Question

The normal to the curve xy = 4 at the point (1,4) meets the curve again atA.(âˆ’4,âˆ’1)B.(âˆ’8,$\dfrac{1}{2}$â€‹)C.(âˆ’16,$- \dfrac{1}{4}$â€‹)D.(âˆ’1,âˆ’4)

Hint â€“ In order to solve this problem of, firstly find the slope of the normal to the curve and obtain the equation of normal with the help of a given point then solve the equations to get the coordinates. Doing this will solve your problem.

The given equation is xy = 4 â€¦â€¦(1)
We can write ${\text{y = }}\dfrac{{\text{4}}}{{\text{x}}}$â€¦â€¦â€¦.(2)
On differentiating (2) with respect x we get the new equation as:
$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{{\text{ - 4}}}}{{{{\text{x}}^{\text{2}}}}}{\text{ = y'}}$
${\text{y}}{{\text{'}}_{{\text{x = 1}}}}{\text{ = - 4}}$
This is the slope of tangent of the curve at x = 1.
So the slope of normal will be = $\dfrac{1}{4}$
(since slope of normal is equals to $\dfrac{{ - 1}}{{{\text{slope of tangent}}}}$)
So the equation of normal at the point (1,4) can be written as:
${\text{(y - 4) = }}\dfrac{{\text{1}}}{{\text{4}}}{\text{(x - 1)}}$
4y â€“ 16 = x â€“ 1
4y â€“ x - 15 = 0 â€¦â€¦(3)
From (2) we can put the value of y in (3) we get the equation as:
$\dfrac{{16}}{x} - x - 15 = 0 \\ 16 - {x^2} - 15x = 0 \\ {x^2} + 15x - 16 = 0 \\$
On solving the above equation by splitting the middle term we get the equation as:
${x^2} + 16x - x - 16 = 0 \\ x(x + 16) - 1(x + 16) = 0 \\ (x + 16)(x - 1) = 0 \\$
We get x = -16 as x = 1 is already provided in the question.
On putting x = -16 in equation (2) we get the value of y as:
y = $- \dfrac{1}{4}$
Hence, the normal meets again with the curve at $\left( { - 16,\, - \frac{1}{4}} \right)$.

Note â€“ Whenever you face such types of problems of tangents and normal with one of the points from which the line passes is given and the curve is also given then first you find the slope of tangent and normal at the given point then generate the equation. Here we have asked to find the other point at which the same normal cuts the curve again then we have solved the equation of normal with that of curve to get the other coordinates at which the normal cuts the curve again. Proceeding like this you will get your solution correct.