2 parabola questions

11. The line with equation y=m(x+a), where m can vary but a is constant, meets the parabola with equation y²=4ax in two points P and Q. Find, in terms of a and m the mid-point R of PQ. (ans: [a(2-m²)/m², 2a/m])
As m varies show that R lies on the curve with equation y²=2a(x+1).

12. The points P(ap², 2ap) and Q(aq², 2aq) lie on the parabola with equation y²=4ax. Show that the tangents to the parabola at P and Q meet at R[apq, a(p+q)]. (i can do this)
Show further that the area of PQR is (a²/2)|p-q|³.

thanks :smile:

Reply 1

Fermat

11.
Denote the points as,
P(x1,y1), Q(x2,y2), R(x3,y3)

Since R is the mid-pt of PQ, then
x3 = ½(x1+x2)

y = m(x+a)
y² = 4ax

m²(x+a)² = 4ax
m²x² + (2m²a - 4a)x + m²a² = 0

For any quadratic the sum of the rooots is given by,

r1 + r2 = -b/a (where a and b come from the standard quadratic, ax² + bx + c)

P and Q are the points of intersection of the curve and the straight line,
Therefore, X1 and x2 are solutions of m²x² + (2m²ax - 4a)x + m²a² = 0
i.e. r1 = x1, r2 = x2
and,
x1+x2 = r1 + r2 = -b/a = -(2m²a - 4a)/m² = -2a(m² - 2)/m²
x1+x2 = 2a(2 - m²)/m²
x3 = ½(x1+x2) = a(2 - m²)/m²
y3 = m(x3 + a) - using the eqn of the straight line
y3 = m{a(2 - m²)/m² + a}
y3 = m{2a/m² -a + a}
y3 = 2a/m
giving,
R(x3,y3) = [a(2 - m²)/m², 2a/m]
=======================

xR=a(2 - m²)/m², yR = 2a/m
therefore,
m = 2a/yR

substituting for m in the expression for xR,

xR = 2a/m² - a
xR = 2a/[2a/yR]² - a
xR = (2a/4a²)(yR)² - a
xR + a = yR²/2a
yR² = 2a(xR + a)
so, the locus of R is,

y² = 2a(x + a)
==========

Reply 2

Fermat

12.
The three points are,

P(ap²,2ap), Q(aq²,2aq), R(apq,a(p+q))

The area of a triangle ABC, with coords A(x1,y1), B(x2,y2), C(x3,y3) is given by,

2A = x1(y3 - y2) + x2(y1 - y3) + x3(y2 - y1)

Puting,

(x1,y1), = (ap²,2ap),
(x2,y2), = (aq²,2aq),
(x3,y3), = (apq,a(p+q)),

and substituting into the expression for 2A,

2A = ap²(a(p+q) - 2aq) + aq²(2ap - a(p+q)) + apq(2aq - 2ap)
2A = a²p²(p - q) + a²q²(p - q) + 2a²pq(q - p)
2A = a²(p-q)(p² - 2pq + q²)
2A = a²(p-q)(p-q)²
2A = a²(p-q)³
A = ½a²|p-q|³ - considering the area to be +ve only
===========

Reply 3

shift3

Fermat

The area of a triangle ABC, with coords A(x1,y1), B(x2,y2), C(x3,y3) is given by,

2A = x1(y3 - y2) + x2(y1 - y3) + x3(y2 - y1)

where did you get that from?

Reply 4

Fermat

shift3

where did you get that from?

I just worked it out!
It's not too difficult. Draw a triangle in the positive quadrant only (x +ve, y +ve)
Label it ABC. Drop perpindiculars, to the x-axis, from the vertices to D,E,F respectively.
The area of the triangle is given by the area of the two trapeziums ABED plus BCFE minus the area of the trapezium ACFD.
Plug in the coordinates of the vertices and you will end up with that formula.

Another formula for the area of a triangle, involving the lengths of the sides of the tiangle only, is given by,

Ar = ½√{4b²c² - (b² - c² + a²)²}

You can work this out yourself by looking at the formula for the area of a triangle given by,

Ar = ½absinC

Now use the cosine rule and the identity sin² + cos² = 1 to eliminate the trig functions.

Reply 5

shift3

brilliant! thanks. :smile:

Reply 6

shift3

i'm done with the exercise, but there are two questions i can't do:

13. Show that the normals at P(ap², 2ap) and Q(aq², 2aq) o the parabola with equation y²=4ax meet at the point R [a(p²+pq+q²+2), -apq(p+q)]. (i can do this)
Given that the line PQ passes through the focus S(a,0), show that as p and q vary, R lies on the curve with equation y²=a(x-3a).

16. The variable chord PQ on the parabola with equation y²=4x substends a right angle at the origin O. By taking P as (p², 2p) and Q as (q², 2q), find a relation between p and q. (i can do this -- ans: pq+4=0)
Hence show that PQ passes through a fixed point on the x-axis.

thanks again.

Reply 7

Fermat

13.
P(ap²,2ap), Q(aq²,2aq),
xR = a(p² + pq + q² + 2)
yR = apq(p+q)

xR/a = (p+q)² - pq + 2

The line PQ
=========
(y - 2ap)/(x - ap²) = (2aq - 2ap)/(aq² - ap²)

The focal point S , is on this line, so its coordinates (a,0) satisfy the eqn of the line.

(0 - 2ap)/(a - ap²) = (2aq - 2ap)/(aq² - ap²)
-2ap/(a - ap²) = 2a(q - p)/a(q² - p²)
-p/(1 - p²) = (q - p)/{(q - p)(q+p)}
-p/(1 - p²) = 1/(p + q)
p(p+q) = p² - 1
p² + pq = p² - 1
pq = -1
======

substituting for pq in the expression for xR and yR,

xR/a = (p+q)² + 1 + 2
yR = -a(p+q)

eliminating (p+q) gives us,

xR/a = (-yR/a)² + 3
a.xR = yR² + 3a²
yR² = axR - 3a²
============

So, the locus of the point R is,

y² = ax - 3a²
==========

Reply 8

Fermat

16.
Eqn of PQ
========
(y - 2p)/(x - p²) = (2q - 2p)/(q² - p²) = 2(q-p)/{(q-p)(q+p)} = 2/(q+p)
(y - 2p)/(x - p²) = 2/(p+q)

Let PQ pass through the x-axia at the point M.
Hence M is a point on the lne PQ and its coordinates (h,0) satisfy the eqn of the line, i.e.

(0 - 2p)/(h - p²) = 2/(p+q)
-2p(p+q) = 2h - 2p²
-2p² - 2pq = 2h - 2p²
-2pq = 2h
h = -pq
h = 4 - using the relationship you established in the earlier part of the question.
h = const.

Hence, M is a fixed point
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