Answer to Question #221372 in Analytic Geometry for jceka

Question

1.The points P(ap2 , 2ap) and Q(aq2 , 2aq) lie on the parabola y 2 = 4ax. Prove that if P Q is a focal chord then the tangents to the curve at P and Q intersect at right angles at a point on the directrix.

2. The tangents at the points P(ap2 , 2ap) and Q(aq2 , 2aq) on the parabola y 2 = 4ax intersect at the point R. Given that the tangent at P is perpendicular to the chord OQ, where O is the origin, find the equation of the locus of R as p varies.

3. The coordinates of the ends of a focal chord of the parabola y 2 = 4ax are (x1, y1) and (x2, y2). Show that x1x2 = a 2 and y1y2 = −4a 2 .

4. Prove that the line x − 2y + 4a = 0 touches the parabola y 2 = 4ax, and find the coordinates of P, the point of contact. If the line x − 2y + 2a = 0 meets the parabola in Q, R, and M is the mid-point of QR, prove that PM is parallel to the axis of x, and that this axis and the line through M perpendicular to it meet on the normal at P to the parabola

Answer 1

1) Because (P,Q) is focal hord we have:

${a\cdot p^2-a\cdot q^2 \over 2\cdot a\cdot p-2\cdot a\cdot q)}=$${a\cdot p^2-a\over 2\cdot a\cdot p};$

$p+q=p-{1\over p}$ ;

$p\cdot q=-1$

Let us find the intersection of Z(-a,u(p)) tangent at P with asymptota x=-a:

$4\cdot a\cdot x-y^2=0 - equation \space of parabola$

$2\cdot a\cdot dx-y\cdot dy=0;\space \\ dx=x(P)-x(Z)=a\cdot p^2+a;\\ dy=y(P)-y(Z)=2\cdot a\cdot p-u(p);\\ 2\cdot a\cdot(a\cdot p^2+a)-2\cdot p\cdot a\cdot (2\cdot a\cdot p-u(p); \\a\cdot p^2+a-2\cdot a\cdot p^2+p\cdot u(p)=0;\\ u(p)=\frac{a\cdot p^2-a}{p}=a\cdot p-a\cdot \frac{1}{p}=a\cdot (p+q);$

Similarly u(q)=a(q+p)

Thus u(p)=u(q) therefore tangents at P and Q intersecs at Z(-a,a(p+q));

k(P)=dy(p)/dx(P)=${2\cdot a\cdot p-a\cdot (p-\frac{1}{p})\over a\cdot p^2+a }=\frac{1}{p}$  is the slope of tangent at P;

Similarly k(Q)=$\frac{1}{q}$ .

Therefore k(P)$\cdot$ k(Q)=-1 and tangents at P and at Q are ortogonal in Z.

Part 1 is done.

2) Solution of part 2.

Let us find the equation of QO:

Its equation is given by Y=$\frac{2}{q}\cdot X\space(q<0)$

From the part 1 Y=ap+X/p is the equation of tangent at P;

Because of OQ orthogonal to tangent at P we have

$\frac{2}{q}\cdot \frac{1}{p}=-1$

Therefore q=-2/p.

Point R(p) satisfies the system:

$-\frac{1}{q}\cdot X+Y=a\cdot q;\\$

$-\frac{1}{p}\cdot X+Y=a\cdot p;$

We solve this system by the method of determinants:

$\Delta=-\frac{1}{q}+\frac{1}{p}=\frac{q-p}{p\cdot q};\\ \Delta_X=a\cdot(q-p);\\ X(p)=\frac{\Delta_X}{\Delta}=a\cdot p\cdot q=a\cdot p\cdot(-\frac{2}{p})=\\ =-2\cdot a;$

$\Delta_Y=a\cdot\frac{q^2-p^2}{p\cdot q}; Y(p)=\frac{\Delta_Y}{\Delta}=a\cdot(p+q)=a\cdot(p-\frac{2}{p});$

R(p)=(-2a,a(p-2/p)) is the parametric equation of locus, it is vertical line parallel to asymptota.

3) Solution of the third subproblem.

We have

$y_1^2=4\cdot a\cdot x_1\\$

$y_2^2=4\cdot a\cdot x_2\\$

$\frac{y_2-y_1}{x_2-x_1}=\frac{y_2}{x_2-a}$ is the equation of focal hord;

$4\cdot a \cdot \frac{y_2-y_1}{y_2^2-y_1^2}=\frac{y_2}{\frac{y_2^2}{4\cdot a }-a}\\ \frac{1}{y_2+y_1}=\frac{y_2}{y_2^2-4\cdot a^2}\\$

$y_2^2-4\cdot a^2=y_1\cdot y_2+y_2^2\\$

$y_1\cdot y_2=-4\cdot a^2\\$, the first statement is proved

$|y_1\cdot y_2|=4\cdot a^2\\$

$4\cdot a\cdot \sqrt{x_1\cdot x_2}=4\cdot a^2\\$

$\sqrt{x_1\cdot x_2}=a\\$

$x_1\cdot x_2=a^2\\$, the second statement is proved

4) Solution of the fourth part:

Let P(ap²,2ap) be a point of contact of some tangent to parabola.

We saw in the first task that equation of any tangent is

$Y=a\cdot p+\frac{1}{p}\cdot X\\$

For given line we have y=1/2∙x+2a 

Therefore it is tangent line with p=2 and P(4a,4a)- the point of contact.

Let us find intersections Q,R of line x − 2y + 2a = 0 with parabola y ² = 4ax.

x=2y-2a;

y²-8ay+8a²=0;

$y_1=(8\cdot a-4\cdot \sqrt{2}\cdot a)/2=4\cdot a-2\cdot \sqrt{2}\cdot a;\\$

$x_1=\frac{y_1^2}{4\cdot a}=\frac{4\cdot a^2\cdot (6-2\cdot\sqrt{2}}{4\cdot a}=(6-2 \cdot \sqrt{2})\cdot a;\\$

$y_2=(8\cdot a+4\cdot \sqrt{2}\cdot a)/2=4\cdot a+2\cdot \sqrt{2}\cdot a;\\$

$x_2=(6+2\cdot \sqrt{2})\cdot a;\\$

$M=\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)=(6a,4a);\\$

We see that y_M=y_P=>MP||OX

Let us prove that ZP, where Z(6a,0), P(4a,4a), is a normal to parabola.

Normal to the line x − 2y + 4a = 0 at P(4a,4a) has an equation

2x+y-12a=0. By substituting coordinates of Z(6a,0) in the last equation we

have 8a+4a-12a=0=>ZP is normal to parabola.