Question

find the asymptotes parallel to x-axis of the curve y^4+x^2y^2+2xy^2_4x^2_y+1=0

Answer 1

Answer:

y = -2 and y = +2.

Step-by-step explanation:

Equation of the curve is y⁴ + x² y² + 2 x y² - 4 x² - y + 1 = 0.

=> (y²-4) x² + 2y² x + (y⁴ - y + 1) = 0.

=> x = [ -y² +- √{y⁴ - (y²-4)(y⁴ -y+1)} ] / [y² - 4).

=> x =  -y² /(y² - 4) + - √(5y⁴- y⁶+y³-y²-4y+4) / (y² - 4)

The asymptotes obtained from the two factors in the denominator are y = 2 and y = -2.

The term under the root becomes negative for large values of y. So there are no asymptotes for y = infinity or - infinity.

Answer 2

ANSWER:------

Find the equations of the tangent lines to the curvey

=x−1x+1that are parallel to the linex−2y=2.

There is a bit of algebra and arithmetic for this. Let's focus on the reasoning and the calculus.

One

A line parallel tox−2y=2must have the same slope.

The slope of this line is12.

So we want the slope of the tangent line to be12Two

How do we find the slope of the tangent line?

-- The derivative. So, we want the derivative to be12.

What is the derivative ofy=x−1x+1?

Use the quotient rule:y'=x(x+1)−(x−1)⋅1(x+1)2=2(x+1)

2ThreeFindxto

makey'=122(x+1)2=12if and only

if(x+1)2=4Sox+1=±2andx=1,−3Four

Find theyvalues atx=1(y=0)and

atx=−3(y=2)

#FiveFind the equations of the lines:through(1,0)with

slopem=12and through(−3,2)with

slopem=12. #-2

hope it helps:)

T!—!ANKS!!!!