anonymous
  • anonymous
Show that the line y=-3x+2 is tangent to the parabola y^2=-24x, and find the point of contact
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@mathmate
anonymous
  • anonymous
@ganeshie8
mathmate
  • mathmate
hints: 1. calculate the derivative of the parabola (using implicit differentiation). The result could be in terms of x, or y, or both, call it y'. 2. Calculate the point of intersection of the line and the parabola. you can do that by substitution. y^2=-24x..........(1) y=-3x+2...........(2) You can substitute y from (2) into (1), but with some algebraic effort, or solve for x in (2) and substitute the value of x (in terms of y) into (1), which will take a little less algebraic work. If you find a double root, it is sign that the line is tangent to the parabola. Back substitute x or y into equation (2) to find the other variable, thus obtain the point of tangency (contact) as (xt, yt). 3. Substitute (xt,yt) into the expression of y' in step 1 and compare with the slope of the line. If they are the same, the line is tangential to the parabola.

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anonymous
  • anonymous
2yy'=-24 y'=-24/2y y'=-12/y
anonymous
  • anonymous
y'=-12/sq.rt -24x
mathmate
  • mathmate
You don't need to solve in terms of x, because you will know the value of y after solving the next step.
anonymous
  • anonymous
(-3x+2)^2=-24x 9x^2-12x+4=-24x 9x^2+12x+4=0 9(x+2/3)^2-4+4=0 9(x+2/3)^2=0 (x+2/3)^2=0 x+2/3=0 x=-2/3
anonymous
  • anonymous
y^2=-24x y^2=16 y=4 or y =-4
anonymous
  • anonymous
y'=-12/y y'=-12/4 y'=-3 and y'=-12/-4 y'=3
anonymous
  • anonymous
is that correct so far?
anonymous
  • anonymous
r u there?
anonymous
  • anonymous
@mathmate
mathmate
  • mathmate
I'll take a look, was on the phone, sorry.
anonymous
  • anonymous
ok
mathmate
  • mathmate
Yep, so far so good. But you're on thin ice from here on! :)
mathmate
  • mathmate
Would you like to continue?
anonymous
  • anonymous
yes but im not sure what i just find.. im thinking its the point of intersection .. where the line and the parabola function meet ..am i right?
mathmate
  • mathmate
yes, but remember, the curves look like this: |dw:1446773311689:dw| So you have to choose your valid solution carefully.
anonymous
  • anonymous
would u say that the point y=-4 and y=4 are double root? or would it be y'=3 and y'=3 the double root ?
mathmate
  • mathmate
|dw:1446773439742:dw|
anonymous
  • anonymous
whats the difference between y and y' .. whats the y' ?
mathmate
  • mathmate
There are two y' because you had two values of y. You have only one intersection point, so choose the correct one.
anonymous
  • anonymous
would i need to find a x' also?
mathmate
  • mathmate
y'=slope. If the line y=-3x+2 is a tangent line, it should have the same slope as the point of tangency, by the definition of tangent. No you don't need to find x'. By the way, have you done implicit differentiation? y' is a different way of writing \(\frac{dy}{dx}\)
anonymous
  • anonymous
o yea dy/dx which is the slope
mathmate
  • mathmate
It turns out that if you would have substituted x into the equation and solve for y, you would have got a double root of (y-4)^2=0, which is an indication of tangency. Since y was substituted, so you solved for x, and necessarily you have two solutions of y, from which you have to reject one.
mathmate
  • mathmate
|dw:1446773851724:dw|
anonymous
  • anonymous
ok so the point of contact would be (-2/3,4)
mathmate
  • mathmate
exactly. Because the other solution (-2/3,-4) will not be on the line y=-3x+2 (-3(-2/3)+2)=4 \(\ne\) -4, the other solution has to be rejected
anonymous
  • anonymous
ok cool thanks
mathmate
  • mathmate
no problem! :)

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