Question

Find the slope of the normal to the curve $4x^3-3x{y}^2+6x^2-5xy-8y^2+9x+14=0$ at the point $(-2,3)$.

• A. $\infty$
• B. $1$
• C. $\frac{9}{2}$
• D. $\frac{2}{9}$

Answer 1

The correct option is D

$\frac{2}{9}$

Explanation for correct option:

Step:1 Find out the differentiation of both sides the given equation

The given function is

$4x^3-3x{y}^2+6x^2-5xy-8y^2+9x+14=0$

$\Rightarrow$$12x^2-3y^2-6xy\frac{dy}{dx}+12x-5y-5x\frac{dy}{dx}-16y\frac{dy}{dx}+9=0$

$\Rightarrow$ $\frac{dy}{dx}(6xy+5x+16y)=(12x^2-3y^2+12x-5y+9)$

$\Rightarrow$ $\frac{dy}{dx}=\frac{(12x^2-3y^2+12x-5y+9)}{(6xy+5x+16y)}$

Step:2 Find out the slope to the normal to the curve at the given point.

$\begin{array}{rcl}{\left[\frac{dy}{dx}\right]}_{(-2,3)}& =& \frac{(12{(-2)}^2-3\times 3^2+12(-2)-5\times 3+9)}{(6\times (-2)\times 3+5\times (-2)+16\times 3)}\\ & =& -\frac{9}{2}\end{array}$

$\therefore$Slope to the normal of the curve $=\left[\frac{-1}{\left(-{\displaystyle \frac{9}{2}}\right)}\right]$

$=\frac{\mathbf{2}}{\mathbf{9}}$

Hence, Option (D) is correct .