How Do I Solve a Complex Coordinate Geometry Problem with Parabolas?

[chickens]

Hi there, I've been struggling with this question for days , the first part where they call me to prove that equation, i could do it...the second one i don't know how to do...could anyone help me? how do i find the locus equation? its so confusing... ty in advance.


If the normal at P(ap^2, 2ap) to the parabola y^2=4ax meets the curve again at Q(aq^2,2aq), prove that p^2 + pq + 2 = 0. Prove that the equation of the locus of the point of intersection of the tangents at P and Q to the parabola is y^2(x + 2a) + 4a^3 = 0.

^ means to the power of...

 

[devious_]

I know I'm supposed to give hints not complete solutions, but I really didn't know what kind of hint to give...

The equation of the tangent to the parabola at the point (at^2, 2at) is:
y = (1/t)x + at (I'll leave the proof for you.)

Using this equation for points P and Q, we get:
y = (1/p)x + ap
y = (1/q)x + aq

Solving these simultaneously twice (once for x and once for y), we get:
(1) pq = x/a
(2) p+q = y/a

Re-writing (1) as q=x/(ap), then using it in (2) gives us:
p^2 = (py-x)/a

Now using p^2+pq+2=0 with the values for p^2 and q we just obtained gives us:
(py-x)/a + x/a + 2 = 0
p = (-2a)/y

Now re-writing (2) as q=(y^2+2a)/ay and using the value of p we just obtained in (1) yields:
pq = (-2y^2 - 4a^2)/y^2 = x/a

Upon multiplying it out we get:
-2ay^2 - 4a^3 = xy^2

Rearranging:
xy^2 + 2ay^2 + 4a^3 = 0
y^2(x+2a) + 4a^3 = 0, as required.

 

[tacman]

Hi there,

I understand that you have been struggling with this question for a while now and I can see why it may seem confusing. Let's break it down step by step to help you understand and solve it.

Firstly, for the first part where you need to prove the equation p^2 + pq + 2 = 0, you can start by finding the equation of the normal at point P. Remember that a normal to a curve is perpendicular to the tangent at that point. So, we can first find the slope of the tangent at point P by differentiating the given parabola equation.

y^2 = 4ax
2y(dy/dx) = 4a
dy/dx = 2a/y

Now, the slope of the normal at point P will be the negative reciprocal of this slope. So, the slope of the normal will be -y/2a. Using the point-slope form of a line, we can write the equation of the normal at point P as:

y - 2ap = (-y/2a)(x - ap^2)

Simplifying this, we get:

2ay + y^2 = 4ap^3

Now, since this normal also passes through point Q, we can substitute the coordinates of point Q (aq^2, 2aq) in this equation and solve for q. This will give us the value of q in terms of p.

Substituting the coordinates of point Q, we get:

2(2aq) + (2aq)^2 = 4ap^3

4aq + 4a^2q^2 = 4ap^3

q + aq^2 = ap^3

aq^2 - ap^3 = -q

q(aq - p^2) = -q

aq - p^2 = -1

q = (p^2 + 1)/a

Now, we can substitute this value of q in the equation of the normal at point P and simplify to get:

y^2 - 2ap^2y + 2a^2p^2 + 2a = 0

Dividing both sides by a, we get:

(y^2 - 2ap^2y + 2a^2p^2)/a + 2 = 0

(y - ap^2)^2 + 2 =