Question

#2- For the curve given by 4x2+y2=48+2xy show that

Answer 1

it is an implicit function, (for high school students)

1)  d(4x2+y2)/dx=d(48+2xy)/dx, after the differentiation, we obtain 
8x + 2y .dy/dx = 2y + 2x .dy/dx
8x - 2y = 2x .dy/dx - 2y .dy/dx,  4x - y = (x .- y ).dy/dx, and then 
dy/dx = y-4x / y - x
2) when x cordinate is 2
we have 4(2^2) + y^2 = 48 + 2 . 2 y, and the y^2 - 4y - 32=0, after solving the quadratic equation we find y = - 8, or y = 4, this the positive y coordinate, the answer is y = 4
3) Let be P (xp, yp) the point, since P is the point at which the line tangent to the curve at P is horizontal, so we can write 4xp2+yp2=48+2xpyp, but xp=2, we have yp^2 - 4yp - 32=0, so P(2, 4)

Answer 2

Answer:

Step-by-step explanation:

A. The given curve is:

$4x^2+y^2=48+2xy$

$8x+2y\frac{dy}{dx}=2(x\frac{dy}{dx}+y)$

$8x+2y\frac{dy}{dx}=2x\frac{dy}{dx}+2y$

$8x-2y=2(x-y)\frac{dy}{dx}$

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

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

Hence proved.

B. The given curve is:

$4x^2+y^2=48+2xy$

Put x=2, we get

$4(2)^2+y^2=48+2(2)y$

$16+y^2=48+4y$

$y^2-4y-32=0$

$(y+4)(y-8)=0$

Thus, y=-4,8

C. The given curve is:

$4x^2+y^2=48+2xy$

To show point P with x-coordinate 2 at which the line tangent to the curve at P is horizontal, consider $\frac{dy}{dx}=0$

⇒$\frac{y-4x}{y-x}=0$

⇒$y=4x$

At y=8,

⇒8=4x

⇒x=2

The equation of the tangent is:

y-8=0(x-2)

⇒y=8.

Thus, $\frac{dy}{dx}=0$ at (2,80, therefore with  x-coordinate 2 the line tangent to the curve at P is horizontal.