Which Curve Intersects Every Curve of the Family y = 1/x + k at Right Angles?

[tandoorichicken]

Which of the following is an equation of a curve that intersects at right angles every curve of the family y = 1/x + k where k takes all real values?
a)y= -x b)y= -x^2 c)y= -x^3/3 d)y=x^3/3 e)y=ln x

 

[NateTG]

Are you sure that you don't mean
[tex]y=\frac{1}{x} + k[/tex]

then [tex]\frac{dy}{dx}=\frac{-1}{x^2}[/tex]

any suitable function would have the negative reciprocal slope, so
[tex]\frac{df}{dx}=\frac{x^2}{1}=x^2[/tex]
sp
[tex]f(x)=\frac{x^3}{3}+C[/tex]

and the answer is d.

 

[M98Ranger]

The correct answer is b) y = -x^2. This is because for any value of k, the curve y = 1/x + k will intersect the curve y = -x^2 at right angles. This can be seen by taking the derivative of both curves and setting them equal to each other, which results in the equation 1/x^2 = -2x. Solving for x gives the intersections as x = ±1, which are indeed at right angles. Therefore, any curve of the form y = -x^2 will intersect all curves of the family y = 1/x + k at right angles.