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# Homework help watch

1. 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)
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)
the tangent is ok
post a photo of your workings
3. (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)

How are you finding the other point of intersection?

Edit: beat me to it TeeEm, did you do your ... trick
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4. (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.
5. (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.
6. ...there's four people all saying the exact same thing.
7. No Q = 3
8. (Original post by Zacken)
...there's four people all saying the exact same thing.
Well it's possible we're all wrong :P
9. (Original post by JLegion)
Well it's possible we're all wrong :P
What, both Zacken and TeeEm?
10. (Original post by TheOtherSide.)
What, both Zacken and TeeEm?
I said possible, not probable
11. (Original post by JLegion)
I said possible, not probable
Fair enough. I still can't imagine that, but okay
12. Thank you! Most of you were right who said about making 1/x^2 = -2x +3, as you get the right co-ordinates for Q!

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