Question

Help and assurance, once more :)

P is a point on the parabola given parametrically by x=at^2, y=2at, where a is a constant. Let S be the point (a,0), Q be the point (-a, 2at) and T be the point where the tangent at P to the parabola crosses the axis of symmetry of the parabola.
(a) Show that SP=PQ=QT=ST=at^2+a

Answer 1

x'=2at
y'=2a

dy/dx= 1/t

Answer 2

equation of line =

y= 1/tx+b

2at=(1/t)(at^2) +b
2at=(at) +b
b=at

y=(1/t)x +at

Answer 3

Now what?

Answer 4

Find the intersection of axis of symmetry and Tangent

Answer 5

How do I do that?

Answer 6

Axis of symmetry is nonetheless the x-axis or y=0.
y = (1/t)x + at, y =0
x = -at^2 => point of intersection is (-at^2,0)

Answer 7

Now use the distance formula to get the proof.

Answer 8

Ok, thank you. There's a second part to the question, which I'll post up.
Can you just show me the proof?
I know the distance formula... but just need some assurance

Answer 9

@Ishaan94

Answer 10

help still needed??.... proof for distance formula?

Answer 11

Yes please

Answer 12

"... S be the point (a,0), Q be the point (-a, 2at) and T ... "
"P is a point on the parabola given parametrically by x=at^2, y=2at ..."
"... Show that SP=PQ=QT=ST=at^2+a"

We got T as (-at^2,0).
SP = \(\sqrt{(a -at)^2 + (2at)^2}\), PQ = \(\sqrt{(at^2 +a)^2 +(2at -2at)^2}\), QT = \(\sqrt{(-a + at^2)^2 +(2at)^2} \), ST =\(\sqrt{(a + at^2)^2+(0)^2}\)

Answer 13

Oh Ok. Thank you! :)
I'll post the second part up in a while.

Answer 14

i think one part of the above question, where SP=PQ is to be done, can be done straight away by seeing the fact that the basic principal of any point on a parabola is that it is equidistant from the directrix and the focus. (and here we know S(a,0) will be the focus and Q(-a,2at) lies on the directrix. So by the basic definition of the conic, SP=PQ by default.