Why Does Substituting Into y^2 = 16x Yield t = 0?

[synkk]

The point P(4t^2, 8t) lies on the parabola C with equation y^2 = 16x. The point P also lies on the rectangular hyperbola H with equation xy = 4

a) Find the value of t, and hence find the co-ords of P.

working:
so x = 4t^2 and y = 8t
i sub these into xy = 4 and get t = 1/2 and can then find the points of P and thus answer the question

however if i sub x = 4t^2 and y = 8t into y^2 = 16x then I get:
(8t)^2 = 16(4t^2)
64t^2 = 64t^2
t = 0?

Could anyone tell me what I am doing wrong when subbing it into y^2 = 16x?

 

[rock.freak667]

You did nothing wrong, you just ended up with 0 = 0 which is true. So you only real solution it t = 1/2.

 

[synkk]

You did nothing wrong, you just ended up with 0 = 0 which is true. So you only real solution it t = 1/2.

I don't understand, shouldn't I get t = 1/2 no matter where I substitute the values into?

 

[rock.freak667]

I don't understand, shouldn't I get t = 1/2 no matter where I substitute the values into?

Well for the parabola C, the parameter x=4t², y = 8t is valid. However for the hyperbola, it is not. So you putting it back into the equation for C would just be verifying the parameters are correct.

 

[inkjet]

When we arrange for t, the answer will be zero.

 

[SammyS]

The point P(4t^2, 8t) lies on the parabola C with equation y^2 = 16x. The point P also lies on the rectangular hyperbola H with equation xy = 4

a) Find the value of t, and hence find the co-ords of P.

working:
so x = 4t^2 and y = 8t
i sub these into xy = 4 and get t = 1/2 and can then find the points of P and thus answer the question

however if i sub x = 4t^2 and y = 8t into y^2 = 16x then I get:
(8t)^2 = 16(4t^2)
64t^2 = 64t^2
t = 0?

Could anyone tell me what I am doing wrong when subbing it into y^2 = 16x?

The solution to 64t² = 64t² is that t can be any real number, not just t=0.