What the heck does this math question mean? What are they asking for? How is it solved?

Question

Find all plane curves such that for each point on the curve, the y-axis bisects that part of the tangent line between the point of tangency and the x-axis.

Answer 1

So you have a curve that you can graph on the usual x-y Cartesian plane. And for each point, you have a tangent line at that point. For any given point, the tangent line eventually crosses the x-axis. If you connect the tangent point with this point on the x-axis, you get a line segment. The midpoint of this segment is the point where it crosses the y-axis.

They ask you to find all the different curves that have this property. Again, it's a curve where each tangent line has a segment that starts at the tangent point, ends at the x-axis, and exactly halfway through crosses the y-axis.

You know that the slope of the tangent line is equal to the derivative. But let's first look for the general equation of lines that follow the behavior described above.

Let (xT, yT) be the tangent point of such a line. Let (x1, 0) be the point where it hits the x-axis. We know it hits the y-axis exactly half way through. So the midpoint is (0, (yT+0)/2), or just (0, yT / 2). Taking just the tangent point and this midpoint, we have two points in terms of just xT and yT.

The slope is (yT - yT/2) / (xT - 0) = (yT) / (2 xT). Remember that the tangent point is a point on the curve. So if this is always the slope of any tangent point, then the slope is always y / 2x for a given x. So really, we're looking for curves where
dy/dx = y / 2x

Split this up as (1/y)dy = (1 / 2x)dx and do the integration to find a general expression for y = f(x).

Answer 2

Solving the differential equation 2y'x = y gives the answer y = c√x, where c is a constant, giving you the family of plane curves satisfying the condition. Well, that also includes y = c√(-x), which is the mirror of the first family of plane curves. To understand the question, draw a parabola y = √x. Then draw some lines tangent to the parabola, and extend them until they intersect the x-axis. You will see that the y-axis intersect the tangent lines halfway between the point of tangency with the parabola and the point of intersection with the x-axis.

Once you understand the question, then it's just a matter of basic coordinate geometry to figure out the equation of the tangent lines:

y = f'(k) x + f(k) - f'(k) k

where for clarity, f(k) is the equation of the curve, and k is the x location of the tangent point. At the point of intersection with the x-axis, y = 0, so we have

x = (f'(k) k -f'(k)) / f'(k)

but this x = -k if the y-axis bisects the tangent line. So, therefore, we have

-k = (f'(k) k -f(k)) / f'(k)

and that leads to the differential equation to be solved

2 f'(k) k = f(k) , or

2y'x = y

Answer 3

The question says that there are curves on xy-plane whose equations we have to find. The curves are such that if a tangent is drawn at any point on the curve, the point of intersection of y-axis with the tangent is the midpoint of the contact point of the tangent and the point where the tangent intersects x-axis.

The discussion by the repondents above is good and perfect and there is nothing more for me to add.

Answer 4

For the additional detail 3x=126 x=126/3 x=42 to check substitute x with 42 42*3 = 126 x=42