Locus and the Parabola (1 Viewer)

[FDownes]

I'm having some difficulties with this question simply due to the way its worded. Could someone help me out?

a) Find the coordinates of P on the parabola x = 4t, y = 2t², where t = 2.

This first part is easy, the answer is obviously (8, 8) This second part however...

b) Find the equation of the tangent at P.

... I'm not so sure on.

 

[cwag]

find cartesian equation by solving simultaneously..then diferentiate and sub in x=8 for gradient. then use y - y₁ = m (x - x₁)

 

[watatank]

I'm having some difficulties with this question simply due to the way its worded. Could someone help me out?

a) Find the coordinates of P on the parabola x = 4t, y = 2t², where t = 2.

This first part is easy, the answer is obviously (8, 8) This second part however...

b) Find the equation of the tangent at P.

... I'm not so sure on.

x = 4t, t = x/4

y = 2t²
= 2 * (x/4)²
= x²/8

dy/dx = x/4

sub x value of P to find the gradient = 8/4 = 2

then use point gradient formula y - y₁ = m(x-x₁)

y-8 = 2(x - 8)

2x - y - 8 = 0

fixed now.

 

[FDownes]

I think you've made a mistake there... y = x²/8, not x²/16, so the equation you eventually end up with is 2x - y - 8 = 0.

Thanks for clearing that up though.

EDIT: Beat'd

 

[FDownes]

Another simple question that I'm just not quite sure how to approach;

Find the equation of the locus of a point that is always 3 units away from the line 4x - 3y - 1 = 0

 

[tacogym27101990]

you have to use the perpendicular distance formula to fins the new equation

3=|4x-3y-1|
sqr(4^2+3^2)

then figure that out and you8'll have ure equation.

remember therell be two new equations