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

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However, for the point P to lie on both the parabola and the hyperbola, t must also satisfy the equation xy = 4. By substituting x=4t2 and y=8t into this equation, we get t=1/2 as the only valid solution. Therefore, the coordinates of P are (2,4) and in summary, the value of t is 1/2 and the coordinates of P are (2,4).
  • #1
synkk
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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?
 
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  • #2
You did nothing wrong, you just ended up with 0 = 0 which is true. So you only real solution it t = 1/2.
 
  • #3
rock.freak667 said:
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?
 
  • #4
synkk said:
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.
 
  • #5
When we arrange for t, the answer will be zero.
 
  • #6
synkk said:
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.
 

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

1. How do you find the vertex of a parabola?

The vertex of a parabola can be found by using the formula x = -b/2a, where a and b are the coefficients of the quadratic equation in standard form (ax^2 + bx + c). This will give you the x coordinate of the vertex. To find the y coordinate, simply plug in the x value into the original equation and solve for y.

2. How do you determine if a point is on a parabola?

To determine if a point is on a parabola, you can plug in the x and y coordinates of the point into the standard form of a quadratic equation (ax^2 + bx + c). If the resulting equation is true, then the point lies on the parabola. If the equation is false, then the point is not on the parabola.

3. What is the focus of a parabola and how do you find it?

The focus of a parabola is a point on the interior of the parabola that is equidistant from the directrix and the vertex. To find the focus, you can use the formula F = (a, b + 1/4c), where a and b are the coefficients of the quadratic equation in standard form and c is the constant term. This will give you the coordinates of the focus point.

4. How do you find the axis of symmetry of a parabola?

The axis of symmetry of a parabola is a vertical line that passes through the vertex and divides the parabola into two symmetrical halves. To find the axis of symmetry, you can use the formula x = -b/2a, where a and b are the coefficients of the quadratic equation in standard form. This will give you the x coordinate of the axis of symmetry.

5. Can a parabola have more than one x-intercept?

Yes, a parabola can have zero, one, or two x-intercepts. The number of x-intercepts depends on the position of the parabola in relation to the x-axis. If the parabola intersects the x-axis at two points, it will have two x-intercepts. If the parabola does not intersect the x-axis at all, it will have zero x-intercepts. If the parabola is tangent to the x-axis, it will have one x-intercept.

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