1995 Question 8 (hyperbola version)) (1 Viewer)

[OLDMAN]

Here's the hyperbola morph of the ellipse question 8, 1995.

a)Consider the line y=mx+c and the hyperbola H,
x^2/a^2-y^2/b^2=1.
Show that the conditions for cutting, touching and avoiding are
c^2>(am)^2-b^2, c^2=(am)^2-b^2, and c^2<(am)^2-b^2 respectively.
b)The point M(X_0,Y_0) lies "inside" H when
X_0^2/a^2-Y_0^2/b^2>1.
The line L is given by the equation
xX_0/a^2-yY_0/b^2=1.
(i) Using the result of (a) show that the line L lies entirely "outside" H. That is if (X_1,Y_1) is any pt. on L, then X_1^2/a^2-Y_1^2/b^2<1.
(ii) The chord of contact to the hyperbola from any pt. (X_2,Y_2)
"outside" H has equation
xX_2/a^2-yY_2/b^2=1.
Show that (X_0,Y_0) lies on the chord of contact to H from any point on L. That is if (X_2,Y_2) lies on L, then (X_0,Y_0) will lie on the chord of contact from (X_2,Y_2).

 

[ND]

Solving y=mx+c and x^2/a^2-y^2/b^2=1 simultaneously:
x^2/a^2 - (mx + c)^2/b^2 = 1
x^2.b^2 - a^2.m^2.x^2 -2a^2.m.x.c -a^2.c^2 -a^2.b^2 = 0
(b^2-a^2.m^2)x^2 -2a^2.m.c.x -(a^2.c^2+a^2.b^2)

/\ = 4a^4.m^2.c^2 + 4(m^2.c^2+a^2.b^2)(b^2-a^2.m^2)
= 4(a^2.m^2.c^2 + a^2.b^2.c^2 - a^4.m^2.c^2 + 4a^2.b^4 - a^4.b^2.m^2)
= 4a^2.b^2(c^2+b^2-(am)^2)

for the line to touch, /\ = 0,
.'. c^2+b^2-(am)^2 = 0
c^2 = (am)^2-b^2

for line to intersect, /\ > 0,
.'. c^2 > (am)^2-b^2 [as 4a^2.b^2 > 0]

for line to miss, /\ < 0
.'. c^2 < (am)^2-b^2

b) I downloaded the '95 paper from this site, and was able to do part (i) using pq \< (p^2 + q^2)/2 as given, but would i be allowed to use that rule on here(if i prove it) considering it says "using the result of (a)"? I can imagine there would be another soln with the discriminant too, but that would be long and tedious.

ii) i've forgotten everything about the chord of contact.

 

[OLDMAN]

ND : unfortunately, because of the +positive-negative ie.
x^2/a^2-y^2/b^2 of the hyperbola, the p,q inequality is useless- the reason for using the discriminant instead.

However, I'll be interested in an alternative to the discriminant for this question.

 

[ND]

Ah yeh i see. Let's see if spice girl can find a nice soln for it.

 

[spice girl]

Well, there's one I can think of so far, but it's a hassle and probably not worth considering.

Using the property that triangle formed by the tangent and the two asymptotes has area = ab.

Now, consider a tangent to the hyperbola in the 1st quadrant. If this was y=mx + c, then m > 0, c < 0.

The tangent intersects the y-axis at Z(0,c). Now if it intersects the tangent at X and Y

finding x-coords for X: y = mx+c & y=bx/a
bx/a = mx+c
bx=amx + ac
x=-ac/(am-b)

finding x-coords for Y: y = mx+c & y=-bx/a
-bx/a = mx+c
-bx=amx + ac
x=-ac/(am+b)

Now, from a diagram it's clear that the x-coord for X is larger than that of Y.

Notice that area(XOY) = area(XOZ) - area(YOZ) (*)
for triangles XOZ and YOZ, we have a common base given by |OZ| = (0 - c) = -c, and we have heights given by x-coords of X and Y.

thus from (*), ab = -1/2 c ({-ac/(am-b)} - {-ac/(am+b)})
ab = -1/2 c (-2abc / ({am}^2 - b^2) )
= ab*c^2 / ({am}^2 - b^2)
thus ({am}^2 - b^2) = c^2

if we pivot the line y=mx+c as to keep c constant, such that the line is no longer a tangent, it's clear that for cutting hyperbola, LHS < RHS, whilst for missing the hyperbola, LHS > RHS.

If c is positive, we have the tangent point in the 3rd or 4th quadrant, and x-coord for X is smaller than that of Y. The negative signs just get switched around.

 

[OLDMAN]

Thanks Spice. Also thought along those lines but decided it was too fiddly. Hope we are not driving the students too quickly towards question 8. How were you at this time last year? Don't tell me -you've been seeking out q7/8's since Yr.11! Care to help ND out on the Ellipse question?

 

[ND]

Cool solution spice girl, very interesting.

 

[OLDMAN]

For part b) i), using the discriminant isn't really that tedious, considering it is a q8.

write the line in y=mx+c ie. y=((X_0b^2)/(Y_0a^2))x-b^2/Y_0
We want to prove that c^2<(am)^2-b^2 ie.
b^4/Y_0^2 < ((X_0b^2)/(Y_0a))^2-b^2 (****)
How does one prove an inequality (or an equality, for that matter) efficiently, so that one does not appear to be assuming it in the first place? One way is to write it backwards. eg.
X_0^2/a^2-Y_0^2/b^2>1 given, X_0,Y_0 "inside" H
(X_0^2b^4)/(Y_0^2a^2)-b^2>b^4/Y_0^2 times by b^4/Y_0^2

and thus,

b^4/Y_0^2 < ((X_0b^2)/(Y_0a))^2-b^2 (****I really started with this) thus, L lies entirely outside H.
Part ii) follows a very similar line of reasoning as the "logical" proof of the equation of a chord of contact from an external pt. for a parabola, ellipse and hyperbola.