Question

If a chord of the parabola y^2 = 4ax is a normal at one of its ends, show that its mid-point lies on the curve 2(x−2a) = (y^2 /a) +( 8a^3/ y^2) . Prove that the shortest length of such a chord is 6a√3

Answer 1

If the normal at the point $P(at_1^2, 2at_1)$ meets the parabola $y^2=4ax$ again at $Q(at_2^2, 2at_2)$ then

$$t_2=-t_1-{2\over t_1}$$

$$Q=(a(t_1+{2\over t_1})^2, -2a(t_1+{2\over t_1}))$$

Let $M(x,y)$ be the midpoint of $PQ$

$$(x, y)=\big(\dfrac{at_1^2+a(t_1+\dfrac{2}{t_1})^2}{2}, \dfrac{2at_1-2a(t_1+\dfrac{2}{t_1})}{2}\big)$$

$t_1=-\dfrac{2a}{y}$

$x=at_1^2+2a+\dfrac{2a}{t_1^2}$

$x-2a=\dfrac{4a^3}{y^2}+\dfrac{y^2}{2a}$

$$2(x-2a)=\dfrac{y^2}{a}+\dfrac{8a^3}{y^2}$$

$$PQ^2=(-at_1^2+a(t_1+\dfrac{2}{t_1})^2)^2+(-2a(t_1+{2\over t_1})-2at_1)^2$$

$$PQ^2=(4a+\dfrac{4a}{t_1^2})^2+(4at_1+\dfrac{4a}{t_1})^2$$

$$PQ^2=16a^2(1+\dfrac{2}{t_1^2}+\dfrac{1}{t_1^2}+t_1^2+2+\dfrac{1}{t_1^4})$$

$$PQ^2=16a^2(t_1^2+\dfrac{3}{t_1^2}+\dfrac{1}{t_1^4}+3)$$

Consider the function

$$f(u)=u+\dfrac{3}{u}+\dfrac{1}{u^2}+3, u>0$$

$$f'(u)=1-\dfrac{3}{u^2}-\dfrac{2}{u^3}$$

$$f'(u)=0=>\dfrac{u^3-3u-2}{u^3}=0$$

$$u^2(u-2)+2u(u-2)+(u-2)=0$$

$$(u-2)(u^2+2u+1)=0$$

$$(u-2)(u+1)^2=0$$

Since $u>0,$ we take $u=2$

$0<u<2, f'(u)<0, f(u) \ decreases$

$u>2, f'(u)>0, f(u) \ increases$

$f(2)=2+\dfrac{3}{2}+\dfrac{1}{4}+3=\dfrac{27}{4}$

$PQ^2\geq16a^2(\dfrac{27}{4})$

$PQ^2\geq108a^2$

$$PQ\geq6\sqrt{3}a, a>0$$