Question

If the line $ax+by+c=0$ is a normal to the curve $x^{3}y=1$, then ab is?

• A. Always positive
• B. Always negative
• C. May be positive or negative
• D. Cannot say anything

Answer 1

The correct option is B Always negative

convert the equation in the form of y = mxtc

when m is the slope & c is the y interrupt

ax + by + c = 0

ax + by = -c

by = -c - ax

y = $\frac{-ax}{b}-c$

slope of the line is $\frac{-a}{b}$

equation of curve is $x^{3}y=1$

differentiating the equation of cover w r.t x we

have

$\frac{d}{dx}\left(x^{2}y\right)=\frac{d}{dx}\left(1\right)$

$x^3\frac{dy}{dx}+3n^{2}y=0$

$\frac{dy}{dx}=\frac{-3x^{2}y}{x}=\frac{-3y}{x}=\frac{-3y}{x}$ =slope of tang

slope of normal will be $\frac{x}{3y}$

Now since the line is normal to curve

$\frac{x}{3y}=\frac{-a}{b}$ [Always -ve]