Question

The equation of the common tangent to the curves $y^2=8x$ and $xy=-1$is

• A. $3y=9x+2$
• B. $y=2x+1$
• C. $2y=x+8$
• D. $y=x+2$

Answer 1

The correct option is D

$y=x+2$

Explantion for the correct options:

Step 1: Analysing the given equation and condition

We know that the equation of a tangent to the parabola $y^2=4ax$ is $y=mx+\frac{a}{m}$, where $m$is the slope of the tangent.

Here, in this question

$$\begin{gathered} y^2=8x \\ \Rightarrow y^2=4\times 2\times x \end{gathered}$$

Thus, $a=2$.

Thus, the equation of a tangent to the curve $y^2=8x$ is $y=mx+\frac{2}{m}$, where $m$is the slope of the tangent.

Now the tangent obtained is common to both the curves $y^2=8x$ and $xy=-1$.

So, the equation $y=mx+\frac{2}{m}$ must satisfy $xy=-1$.

Thus, we have

$$\begin{gathered} \Rightarrow xy=-1 \\ \Rightarrow x\left(mx+\frac{2}{m}\right)+1=0 \\ \Rightarrow m{x}^2+\frac{2}{m}x+1=0 \end{gathered}$$

Step 2: Finding the discriminant

This is a quadratic equation with $x$. As the tangent is common to $y^2=8x$ and $xy=-1$ and as the tangent can touch the curves at only one point, we must have the discriminant $D=0$.

Now the discriminant of the above quadratic equation $m{x}^2+\frac{2}{m}x+1=0$ is given by

$$\begin{gathered} \Rightarrow D={\left(\frac{2}{m}\right)}^2-4\times m\times 1 \\ \Rightarrow =\frac{4}{m^2}-4m \\ \Rightarrow =\frac{4-4m^3}{m^2} \end{gathered}$$

Substituting $D=0$

$$\begin{gathered} \Rightarrow \frac{4-4m^3}{m^2}=0 \\ \Rightarrow 4-4m^3=0 \\ \Rightarrow m^3=1 \\ \Rightarrow m=1 \end{gathered}$$

Step 3: Finding the equation of the common tangent

Thus, the required common tangent can be obtained by putting $m=1$ in the equation $y=mx+\frac{2}{m}$.

So, the equation of the required tangent is given by

$$\begin{gathered} \Rightarrow y=mx+\frac{2}{m} \\ \Rightarrow y=1\times x+\frac{2}{1} \\ \Rightarrow y=x+2 \end{gathered}$$

Hence, option (D) is the correct answer.