HELP ME!!! find the length of the chord common to the parabolas y^2 = 2x + 4y + 6 and y^2 = 3x + 3y +1
I drew your graphs in desmos.com and they don't seem to have a common chord, unless I misunderstand your question?
it is really written on the book. hehe
Don't mind me, I'm confused lol.
you may isolate x in both equations and find intersect points with equating x's in terms of y
find the 2 intersection points of 2 parabolas, then use distance formula to get length of chord
thank you :)
actually the 2 parabola's intersect the easy way to find the equation of the chord if to equate the 2 equations are they are both equal to y^2 then simplify and you'll get the equation of the chord
you don't need to find points of intersection... or anything else... is a difficult problem to understand by simple to get the solution
can you show it to me step by step?
ok so the equations are \[y^2 = 2x + 4y + 6... and..... y^2 = 3x + 3y + 1\] so you end up with \[2x + 4y + 6 = 3x + 3y + 1\] its that simple
\[10\sqrt{2}\]
is the answer
here is the graph
ops I read equation rather than length...sorry about that
so just find where the chord intersects... and you'll get the 2 points for the distance
@zith how did you get that 10 square root blabla? :)
so you are looking at \[(x -5)^2 = 3x + 3(x -5) + 1\] just work on that...
@campbell_st how?
Simplify campbell's last equation into the form y = mx + b then use that equation and the equation of one of the parabolas to solve for x and y. You should end up with 2 points (x, y). Use those points and pythagoras to calculate the length.
\[x^2 -10x + 25 = 3x + 3x -15 + 1... or .... x^2 -16x + 39 = 0\] which factors to (x - 3)(x - 13) = 0 so the x values are x = 3 and x = 13 so the y values are y = -2 and y = 8 so the points are (3, -2) and (13, 8) find the distance between them
so i have already told u about distance formula it is
Thank you so much @Mandre :)
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