Question

Find the equation of the tangent of x^2- xy + y^2 =7 at (-1, 2)

Answer 1

Answer:

It is 3y = 4x + 10

Step-by-step explanation:

Let's first get the slope of the curve.

[ slope is the derivative of the equation ]

${x}^{2} - xy + {y}^{2} = 7$

introduce dy/dx :

$\frac{d}{dx} ( {x}^{2} - xy + {y}^{2} ) = \frac{d}{dx} (7) \\ \\ 2x - (y + \frac{dy}{dx} ) + 2y \frac{dy}{dx} = 0$

make dy/dx the subject:

$2x - y - \frac{dy}{dx} + 2y \frac{dy}{dx} = 0 \\ \\ 2y \frac{dy}{dx} - \frac{dy}{dx} = y - 2x \\ \\ \frac{dy}{dx} (2y - 1) = y - 2x \\ \\ \frac{dy}{dx} = \frac{y - 2x}{2y - 1}$

At point (-1, 2):

$\frac{dy}{dx} = \frac{2 - 2( - 1)}{2(2) - 1} \\ \\ slope = \frac{4}{3}$

but a tangent has the same slope as the curve:

$y = mx + c$

m is the slope

c is the y-intercept

At (-1, 2):

$2 = ( - 1 \times \frac{4}{3} ) + c \\ \\ c = 2 + \frac{4}{3} \\ \\ c = \frac{10}{3}$

equation:

$y = \frac{4}{3} x + \frac{10}{3} \\ \\ { \boxed{3y = 4x + 10}}$

Answer 2

Answer:

Step-by-step explanation:

Here is an other way:

z=f(x,y)=x²-xy+y²-7=0

$\dfrac{\partial f(x,y)}{\partial x} =2x-y\\\\\dfrac{\partial f(x,y)}{\partial y} =-x+2y\\\\\\\dfrac{dy}{dx} =-\dfrac{\dfrac{\partial f(x,y)}{\partial x} }{\dfrac{\partial f(x,y)}{\partial y}} =-\dfrac{2x-y}{-x+2y} \\\\(-1;2)\\Slope=-\dfrac{2*(-1)-2}{1+2*2} =\dfrac{4}{5} \\\\y-2=\frac{4}{5} (x+1)\\\\\\\boxed{y=\dfrac{4x}{5} +\dfrac{14}{5} }$