Homework help

The tangent at point P(1,1) to the curve x= 1/t, y=t^2 intersects the curve at point Q.
Find the equation of the tangent at P and the co-ordinates of Q.

So using dy/dx = dy/dt divided by dx/dt, I have got the gradient (m) as -2 then subbing in x=1 and y =1 I have got the equation y = -2x + 3. But I do not think this is right as you do not get the co-ordinates of Q which are (-1/2, 4)

Reply 1

TeeEm

19

Original post by Improvement

The tangent at point P(1,1) to the curve x= 1/t, y=t^2 intersects the curve at point Q.
Find the equation of the tangent at P and the co-ordinates of Q.

So using dy/dx = dy/dt divided by dx/dt, I have got the gradient (m) as -2 then subbing in x=1 and y =1 I have got the equation y = -2x + 3. But I do not think this is right as you do not get the co-ordinates of Q which are (-1/2, 4)

the tangent is ok
post a photo of your workings

Reply 2

gdunne42

21

Original post by Improvement

The tangent at point P(1,1) to the curve x= 1/t, y=t^2 intersects the curve at point Q.
Find the equation of the tangent at P and the co-ordinates of Q.

So using dy/dx = dy/dt divided by dx/dt, I have got the gradient (m) as -2 then subbing in x=1 and y =1 I have got the equation y = -2x + 3. But I do not think this is right as you do not get the co-ordinates of Q which are (-1/2, 4)

Your tangent equation is correct.

How are you finding the other point of intersection?

Edit: beat me to it TeeEm, did you do your ... trick :wink:

Posted from TSR Mobile

(edited 8 years ago)

Reply 3

JLegion

2

Original post by Improvement

The tangent at point P(1,1) to the curve x= 1/t, y=t^2 intersects the curve at point Q.
Find the equation of the tangent at P and the co-ordinates of Q.

So using dy/dx = dy/dt divided by dx/dt, I have got the gradient (m) as -2 then subbing in x=1 and y =1 I have got the equation y = -2x + 3. But I do not think this is right as you do not get the co-ordinates of Q which are (-1/2, 4)

[I'm assuming you are asking how to get the coordinates of Q.]

You have the tangent correct. The tangent intersects your curve at the point Q, so you are looking for when the curve intersects the tangent.

Because your tangent has equation y= -2x +3, you can put it equal to y=t^2.

You can also use the equation x=1/t to find t in terms of x; and substitute accordingly, so you have an equation entirely populated by Xs. Work out the value of x this way, and then find y by substituting into your tangent equation.

I hope my explanation made sense... I'm new to this.

Reply 4

ztibor

10

Original post by Improvement

The tangent at point P(1,1) to the curve x= 1/t, y=t^2 intersects the curve at point Q.
Find the equation of the tangent at P and the co-ordinates of Q.

So using dy/dx = dy/dt divided by dx/dt, I have got the gradient (m) as -2 then subbing in x=1 and y =1 I have got the equation y = -2x + 3. But I do not think this is right as you do not get the co-ordinates of Q which are (-1/2, 4)

The tangent is right and from that you can get the point Q

as x=1/t and y=t^2 so y=1/x^2

1/x^2=-2x+3 will be a cubic equation but it can be factorized

-2x^3+3x^2-1=0 => -2x^3+2x^2 +x^2 -1=0 => -2x^2(x-1) + (x+1)(x-1)=0
Taking out the common factor you can solve the equation
considering that any of the factors may be zero.