Parametric Q (1 Viewer)

[Aysce]

Haii question is:

Find the coordinates of point P on the parabola x = t^2, y=-2t where t=2. Find the equation of the line PS where S is the focus of the parabola. Fairly simple question, P is (4,-4). I redid the question twice and still got the wrong answer

 

[nightweaver066]

Haii question is:

Find the coordinates of point P on the parabola x = t^2, y=-2t where t=2. Find the equation of the line PS where S is the focus of the parabola. Fairly simple question, P is (4,-4). I redid the question twice and still got the wrong answer

x = t^2 , y = -2t

When t = 2,
x = 4, y = -4

So, P(4, 4)

From x = t^2, y = -2t
t = -y/2

So, x = (-y/2)^2
x = y^2/4
4x = y^2

So focal length is 1 with vertex at (0,0) concave to the right.

S(1, 0)

Equation of line PS:
y - 0 = (0 + 4)/(1 - 4) (x - 1)
y = -4/3 (x-1)
3y = -4x + 4
4x + 3y - 4 = 0

 

[rolpsy]

 

[Aysce]

But isn't the focus for x^2 = 4ay always (0,a) ?

 

[nightweaver066]

But isn't the focus for x^2 = 4ay always (0,a) ?

That is correct for a concave up parabola with the vertex at the centre.

For a parabola concave to the right, it is y^2 = 4ax where a is the focal length which means the focus is (a, 0)

Keep in mind that this parabola is concave to the right and the focus is always inside the cup of the parabola.

 

[Aysce]

AHH YOU LEGEND !! ;D thanks so much.