Question

What is the slope of the line tangent to the curve 3y2 - 2x2 = 6 - 2xy at the point (3,2)?

Answer 1

Answer:  The slope of the tangent line at the point (3, 2) is $\dfrac[4}{9}.$

Step-by-step explanation:  We are given to find the slope of the line tangent to the following curve at the point (3, 2).

$3y^2-2x^2=6-2xy~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We know that the derivative of a function at any given point is equal to the slope of the tangent at that point.

Now, differentiating both sides of equation (i) with respect to x, we have

$\dfrac{d}{dx}(3y^2-2x^2)=\dfrac{d}{dx}(6-2xy)\\\\\\\Rightarrow 6y\dfrac{dy}{dx}-4x=0-2\left(y+x\dfrac{dy}{dx}\right)\\\\\\\Rightarrow 2x\dfrac{dy}{dx}+6y\dfrac{dy}{dx}=4x-2y\\\\\\\Rightarrow \dfrac{dy}{dx}=\dfrac{4x-2y}{2x+6y}\\\\\\\Rightarrow \dfrac{dy}{dx}=\dfrac{2x-y}{x+3y}.$

So, at the point (3, 2), we get

$\dfrac{dy}{dx}=\dfrac{2\times3-2}{3+3\times2}=\dfrac{6-2}{3+6}=\dfrac{4}{9}.$

Thus, the slope of the tangent line at the point (3, 2) is $\dfrac{4}{9}.$

Answer 2

The slope of the tangent at point (3,2) will be ($5\div 18$) and it can be evaluated by differentiating the given equation of curve.

Given :

• Curve - $3y^2-2x^2=6-2xy$ ---- (1)

• Point - (3,2)

To determine the slope of the tangent, differentiation of the given equation of curve is carried out.

Differentiating equation (1) with respect to x.

$\dfrac{d}{dx}(3y^2-2x^2) = \dfrac{d}{dx}(6-2xy)$

$6y \dfrac{dy}{dx}-4x = 0-2y-2x\dfrac{dy}{dx}$

Now, further simplification of the above equation gives:

$\dfrac{dy}{dx}(6y+2x)=4x-2y$

$\dfrac{dy}{dx}= \dfrac{4x-2y}{6y+2x}$

Now, at point (3,2) the slope of the tangent will be:

$m = \dfrac{4\times 3-2\times 2}{6\times 2+2\times3}$

$m = \dfrac{5}{18}$

So, the slope of the tangent at point (3,2) will be ($5\div 18$).

For more information, refer the link given below:

https://brainly.com/question/20036619