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# The normal to the curve xy = 4 at the point (1,4) meets the curve again atA. (−4,−1)B. (−8,$\dfrac{1}{2}$​)C. (−16,$- \dfrac{1}{4}$​)D. (−1,−4)

Last updated date: 17th Jul 2024
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Hint: To solve this problem, first determine the slope of the normal to the curve and obtain the normal equation with the help of a given point, then solve the equations to obtain the coordinates.

The given equation is xy = 4 ……(1)
We can write ${\text{y = }}\dfrac{{\text{4}}}{{\text{x}}}$……….(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 the 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 – 16 = x – 1
4y – x - 15 = 0 ……(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: When dealing with tangents and normals, first find the slope of the tangent and normal at the given point before generating the equation. We asked to find the other point at which the same normal cuts the curve again, and then we solved the normal equation with the curve equation to get the other coordinates at which the normal cuts the curve again. If you continue in this manner, you will arrive at the correct solution.