Answer to Question #124544 in Geometry for jim

Question

1. (a) 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.

(b) Find the asymptotes of the hyperbola
x^2 − y^2 + 2x + y + 9 = 0.

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​=−t₁-2/t₁

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

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

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

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

$PQ^2\geq108a^2$

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

​

b) A hyperbola with a horizontal transverse axis and center at (h, k) has one asymptote with equation y = k + a(x - h) and the other with equation y = k - a(x - h).a = distance from vertices to the center$y=\frac{\frac{\sqrt{33}}{2}}{\frac{\sqrt{33}}{2}}\left(x-\left(-1\right)\right)+\frac{1}{2},\:\quad \:y=-\frac{\frac{\sqrt{33}}{2}}{\frac{\sqrt{33}}{2}}\left(x-\left(-1\right)\right)+\frac{1}{2}$

$y=x+1+\frac{1}{2},\:\quad \:y=-\left(x+1\right)+\frac{1}{2}$