Question

Find the equation of the line tangent to y^2 = -16x and parallel to x+y=1

Answer 1

@satellite73

Answer 2

Please find the slope of the line that sets the standard of parallel.

Answer 3

y=-x+1 so the slope is -1

Answer 4

Good. Can you find a general expression for dy/dx from the parabola?

Answer 5

wait, we are not supposed to use calculus. only analytic geometry expressions are allowed. :)

Answer 6

I wish I knew how to tell what section posts were in. You haven't a plan for that? Did you first get a picture?

Answer 7

i have seen these before, the non calculus version
you have get a quadratic equation and assert that there is only one solution, set the discriminant equal to zero and solve
it is a pain
lets see if i can remember it

Answer 8

nope, but i have the answer, here it is:

x + y=4

Answer 9

@satellite73 yeah, it is the right procedure

Answer 10

@tkhunny have you seen this? i have seen it here a couple times, always forget, but maybe we can recreate it

Answer 11

oh yeah it is not that bad i think

Answer 12

Workign on it. It's only quadratic, so it shouldn't be too bad. They were doing this kind of thing long before the calculus was invented.

Answer 13

\[y^2=-16x\\
y=-x+b\] set them equal
\[(-x+b)^2=-16x\] and you want only one solution to this, solve for \(b\)

Answer 14

i better do it with pencil an paper first but the idea is that
\[b^2-2bx+16x+b^2=0\] should have only one solution
\[x^2+(16-2b)x+b^2=0\] set the discriminant equal to zero and solve for \(b\)

Answer 15

got it

Answer 16

i already solve b=4

Answer 17

ok then you beat me to it

Answer 18

so y=-x + b
y=-x + 4
x+y=4

alright. thanks to you @satellite73 and @tkhunny for helping! :)

Answer 19

wait. another question for the two of you.
find the equations of the lines tangent to y^2= 4x containing (-2,1)

Answer 20

@satellite73

Answer 21

No calculus rule, again?

Answer 22

yup

Answer 23

We should be able to get this far without much trouble. Pick an arbitrary point on the parabola, (a,b), and write the equation of the line associated with it.

\(y-1 = \dfrac{b-1}{a+2}(x+2)\)

Enforcing the relationship between a and b, we get.

\(y-1 = \dfrac{b-1}{\dfrac{b^{2}}{4}+2}(x+2)\)

or

\(y-1 = 4\cdot\dfrac{b-1}{b^{2}+8}(x+2)\)

Did you get that far?

Answer 24

yeah. what should i do next?

Answer 25

equate it to y^2=4x

Answer 26

equate it to y^2=4x?

Answer 27

Then you apply the old geometer's trick.

If the point on the top is \((a_{1},b_{1})\), the equation of that tangent is \(b_{1}y = 2(a_{1} + x)\)

If the point on the bottom is \((a_{2},b_{2})\), the equation of that tangent is \(b_{2}y = 2(a_{2} + x)\)

Since both have to pass through (-2,1), this gives the relationship between the two a and b.

Answer 28

In other words, for the top one, the slope is \(\dfrac{2}{b_{1}}\) and for the bottom one, the the slope is \(\dfrac{2}{b_{2}}\) and this completes the exercise.