Locus of mid point of parabola and straight line

[tony24810]

Homework Statement

A line passing through the point (1,0) cuts the curve Y^2=4x at two points A and B. Find the equation of the locus of the mid-point of AB.

Homework Equations

The Attempt at a Solution

the graph is parabola, the solution should also be a parabola with x-intercept at (1,0).

tried to set up 2 equations are mixed together but didn't work out

wondering if parametric equation should be used

i think i should try to find a general expression for x and y, then use mid point respectively, then mix the parametric equation

please help

 

[phinds]

Write a standard line equation given the information provided and solve it for the two points of intersection and keep going from there. Don't TALK about the math, start to DO the math (and show it here). We can't help you from just your words.

"Didn't work out" is NOT a helpful description of where you are having a problem.

 

[Ray Vickson]

Homework Statement

A line passing through the point (1,0) cuts the curve Y^2=4x at two points A and B. Find the equation of the locus of the mid-point of AB.

Homework Equations

The Attempt at a Solution

the graph is parabola, the solution should also be a parabola with x-intercept at (1,0).

tried to set up 2 equations are mixed together but didn't work out

wondering if parametric equation should be used

i think i should try to find a general expression for x and y, then use mid point respectively, then mix the parametric equation

please help

PF Rules require you to show your work, whether correct or not.

Anyway, you are mis-using the word "locus": a locus (of points) means a curve traced out as one curve moves relative to another. So, if your line moved and the parabola remained stationary, the mid-point would trance out a curve---that would be the locus. If you have just two fixed curves that do not move anywhere, you don't have a "locus". Did you mean "location"?

 

[ehild]

Anyway, you are mis-using the word "locus": a locus (of points) means a curve traced out as one curve moves relative to another. So, if your line moved and the parabola remained stationary, the mid-point would trance out a curve---that would be the locus. If you have just two fixed curves that do not move anywhere, you don't have a "locus". Did you mean "location"?

The straight line is not fixed. It goes through the point (1,0) but its slope m is not given. The intercepts with the parabola depend on m and so is the midpoint. It describes a curve, and the question is the equation of that curve, the equation of the locus of the midpoint.

 

[tony24810]

please find my working out in attached pdf

 

[Ray Vickson]

please find my working out in attached pdf

Your document is messy and not well documented, so I will not even attempt to read it. Generally, PF encourages posters to type out their work whenever possible. And yes, that is a lot of trouble for the student, but it makes it much easier for helpers to give assistance. (Of course, the inclusion of drawings and diagrams is always a thorny issue, but some types of drawings can be pasted directly in a response.)

 

[ehild]

please find my working out in attached pdf

I do not understand your handwriting.
The equation of the parabola is given: y²=4x.
What is the general equation of a straight line which goes through the point (1,0)? What are the points of intersections with the parabola?

 

[SammyS]

please find my working out in attached pdf

Your parametrization, t = 4x, is not particularly helpful.

One parametrization that seems natural to me is to use the slope of the line as the parameter.

The equation of a line with slope, m, passing through the point, (1, 0) is : y = m(x-1).

Solve that simultaneously with y² = 4x .

That should get you a start.
Note: The sum of the two roots of ax² + bx + c = 0 is -b/a .

 

[tony24810]

Thanks Sammy! I think I did it!

is the answer Y^2 = 2x - 2 ?

haha the sum or root really helps!

last night i was thinking it again and again and i thought it has to pair up like t = x and -1/x, it probably is, but it doesn't get me the general point. actually, maybe it does, let me try this method in a moment. but then i cannot prove that they are a pair.

 

[SammyS]

Thanks Sammy! I think I did it!

is the answer Y^2 = 2x - 2 ?

haha the sum or root really helps!

That's what I get, although my procedure was a bit different. $\displaystyle\ \ \ y_{mid}^2=2(x_{mid}-1)\$

I did essentially what you did to get the x-coordinate of the midpoint.

$\displaystyle x_{mid}=1+\frac{2}{m^2}$

However, I used the fact that the mid point of the pair of the points of intersection is on the line $\displaystyle y = m(x-1)$ . Square this, also solve the above for m² and substitute to get $\displaystyle y_{mid}$.

Alternatively, substituting $\displaystyle x_{mid}=1+\frac{2}{m^2}$ directly into the equation of the line gives a parametric form for $\displaystyle y_{mid}$ .

last night i was thinking it again and again and i thought it has to pair up like t = x and -1/x, it probably is, but it doesn't get me the general point. actually, maybe it does, let me try this method in a moment. but then i cannot prove that they are a pair.

By the Way:

I do agree with Ray V. that you should type out (at least the main portions of) your work. I just happened to have the time & patience to go though your pdf renderings of your scanned work. The sketches of your graphs were helpful in showing that you understood the graphs of the given equations, and your second pdf was a bit more readable.

Try to learn some basic LaTeX, or at least learn how to use the superscript/subscript features available here at PF.