Question

If the line $ax+by+c=0$ is a tangent to the curve $xy=4$, then

• A. $a<0,b>0$
• B. $a\le 0,b>0$
• C. $a<0,b<0$
• D. $a\le 0,b<0$

Answer 1

The correct option is C

$a<0,b<0$

Explanation for the correct option

Step 1: Solve for the slopes of line and curve

Given that the line $ax+by+c=0$ is a tangent to the curve $xy=4$

The first order derivative of the equation of the curve at a point gives the slope of the tangent to the curve at that point.

Differentiating the equation of the curve with respect to $x$ we get

$1.y+x\frac{dy}{dx}=0$ $\left[\because \frac{d}{dx}\left(U·V\right)=V·\frac{dU}{dx}+U·\frac{dV}{dx}\right]$

$\Rightarrow$ $\frac{dy}{dx}=\frac{-y}{x}$

From the equation of the curve $xy=4$ we get

$y=\frac{4}{x}$

$\Rightarrow$$\frac{dy}{dx}=\frac{-4}{x^2}$

Slope of a line $=\frac{-\left(coefficientofx\right)}{coefficientofy}$

$\Rightarrow$Slope of line $ax+by+c=0$ $=\frac{-a}{b}$

Step 2: Solve for the required values

As the given line is tangent to the curve, the slope of the tangent and the slope of the given line must be equal

$\Rightarrow$$\frac{-4}{x^2}=\frac{-a}{b}$

$\Rightarrow$ $\frac{a}{b}={\left(\frac{2}{x}\right)}^2$

As ${\left(\frac{2}{x}\right)}^2$ is a square it is $>0$ for all values of $x$

$\Rightarrow$$\frac{a}{b}>0$

This is possible only when, $a>0,b>0$ or $a<0,b<0$.

Hence options(C) i.e. $a<0,b<0$ is the correct answer.