Find the locus of a point question (1 Viewer)

[EViL.GENiUS]

Hi

I would like some help here.

for (I) the equation of the tangent y=px-ap^2
equation of normal y=(-1/p)x + a
point N(ap,0)

i don't know how to do part (II)

thanks

 

[Advv]

You sure your equations are right for part one?

 

[wixxy2348]

use perpendicular distance formula with line of normal and focus of parabola

yepp thats right

 

[Ogden_Nash]

Hi

I would like some help here.

for (I) the equation of the tangent y=px-ap^2
equation of normal y=(-1/p)x + a
point N(ap,0)

i don't know how to do part (II)

thanks

You have found the parametric coordinates of N. ie N(ap,0). Since there is no parameter in the y-coordinate (since y = 0, a constant), this is the locus of N. This is because no value of p can alter the y value.

So the locus of N is y = 0.

(Note: usually when this situation occurs, you might have restrictions on the values of x for which the locus can take, but in this case there is no restriction)

Hope that helps.

 

[shinn]

Yea,

Simultaneously solve for:
y= px-ap^2
y= -(1/p)x +a

you get N (ap ,0)

so the locus of N is just y = 0.

P.s. Wats the perpendicular distance formula used for?..

 

[lolokay]

this is incorrect guys. N is a point on the tangent, not necessarily one which also lies on the parabola. refer to my previous post for the correct method

I don't get what you mean. They didn't assume N was on the parabole - they solved for the point of intersection of the tangent to the parabola, and the normal of this tangent which passes through the focus, which gives N=(ap, 0), so the locus of N is y=0. Why is this incorrect, and how exactly does your method work?