calculus problem...?

Question

At every point of a certain curve, the slope of the tangent equals
-2x/y. Which of the following best describes the curve?

a) straight line
b) circle
c) ellipse
d) hyperbola
e) parabola

im not sure how i would go about this problem. please help me!! thanks so much

Answer 1

c)ellipse because it's equation was implicitly diffeentiated to make this derivative or slope

2x^2 +y^2=1 look coefficient x:Y are not same, therefore not circle

d/dx(fx)=

4x +2yy'=0

2yy'=-4x

y'=-2x/y which is the slope of a ellipse
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if this was associated with a hyperbola then the slope would be positive, and with parabola y would not be in the differentiated equation, and a staright line would also not have y in the differentaited equation, and circle would have been same ration y:x

Answer 2

Differentiate to find the slope. For the straight line and parabola you can take the derivative directly. For the others you need to use implicit differentiation.

The equation of the ellipse is of the form:

x^2/a^2 + y^2/b^2 = 1

Differentiating you get:

2x/a^2 + (2y/b^2)(dy/dx) = 0
dy/dx = (-2x/a^2)/(2y/b^2) = -(b^2)x/((a^2)y)
= -[(b/a)^2](x/y) = -2x/y
for suitable choices of a and b.

That would be:
(b/a)^2 = 2
b^2 = 2a^2

So the equation of the ellipse would be:

x^2/a^2 + y^2/(2a^2) = 1

Answer 3

f (x) = x^3 - 12x + a million . . . the 1st spinoff set to 0 unearths turning or table sure factors f ' (x) = 3x^2 - 12 3x^2 - 12 = 0 3 * (x + 2) * (x - 2) = 0 x = 2 ... x = - 2 . . . the 2nd spinoff evaluated at x = 2 and -2 determines if those factors are min, max, or neither. f ' ' (x) = 6x f ' ' (2) = 6*2 = 12 <== effective fee shows x=2 is a community minimum f ' ' (-2) = 6*(-2) = -12 <== destructive fee shows x=-2 is a community maximum a.) x = - 2 is a maximum, and x=2 is a minimum ... so x = - infinity to -2 is increasing x = -2 to +2 is lowering x = +2 to + infinity is increasing b.) f (-2) = (-2)^3 - 12*(-2) + a million = 17 f (2) = (2)^3 - 12*(2) + a million = - 15 c.) . . . the 2nd spinoff set to 0 unearths inflection factors, or the place concavity modifications 6x = 0 x = 0 <=== inflection element x = - 2 is a maximum, so would desire to be concave down concavity modifications on the inflection element(s) ... so x = - infinity to 0 is concave down x = 0 to + infinity is concave up

Answer 4

well

dy/dx = -2x/y

Separate the variables (Easy to do)

So ydy = -2xdx

Thus ∫ ydy = ∫-2xdx

So ½y² = -x² + k

ie x² + y²/2 = k

Note k MUST be positive as x² and y² are both non-negative.
So let k = c²

Thus x² + ½y² = c²

ie x²/(c)² + y²/(√2c)² = 1

This is an ellipse centre (0, 0) with semi-major axis √2c in the vertical direction (on the y-axis) and semi-minor axis c in the horizontal direction (on the x -axis) ← Option c)

Answer 5

if you have the value of slope =s

you can write -2x/y =s

or y = -2x/s

If s is constant straight line with slope 2/s