Question

Let P be a point on the curve x=t^2 , y= 1/t.
If the tangent to the curve at P meets the x- and y- axes at A and B respectively, prove that PA=2BP

Answer 1

First, I differentiate x and y

x'=2t
y'=1/t^2

Answer 2

Then, I get 1/t^2 / 2t
= 1/2t^3

Now what do I do?

Answer 3

\[={1\over 2t^3}\]

Answer 4

y' = -1/t^2

Answer 5

OH, Yes, I forgot.. so, -1/2t^3

Answer 6

How do I go on from here?

Answer 7

well you know the slope of tangent line at P
now you need equation of line to find intercepts A and B

i think we need to get it in terms of x or unparamaterize it
let me play with it a bit

Answer 8

Hm... but there's no value for t, is there?

Answer 9

wouldn't the equation be

y= 1/2t^3 +1/2t?

Answer 10

\[y= (1/2t^3)x +1/2t\]

Answer 11

ok i think i got it, i was wrong you can leave it terms of t
you are on the right track ^^ but the y_intercept is wrong

point P is (t^2, 1/t)
dy/dx = m = -1/2t^3
tangent line
\[y -\frac{1}{t} =- \frac{1}{2t^{3}}(x-t^{2})\]
\[y =- \frac{1}{2t^{3}}x +\frac{3}{2t}\]
plug in 0 for x and y to get intercepts A and B
\[A = (3t^{2},0)\]
\[B = (0,\frac{3}{2t})\]
to finish proof use distance formula

Answer 12

Ah Ok. Yes, I keep forgetting to add the - sign.

Answer 13

@ishaan94 is there another way to do the problem

Answer 14

That's the way it says to do it in my textbook, so I think I should stick with that method :)

Answer 15

oh ok

Answer 16

I don't think so, the only method I know is the one you did.