Question

If the line ax + by + c = 0 is normal to the curve xy = 1, then the set of values of $\frac{a}{b}$ , is _____________.

Answer 1

The equation of given curve is xy = 1.

Let (x₁, y₁) be the point of contact of normal with the curve.

$\therefore x_1{y}_1=1$ .....(1)

Now,

xy = 1

Differentiating both sides with respect to x, we get

$x\frac{dy}{dx}+y=0$

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

∴ Slope of the normal at (x₁, y₁)

$=-\frac{1}{{\left({\displaystyle \frac{dy}{dx}}\right)}_{\left(x_1,y_1\right)}}$

$=-\frac{1}{\left(-{\displaystyle \frac{y_1}{x_1}}\right)}$

$=\frac{x_1}{y_1}$

$=x_1^2$ [Using (1)]

The given equation of normal to the curve is ax + by + c = 0.

Slope of this line = $-\frac{a}{b}$

$\therefore x_1^2=-\frac{a}{b}$

$\Rightarrow \frac{a}{b}=-x_1^2$

Now, $x_1^2$ is always positive.

∴ $\frac{a}{b}$ < 0 ∀ x ∈ R

Thus, the set of values of $\frac{a}{b}$ is the set of all negative real numbers i.e. (−∞, 0).


If the line ax + by + c = 0 is normal to the curve xy = 1, then the set of values of $\frac{a}{b}$ , is __(−∞, 0)__.