Originally Posted by haboozin
u used DELTA = 0 and got that?
DELTA = 0 came out extremly long... (cannot be done in an exam)
if you used that method can you post your working...
hi haboozin, well actually it wasn't that long and only took about 3/4 pages, so it's certainly doable in an exam and i think it's the formal way of doing it.
i suspect the reason why it takes too long (esp. with expansions and squares of trig expressions, etc) is that you probably attempted to expand everything in your working out. what i did was look for places where you can simplify and factorise, and as it turned out the equation was relatively 'neat' in the end.
i won't type out all the working out i did (because i'm going to show you another method i
just found in the meantime) but this is the 'condensed' equation i eventually got using your initial method:
0 = (eSec@ +1)x^2 - 2a(Sec@ +e)x + (a^2)(e^2 -Sec@^2)/(eSec@ -1) ;
where 'x' is of course the x-coordinate of the point on the circle that the tangent touches. solving for DELTA =0 from that will get you
Sec@ = -e .
Originally Posted by haboozin
also what about perpendicular distance
since the perpendicular distance will be the radius
wouldnt | ax + by + c|/SQRT(a^2+b^2) = aSQRT(e^2 + 1)
ie
|esec@ -1|/SQRT(sec^2@/a^2 + tan^2@/b^2) = aSQRT(e^2 + 1)
but i cant do the algebra
well i can't see how this method will "simplify" the workload needed to get you the answer, it doesn't look that much simpler than your initial method. plus, i personally don't like to work with too many degree 2 powers and expressions in questions of this type since they significantly complicate matters.
anyhow, like i alluded to above, i have just found a new way of doing your question without the need to involve quadratic expressions or equations. (however, i doubt this will actually take less time to do in an exam than the formal method, esp. for those who are good at algebra already anyways.)
but here it is:
the equation of tangent of the
hyperbola is:
xSec@/a - yTan@/b =1
the equation of tangent of the
circle is (you can derive this yourself):
yY +xX -ae(x +X) =a^2 , where {X, Y} is the point on the circle where the tangent touches.
now from the hyperbola tangent, when
y=0,
x=a/Sec@ ;
but since both the tangent to the hyperbola and tangent to the circle are in fact the SAME tangents (or lines), then {a/Sec@, 0} will also satisfy the tangent of circle: (0)Y +aX/Sec@ -ae(a/Sec@ +X) =a^2 ; (notice how the (0)Y just takes care of the 'Y' ^^) -----> so you end up with:
X = a(Sec@ +e)/(1 -eSec@)
now all you need to do is come up with another expression for 'X' and then equate:
the tangent to the circle at {X, Y} is also the tangent to the hyperbola at {aSec@, bTan@}, so that satisfies the circle tangent:
Y(bTan@) +X(aSec@) -ae(X +aSec@) =a^2 ;
but also, the gradient of the hyperbola tangent =
bSec@/aTan@ = gradient of the circle tangent =
(ae-X)/Y ----->
Y =a(ae-X)Tan@/bSec@ ;
and now substitute this expression for 'Y' into the previous equation of tangent:
[a(ae-X)Tan@/bSec@][bTan@] +x(aSec@) -ae(x +aSec@) =a^2
-----> {solving for 'X'}
X =a(Sec@ +e)/(1-e)
and so equating the two expression found so far for 'X':
X = a(Sec@ +e)/(1 -eSec@) = a(Sec@ +e)/(1-e) ----->
0 =e(Sec@ +e)(1 -Sec@) -----> Sec@ = {-e, 1}, but Sec@ can't = 1 as 'e' can't =0 ;
hence,
Sec@ = -e .
hope that helps
i suppose the algebra in this method isn't as hard or prolific as that of the first method.
