Equation of Tangent to Parabola (1 Viewer)

[norez]

"Prove that the line x - 2y - 2 =0 is a tangent to the parabola x^2 = 16y and state the coordinates of the point of contact."

How would I go about doing this?

 

[shinn]

(1) solve simulateneously for (x,y) to find the point of contact.
i.e. rearrange x - 2y -2 = 0 in terms of y and subsubstite into x^2 = 16y.
This will be a quadratic equation and hence you can find the point of contact which turns out to be (4,1).

(2) Check the gradient of the curve y = 1/16 * x^2 evaulated at (4,1) is equal to the gradient of the line which is 1/2.

 

[hscishard]

There are heaps of ways.
Another method would be:

x(x1) = 2a(y+y1)

From x^2=16, a = 4

x(x1) = 8y + 8y1
x(x1) - 8y - 8y1 = 0

Divide by 4 on both sides. This matches the -2y in the given tangent.
(x1)x/4 - 2y -2y1 = 0

(x1)/4 = 1 (coefficient of x in x-2y-2=0)
x1 = 4

-2y1 = -2
y1= 1

This is after you proved that it's a tangent.

 

[random-1006]

There are heaps of ways.
Another method would be:

x(x1) = 2a(y+y1)

From x^2=16, a = 4

x(x1) = 8y + 8y1
x(x1) - 8y - 8y1 = 0

Divide by 4 on both sides. This matches the -2y in the given tangent.
(x1)x/4 - 2y -2y1 = 0

(x1)/4 = 1 (coefficient of x in x-2y-2=0)
x1 = 4

-2y1 = -2
y1= 1

This is after you proved that it's a tangent.

ahh i dont like that method, the easiest way would be to do:

-2y= 2-x
y= -1+(1/2)x

then do x^2= 16y
x^2 = 16 ( -1 +(1/2)x), then solve that quadratic ( you will see the discriminant is zero and you only get one root, ie a tangent)

thats the BEST, EASIEST and QUICKEST

 

[hscishard]

LOL. Do you like this method?

x = 8t y=4t^2 (a = 4)

y= tx - at^2

2y=x-2
y=x/2 -1

t = 1/2

x = 8(1/2) = 4
y = 4(1/2)^2 = 1

 

[random-1006]

LOL. Do you like this method?

x = 8t y=4t^2 (a = 4)

y= tx - at^2

2y=x-2
y=x/2 -1

t = 1/2

x = 8(1/2) = 4
y = 4(1/2)^2 = 1

its not a bad method, but i still think my way was the best, your one involves memorising y= tx -at^2 ( you should be deriving these), and if you did derive it in the exam it would take longer than my way

 

[nat_doc]

how about this METHOD!:

you put the two equations together and then find the discriminate. if it is zero, then the tangent forms a "double root" with that parabola!

 

[random-1006]

how about this METHOD!:

you put the two equations together and then find the discriminate. if it is zero, then the tangent forms a "double root" with that parabola!

that is the exact same as mine