Question

Find the slope of the normal to the curve \[4{x^3} - 3x{y^2} + 6{x^2} - 5xy - 8{y^2} + 9x + 14 = 0\] at the point \[\left( { - 2,3} \right)\]
A. \[\infty \]
B. \[1\]
C. \[\dfrac{9}{2}\]
D. \[\dfrac{2}{9}\]

Answer 1

Hint:In this question, we are asked to find the slope of the normal to the curve at the given point. For that, we first differentiate with respect to x and then substitute the given points in the resultant equation to find the slope of the normal to the given curve.

Formula used:
We have been using the following formulas:
1. \[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\]
2. \[\dfrac{d}{dx}\left ( u.v \right )=v\dfrac{du}{dx}+u\dfrac{dv}{dx}\]

Complete step-by-step answer:
We are given that \[4{x^3} - 3x{y^2} + 6{x^2} - 5xy - 8{y^2} + 9x + 14 = 0...\left( 1 \right)\] at the point \[\left( { - 2,3} \right)\]
Now by differentiating equation (1) with respect to x, we get
\[
  4.3{x^{3 - 1}} - 3{y^2} - 3x \times 2y \times \dfrac{{dy}}{{dx}} + 6 \times 2{x^{2 - 1}} - 5y - 5x\dfrac{{dy}}{{dx}} - 8 \times 2y \times \dfrac{{dy}}{{dx}} + 9 = 0 \\
  12{x^2} - 3{y^2} - 6xy\dfrac{{dy}}{{dx}} + 12x - 5y - 5x\dfrac{{dy}}{{dx}} - 16y\dfrac{{dy}}{{dx}} + 9 = 0 \\
 \]
Now by taking common term \[\dfrac{{dy}}{{dx}}\] out, we get
\[0 = 12{x^2} - 3{y^2} + 12x - 5y + 9 - \dfrac{{dy}}{{dx}}\left( {6xy + 5x + 16y} \right)\]
Now by simplifying, we get
\[\dfrac{{dy}}{{dx}} = \dfrac{{12{x^2} - 3{y^2} + 12x - 5y + 9}}{{6xy + 5x + 16y}}\]
Therefore, the slope of the normal to the curve \[4{x^3} - 3x{y^2} + 6{x^2} - 5xy - 8{y^2} + 9x + 14 = 0\] is \[\dfrac{{dy}}{{dx}} = \dfrac{{12{x^2} - 3{y^2} + 12x - 5y + 9}}{{6xy + 5x + 16y}}\]
Now we substitute the given point in our slope, we get
\[\dfrac{{dy}}{{dx}} = \dfrac{{12{{\left( { - 2} \right)}^2} - 3{{\left( 3 \right)}^2} + 12\left( { - 2} \right) - 5\left( 3 \right) + 9}}{{6\left( { - 2 \times 3} \right) + 5\left( { - 2} \right) + 16\left( 3 \right)}}\]
Now by simplifying the above equation, we get
\[
  \dfrac{{dy}}{{dx}} = \dfrac{{12 \times 4 - 3 \times 9 - 24 - 15 + 9}}{{6\left( { - 6} \right) - 10 + 48}} \\
   = \dfrac{{48 - 27 - 24 - 15 + 9}}{{ - 36 - 10 + 48}} \\
   = \dfrac{{21 - 24 - 15 + 9}}{{ - 46 + 48}} \\
   = \dfrac{{ - 3 - 15 + 9}}{2} \\
 \]
Further simplifying, we get
\[
  \dfrac{{dy}}{{dx}} = \dfrac{{ - 18 + 9}}{2} \\
   = \dfrac{{ - 9}}{2} \\
 \]
Therefore, the slope of the tangent at the normal curve is \[\dfrac{{ - 9}}{2}\]
Now we know that the slope of normal is \[ = - \dfrac{1}{m}\]
Therefore, the slope of normal is
\[
   = - \dfrac{1}{{\left( { - \dfrac{9}{2}} \right)}} \\
   = \dfrac{2}{9} \\
 \]

Hence, option (D) is correct answer.

Note: Students must be aware that differentiating the equation of the line allows one to determine the slope of a curve. students should be careful when differentiating the equation because one error will render the entire problem incorrect. He or she should also keep in mind the normal formula's slope because doing so will enable them to obtain the desired outcome.