SOLUTION: What is the equation formed from the families of straight lines tangent to the parabola {{{y^2=2x}}}

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: What is the equation formed from the families of straight lines tangent to the parabola {{{y^2=2x}}}      Log On


   

Question 1151870: What is the equation formed from the families of straight lines tangent to the parabola y%5E2=2x
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

the equation of tangent line that we seek is y=mx%2Bb
if y%5E2=2x, then
Since both the abscissa (x) and ordinate (y) are positive, we can try to work with the positive half of the function when solving for y:
y=sqrt%282x%29
y=%282x%29%5E%281%2F2%29
To find the line tangent to it, we find the derivative
dy%2Fdx=%281%2F2%29%28sqrt%282%29%2Fsqrt%28x%29%29
Plug in x=2 to get the slope
dy%2Fdx=%281%2F2%29%28sqrt%282%29%2Fsqrt%282%29%29
dy%2Fdx=1%2F2

To get the y-intercept, plug in the numbers in the slope-intercept form linear equation
y=mx%2Bb
when x=2=>y%5E2=2%2A2=>y=sqrt%284%29=2, and tangent will touch parabola at point (2,2)
we have
=>2=%281%2F2%29%2A2%2Bb
=>2=1%2Bb
=>b=1
So the equation is y=x%2F2%2B1







Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The graph of y%5E2=2x is symmetrical with respect to the x-axis. So we can find the equations of the tangent lines for the portion of the graph with y positive; then any tangent line with equation y=mx%2Bb will have a corresponding tangent line with equation y=-%28mx%2Bb%29, or y=-mx-b.

The equation for the portion of the graph with y positive is

y+=+sqrt%282x%29+=+%282x%29%5E%281%2F2%29

Take the derivative, using the power and chain rules, to find the slope:

dy%2Fdx+=%281%2F2%29%282x%29%5E%28-1%2F2%29%282%29+=+%282x%29%5E%28-1%2F2%29

For each value of x, the tangent line will have slope %282x%29%5E%28-1%2F2%29; and the y value will be %282x%29%5E%281%2F2%29

For a given value of x, find the expression for the y-intercept b for the tangent line with y+=+%282x%29%5E%281%2F2%29 and slope %282x%29%5E%28-1%2F2%29.

y+=+mx%2Bb
%282x%29%5E%281%2F2%29+=+%28%282x%29%5E%28-1%2F2%29%29x+%2B+b
sqrt%282%29%2Asqrt%28x%29+=+%281%2F%28sqrt%282%29%2Asqrt%28x%29%29%29%2Ax+%2B+b
b+=+sqrt%282%29%2Asqrt%28x%29+-+%281%2F%28sqrt%282%29%2Asqrt%28x%29%29%29%2Ax
b+=+sqrt%282%29%2Asqrt%28x%29+-+%28sqrt%28x%29%2Fsqrt%282%29%29
b+=+sqrt%28x%29%28sqrt%282%29-1%2Fsqrt%282%29%29
b+=+sqrt%28x%29%282%2Fsqrt%282%29-1%2Fsqrt%282%29%29
b+=+sqrt%28x%29%281%2Fsqrt%282%29%29
b+=+sqrt%28x%2F2%29

So for a given value of x, the tangent line in the first quadrant has slope 1%2Fsqrt%282x%29 and y-intercept sqrt%28x%2F2%29.

The equation is then

y+=+%281%2Fsqrt%282x%29%29x+%2B+sqrt%28x%2F2%29

Choosing some "nice" values of x that make sqrt%282x%29 and sqrt%28x%2F2%29 rational....

x = 1/2: y+=+x%2B1%2F2
x = 2: y+=+%281%2F2%29x%2B1
x = 9/2: y+=+%281%2F3%29x%2B3%2F2
x = 8: y+=+%281%2F4%29%2B2

And then there are corresponding tangent lines on the other branch of the parabola:

x = 1/2: y+=+-x-1%2F2
x = 2: y+=+%28-1%2F2%29x-1
x = 9/2: y+=+%28-1%2F3%29x-3%2F2
x = 8: y+=+%28-1%2F4%29-2

A graph of the parabola and the tangent lines for x=1/2 -- x+1/2 and -x-1/2:


A graph of the parabola and the tangent lines for x=2 -- (1/2)x+1 and (-1/2)x-1:


A graph of the parabola and the tangent lines for x=8 -- (1/4)x+2 and (-1/4)x-2: