Question

slope of the normal to the curve :- X2 - xy + 3y² - 5y = 0 and x = 2 ? ​

Answer 1

$\green{\large\underline{\sf{Solution-}}}$

Given curve is

$\rm :\longmapsto\: {x}^{2} - xy + {3y}^{2} - 5y = 0$

Now, when x = 2, we get

$\rm :\longmapsto\: {2}^{2} - (2)y + {3y}^{2} - 5y = 0$

$\rm :\longmapsto\: 4 - 2y + {3y}^{2} - 5y = 0$

$\rm :\longmapsto\: {3y}^{2} - 7y + 4= 0$

$\rm :\longmapsto\: {3y}^{2} - 3y - 4y + 4= 0$

$\rm :\longmapsto\:3y(y - 1) - 4(y - 1) = 0$

$\rm :\longmapsto\:(y - 1)(3y - 4) = 0$

$\bf\implies \:y = 1 \: \: or \: \: \dfrac{4}{3}$

So, Coordinates are

$\bf\implies \:P(2,1) \: \: and \: \: Q\bigg(2, \: \dfrac{4}{3} \bigg)$

Now, Consider

$\rm :\longmapsto\: {x}^{2} - xy + {3y}^{2} - 5y = 0$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\: \dfrac{d}{dx}[{x}^{2} - xy + {3y}^{2} - 5y] = 0 \:$

$\rm :\longmapsto\:2x - \bigg[x\dfrac{dy}{dx} + y.1\bigg] + 6y\dfrac{dy}{dx} - 5\dfrac{dy}{dx} = 0$

$\rm :\longmapsto\:2x - x\dfrac{dy}{dx} -y + 6y\dfrac{dy}{dx} - 5\dfrac{dy}{dx} = 0$

$\rm :\longmapsto\:- x\dfrac{dy}{dx} + 6y\dfrac{dy}{dx} - 5\dfrac{dy}{dx} = - 2x+y$

$\rm :\longmapsto\:(-x + 6y - 5)\dfrac{dy}{dx} = - 2x +y$

$\bf\implies \:\dfrac{dy}{dx} = - \: \dfrac{(2x - y)}{-x + 6y - 5}$

So, slope of tangent at P is given by

$\bf\implies \:\dfrac{dy}{dx}_{(2,1)} = - \: \dfrac{(4 - 1)}{-2 + 6 - 5} = 3$

$\rm :\longmapsto\:Slope \: of \: normal = - \dfrac{1}{ \: \: \: \: \: \dfrac{dy}{dx}_{(2,1)}}$

$\bf\implies \:Slope \: of \: normal =- \dfrac{1}{3}$

Now, Slope of tangent at Q

$\bf\implies \:\dfrac{dy}{dx}_{(2, \frac{4}{3} )} = - \: \dfrac{4 - \dfrac{4}{3}}{-2+ 6 \times \dfrac{4}{3} - 5}$

$\bf\implies \:\dfrac{dy}{dx}_{(2, \frac{4}{3} )} = - \: \dfrac{\dfrac{12 - 4}{3}}{-2+ 8 - 5}$

$\bf\implies \:\dfrac{dy}{dx}_{(2, \frac{4}{3} )} = - \: \dfrac{8}{3}$

Hence,

$\bf\implies \:Slope \: of \: normal = \dfrac{3}{8}$

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Additional Information :-

Let y = f(x) be any curve, then line which touches the curve y = f(x) exactly at one point say P is called tangent and that very point P, if we draw a perpendicular on tangent, that line is called normal to the curve at P.

2. If tangent is parallel to x - axis, its slope is 0.

3. If tangent is parallel to y - axis, its slope is not defined

4. Two lines having slope M and m are parallel, iff M = m

5. If two lines having slope M and m are perpendicular, iff Mm = - 1.

Answer 2

Given:

equation of the curve: $x^2-xy+3y^2-5y=0$

equation of the line: $x=2$

To find:

The slope of the normal to the curve

Solution:

Value of $y$ when $x=2$ is:

$2^2-2y+3y^2-5y=0$

⇒$3y^2-7y+4=0$

⇒$3y^2-3y-4y+4=0$

⇒$3(y-1)-4(y-1)=0$

⇒$(y-1)(3y-4)=0$

⇒$y=1, \frac{4}{3}$

So we get two co-ordinates $(2,1)$ and $(2, \frac{4}{3})$

Now, to get the slope we differentiate the given equation,

$\frac{d}{dx}[x^2-xy+3y^2-5y]=0$

⇒$2x-[xy'+y]+6yy'-5y'=0$

⇒$-xy'+6yy'-5y'=-2x+y$

⇒$y'=\frac{-2x+y}{-x+6y-5}$

So, we will get two normals and their respective slopes are as follows:

For point $(2,1)$ , slope of tangent is

$y'_{(2,1)}=-\frac{4-1}{-2+6-5}$

$=3$

So, the slope of the normal is $-\frac{1}{3}$.

For point $(2, \frac{4}{3})$, slope of tangent is

$y'_{(2,\frac{4}{3})}=-\frac{4-\frac{4}{3} }{-2+8-5}$

$=-\frac{8}{3}$

So, the slope of the normal is $\frac{3}{8}$.

Hence, the required slope values are $-\frac{1}{3}$ and  $\frac{3}{8}$.