Question

The normal curve $xy=4$ at the point $(1,4)$ meets the curve again at

• A. $(-4,-1)$
• B. $(-8,\frac{1}{2})$
• C. $(-16,-\frac{1}{4})$
• D. $(-1,-4)$

Answer 1

Answer: (C)

$(-16,-\frac{1}{4})$

Differentiating the equation of the curve with respect to x, we get
$y+x{y}^{\prime }=0$
Or
$y^{\prime }=\frac{-y}{x}$
$=\frac{-4}{x^2}$
$y_{x=1}^{\prime }=-4$
Hence slope of normal is
$=\frac{1}{4}$.
Hence equation of normal will be
$y-4=\frac{1}{4}(x-1)$
Or
$4y-16=x-1$
Or
$4y-x=15$ ...(i)
$xy=4$ ...(ii)
Solving the above two equations
$4y-\frac{4}{y}=15$
Or
$4y^2-4=15y$
$4y^2-15y-4=0$
$y=4$ and $y=\frac{-1}{4}$
Hence
$y=\frac{-1}{4}$.
Thus $x=\frac{4}{y}=-16$.
Hence the required point is
$(-16,\frac{-1}{4})$.