Question

The graph of the equation 2x2 + xy + y2 = 4 is the tilted ellipse pictured below; i.e. the points (x,y) in the plane that satisfy the equation yield the pictured ellipse. The point P=(1,-2) on the ellipse and the tangent through P are pictured below. A line that is tangent to an ellipse does not intersect the ellipse in any other point. We can use this fact to determine the slope of the tangent line.

Answer 1

The slope of the tangent at point $P = \left( {1, - 2} \right)$ is $\boxed{\frac{2}{3}}.$

Further explanation:

Given:

The equation of the ellipse is $2{x^2} + xy + {y^2} = 4.$

The point $P = \left( {1, - 2} \right)$ is on the ellipse and the tangent.

Explanation:

The given equation of the ellipse is $2{x^2} + xy + y = 4.$

Now differentiate the equation of ellipse with respect to $x$.

$\begin{aligned}\frac{d}{{dx}}\left( {2{x^2} + xy + {y^2}} \right) &= \frac{d}{{dx}}\left( 4 \right)\\\frac{d}{{dx}}\left( {2{x^2}} \right) + \frac{d}{{dx}}\left( {xy} \right) + \frac{d}{{dx}}\left( {{y^2}} \right)&= \frac{d}{{dx}}\left( 4 \right)\\4x + x \times \frac{{dy}}{{dx}} + y\frac{d}{{dx}}\left( x \right) + 2y\frac{{dy}}{{dx}}&= 0\\4x + y + \left( {x + 2y} \right)\frac{{dy}}{{dx}}&= 0\\4x + y + \left( {x + 2y} \right)\frac{{dy}}{{dx}} &= 0 \\\end{aligned}$

Further solve the above equation.

$\begin{aligned}\left( {x + 2y}\right)\frac{{dy}}{{dx}}&=- 4x - y\\\frac{{dy}}{{dx}}&=-\frac{{4x + y}}{{\left( {x + 2y} \right)}}\\\end{aligned}$

Substitute $1$ for $x$ and $-2$ for y in equation $\dfrac{{dy}}{{dx}}= - \dfrac{{4x + y}}{{x + 2y}}.$

$\begin{aligned}\frac{{dy}}{{dx}} &= - \frac{{4\left( 1 \right) + \left( { - 2} \right)}}{{1 + 2\left( { - 2} \right)}} \\&= - \frac{{4 - 2}}{{1 - 4}}\\&= - \frac{2}{{ - 3}}\\&= \frac{2}{3} \\\end{aligned}$

The slope of the tangent at point $P = \left( {1, - 2} \right)$ is $\boxed{\frac{2}{3}}.$

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Application of derivatives

Keywords: Derivative, graph of the equation, 2x2+xy+y2=4, ellipse, tangent, equation yield, pictured ellipse, line, intersect, slope, normal, slope of tangent line.

Answer 2

Answer:

The slope of tangent at point (1,-2) is $\frac{2}{3}$.

Step-by-step explanation:

The given equation of ellipse is

$2x^2+xy+y^2=4$

Differentiate both sides with respect to x.

$\frac{d}{dx}(2x^2+xy+y^2)=\frac{d}{dx}(4)$

$\frac{d}{dx}(2x^2)\frac{d}{dx}(xy)+\frac{d}{dx}(y^2)=\frac{d}{dx}(4)$

$4x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0$

$(4x+y)+(x+2y)\frac{dy}{dx}=0$

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

Calculate the value of \frac{dy}{dx} at (1,-2).

$\frac{dy}{dx}_{(1,-2)}=-\frac{4(1)+(-2)}{(1)+2(-2)}$

$\frac{dy}{dx}_{(1,-2)}=-\frac{2}{-3}$

$\frac{dy}{dx}_{(1,-2)}=\frac{2}{3}$

Therefore the slope of tangent at point (1,-2) is $\frac{2}{3}$.