Question

Find the equation of all lines having slope 2 and being tangent to the curve $y+\dfrac{2}{x-3}=0$.

Answer 1

Differentiate the given function with respect to x using the formula $\dfrac{d\left( \dfrac{1}{x-a} \right)}{dx}=\dfrac{-1}{{{\left( x-a \right)}^{2}}}$, where a is any constant, and substitute the value $\dfrac{dy}{dx}=2$. Find the values of x and the respective values of y by substituting the values of x in the given curve. Assume the equation of the tangent line as $y=mx+c$ where m is the slope and c is the intercept. Now, consider the value of m = 2 and find the value of c for each set of solutions of (x, y) by considering different cases. Substitute the value of c in the equation of tangent lines to get the answer.

Complete step by step answer:
Here we have been provided with the curve $y+\dfrac{2}{x-3}=0$ and we are asked to find the equation of all the tangent lines to the curve whose slope is equal to 2.
Now, we know that the slope of a curve is the value of $\dfrac{dy}{dx}$, so differentiating the given curve with respect to x and using the formula $\dfrac{d\left( \dfrac{1}{x-a} \right)}{dx}=\dfrac{-1}{{{\left( x-a \right)}^{2}}}$, where a is any constant, we get,
$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}+\dfrac{d\left( \dfrac{2}{x-3} \right)}{dx}=0 \\\ & \Rightarrow \dfrac{dy}{dx}-\dfrac{2}{{{\left( x-3 \right)}^{2}}}=0 \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{2}{{{\left( x-3 \right)}^{2}}} \\\ \end{aligned}$
Substituting the given value of the slope $\dfrac{dy}{dx}=2$ we get,
$\begin{aligned} & \Rightarrow 2=\dfrac{2}{{{\left( x-3 \right)}^{2}}} \\\ & \Rightarrow {{\left( x-3 \right)}^{2}}=1 \\\ \end{aligned}$
Taking square root both the sides we get,
$\begin{aligned} & \Rightarrow x-3=\pm 1 \\\ & \Rightarrow x=3\pm 1 \\\ \end{aligned}$
(1) Considering the positive sign we have,
$\begin{aligned} & \Rightarrow x=3+1 \\\ & \Rightarrow x=4 \\\ \end{aligned}$
So substituting the value of x in the curve we get,
$\begin{aligned} & \Rightarrow y+\dfrac{2}{4-3}=0 \\\ & \Rightarrow y=-2 \\\ \end{aligned}$
That means one of the tangents to the curve is passing through the point $\left( 4,-2 \right)$. Assuming the equation of the tangent line as $y=mx+c$, where m is the slope and c is the y intercept, we have the m = 2 and the line passes through the point $\left( 4,-2 \right)$, so the point must satisfy the equation of the line, therefore we get,
$\begin{aligned} & \Rightarrow y=2x+c \\\ & \Rightarrow -2=2\left( 4 \right)+c \\\ & \Rightarrow c=-10 \\\ \end{aligned}$
Therefore the equation of one of the tangent lines is $y=2x-10$.
(2) Considering the negative sign we have,
$\begin{aligned} & \Rightarrow x=3-1 \\\ & \Rightarrow x=2 \\\ \end{aligned}$
So substituting the value of x in the curve we get,
$\begin{aligned} & \Rightarrow y+\dfrac{2}{2-3}=0 \\\ & \Rightarrow y=2 \\\ \end{aligned}$
That means the other tangent to the curve is passing through the point $\left( 2,2 \right)$. Again assuming the equation of the tangent line as $y=mx+c$ we have the m = 2 and this time the line passes through the point $\left( 2,2 \right)$, so the point must satisfy the equation of the line, therefore we get,
$\begin{aligned} & \Rightarrow y=2x+c \\\ & \Rightarrow 2=2\left( 2 \right)+c \\\ & \Rightarrow c=-2 \\\ \end{aligned}$
Therefore the equation of the other tangent line is $y=2x-2$.

Note: Note that the given curve is a rectangular hyperbola so if you want you can easily draw the graph of the curve with the tangent lines. We obtained two values of x and that is the reason we have obtained two tangents to the hyperbola. The value of x will never be equal to 3 because at x = 3 the value of y will be undefined and therefore the vertical line x = 3 is called an asymptote of the curve.