Conic Section (1 Viewer)

[OLDMAN]

For those who were requesting a conic problem.
An interesting but fairly easy question.
Find the smallest area of the triangle formed by the tangent line to the ellipse (s-major a, s-minor b) with the coordinate axes in the first quadrant.

 

[ND]

For those who were requesting a conic problem.

Oops i didn't see this, and posted the soln.... I've now censored it so the people requesting it can have a go. To see my post, click "Quote" in the bottom right hand corner of this message.

edit: *uncensored*

Let A and B be the point of intersection of the tangent and the x and y axes resp.

eqn of tangent to ellipse at P(acos@, bsin@):

(xcos@)/a + (ysin@)/b = 1

when y = 0:

x = a/cos@

when x = 0:

y = b/sin@

.'. A(a/cos@, 0), B(0, b/sin@)

Now Area = AB/2
= ab/2cos@sin@
= ab/sin2@

because sin2@ varies between -1 and 1, area is minimum when:
sin2@ = 0
2@ = pi/2 (only taking this value because angle has to be in 1st quad)
@ = pi/4

.'. min area = ab/sin(pi/2)
=ab

 

[OLDMAN]

ND its open to all.

 

[ND]

Ah ok.

 

[OLDMAN]

ND : Nice. A few months back you were asking how you might increase your Elegance Quotient. Elegant answers beget elegant questions and so on. In this forum for example, problems have been posed and elegant solutions (particularly Spice Girl's) presented. Look at an elegant solution and see whether you can come up with an alternative elegant question.
Using the parametric form in this problem was an "elegant" step. It behoves (pardon the English) me to come up with another similar question (hopefully elegant) involving the hyperbola -I'll post this in another thread.

Make sure that as time flies inexhorably to the HSC, your list of elegant questions and their corresponding elegant answers grows. Play them over and over again like favourite songs.

 

[OLDMAN]

Remember : one good song is better than 100 lousy songs.

 

[underthesun]

That's a good question, but i have a question too. In exam situation, can you just write this:

For area (ab/sin2@) to be minimum, then the denominator has to be in it's maximum. Because sin2@ has a maximum of 1, area (ab/sin2@) is minimum when sin2@ = 1. Hence we let sin2@ = 1 to find minimum value.

Area = ab/1 = ab

to get full marks?

 

[OLDMAN]

Yes you may. But to be sure, remind the examiner that since it's in the first quadrant sin2@>0.

 

[ND]

I think it would be sufficient to write:

'.' -1 \< sin2@ \< 1, area is minimum when sin2@ = 1.

 

[ND]

Originally posted by OLDMAN

Make sure that as time flies inexhorably to the HSC, your list of elegant questions and their corresponding elegant answers grows. Play them over and over again like favourite songs.

I don't completely understand, do you mean that i should keep going over a question many times finding different solutions?

 

[OLDMAN]

I meant go over the solution, find out what makes it elegant, and usually the structure of the solution will provide you pointers in how an examiner's mind works. eg. if this was shown for an ellipse, what is the corresponding hyperbola question, or if something is valid for n=5, can you answer the same question for any n. Hope I am not talking bs.

 

[ND]

Ah i understand now, thanks.