+5}=1+\frac{4}{x+5}$

$\frac{dy}{dx}$ at $(x_1,y_1)=-\frac{4}{{(x_1+5)}^2}$

Equation of tangent is

$y-y_1=-\frac{4}{{(x_1+5)}^2}(x-x_1)\Rightarrow y-1-\frac{4}{x_1+5}=-\frac{4}{{(x_1+5)}^2}(x-x_1)$

Since it passes through origin $(0,0)$

$-1-\frac{4}{x_1+5}=-\frac{4}{{(x_1+5)}^2}\Rightarrow {(x_1+5)}^2+4(x_1+5)+4x_1=0$

$\Rightarrow {x_1}^2+18x_1+45=0\Rightarrow (x_1+15)(x_1+3)=0\Rightarrow x_1=15$ or $x_1=-3$

So equation of tangent is $y-1-\frac{4}{(-15+5)}=-\frac{4}{{(-15+5)}^2}(x+15)$

$\Rightarrow y-1+\frac{2}{5}=-\frac{1}{25}(x+15)\Rightarrow y-\frac{3}{5}=-\frac{x}{25}-\frac{3}{5}\Rightarrow x+25y=0$