Show that the line y=-3x+2 is tangent to the parabola y^2=-24x, and find the point of contact
@mathmate
@ganeshie8
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.
2yy'=-24 y'=-24/2y y'=-12/y
y'=-12/sq.rt -24x
You don't need to solve in terms of x, because you will know the value of y after solving the next step.
(-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
y^2=-24x y^2=16 y=4 or y =-4
y'=-12/y y'=-12/4 y'=-3 and y'=-12/-4 y'=3
is that correct so far?
r u there?
@mathmate
I'll take a look, was on the phone, sorry.
ok
Yep, so far so good. But you're on thin ice from here on! :)
Would you like to continue?
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?
yes, but remember, the curves look like this: |dw:1446773311689:dw| So you have to choose your valid solution carefully.
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 ?
|dw:1446773439742:dw|
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