I'm still trying to find a clear, easy-to-understand, complete answer, to the following math problem:?

Question

Find the curves where the part of the tangent line cut off by the coordiante axes is bisected by the point of tangency.

Answer 1

The problem describes a particular characteristic and asks the question what curves have this characteristic?

The first thing I thought of was a circle, with a tangent line touching the circle at 45 degrees, so the slope of that tangent line is -1. But this is wrong because we want a curve where ALL the tangent lines are bisected by the point of tangency, not just where there happens to be ONE such tanget line.

Clearly, algebra and calculus are needed here.

For a given curve f(x), the formula for the slope of the tangent line is f'(x). At any point (a,f(a)), the slope is f'(a). If we write the equation of the tangent line, using the point-slope formula, it's y-f(a)=f'(a)*(x-a). If we calculate the x- and y- intercepts of this tanget line, they are (a-f(a)/f'(a),0) and (0,f(a)-f'(a)*a), respectively.

Under what conditions would the point of tangency (a,f(a)) always bisect the line segment which runs from (a-f(a)/f'(a),0) to (0,f(a)-f'(a)*a) ? From this information, can we say what type of curve f(x) is?

If it bisects the line segment, then a is exactly halfway between a-f(a)/f'(a) and 0, hence
a=(a-f(a)/f'(a)+0)/2
2a=a-f(a)/f'(a)
a=-f(a)/f'(a)
a*f'(a)=-f(a)
f'(a)=-f(a)/a
also f(a) is exactly halfway between 0 and f(a)-f'(a)*a, hence
f(a)=(0+f(a)-f'(a)*a)/2
2f(a)=f(a)-f'(a)*a
f(a)=-a*f'(a)
-f(a)/a=f'(a)
the same result.

So, the question is, what category of curves have the feature that f'(a)=-f(a)/a for all values of a? This is a differential equation, y'=-y/x. The solution is y=c/x.

A curve will have the characteristic described above if and only if it is of the form y=c/x, where c is a constant.

Example:
y=2/x is one of those curves. It's derivative is y'=-2/x^2. Let's pick the point on the curve (8,1/4). The slope of the tangent line is at that point is -2/64 = -1/32. Using the point-slope formula, the tangent line is y-1/4=(-1/32)*(x-8). The y intercept of this line is (-1/32)*(-8)+(1/4)=1/2. The x intercept of this line is (-1/4-8/32)/(-1/32)=16. So, the line segment which is the tangent line cut off by the coordinate axes runs from (0,1/2) to (16,0). Is this line segment bisected by (8,1/4)? yes. Alternatively, if we had chosen the point on the curve (1,2), we would find that the tangent line runs from (0,4) to (2,0) and the point of tangency still bisects that line segment. EVERY tangent line for this curve will be bisected by its point of tangency.

Answer 2

Draw any circle with center at the origin. Mark the points on the circle at 45, 135, 225, and 315 degrees. Draw the tangent lines at these 4 points. What do you notice>

The equation of any circle with center at the origin is

x^2 + y^2 = r^2, r>0

The parts of the tangents cut off by the coordinate axes at the points mentioned are bisected by those points as was asked for.

Answer 3

to get carry of the element of the shaded section will want the element of a circle of radius 2 meters minus the element of a circle of radius a million meter. to come to a decision the element of the unshaded fabric, Take the element of a circle of radius 3 minus the shaded section you bought above. I have confidence you be attentive to a thank you to calculate factors of circles and persist with the stairs defined above to derived the answer on your guy or woman. good success

Answer 4

is there an image that goes with this? If there's no image than I don't really get it that much... but the coordinate axes is BISECTED? so... that would be at the the origin... (0,0). wouldn't it?

Answer 5

y=f(x)
y-yo=f´(xo) (x-xo)
Intercept with axis
y=0 so x= xo- yo/f´(xo)
x=o so y = yo-xo*f´(xo)
so
xo= 1/2(xo-yo/f´(xo)
yo= 1/2(yo-xo*f´(xo))
without the subindex
2y=y-x*y´
y = - x dy/dx dy/y = -dx/x so ln Iy I =- ln Ix I = ln I1/xI+C
which leads to xy =K equilateral hyperbola with asymptotes the axis
x= -y/y´ so y´/y= -1/x so lnIyI = -lnIxI+C leads to the same equation