Question

The points P(ap² , 2ap) and Q(aq² , 2aq) lie on the parabola y²= 4ax. Prove that if PQ is a focal chord then the tangents to the curve at P and Q intersect at right angles at a point on the directrix.

Answer 1

Solution:

Let's draw a figure as per given data:

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.

Hence, proved.