help me with conics please urgently (1 Viewer)

fashionista

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i've got this question from the arnold book and i cant get it out
therefore i conclude i hate conics..unless someone can help e change my mind....PLEASSSEE!!!

here goes:
P(asecA, btanA) lies on the hyperbola xx/aa-yy/bb=1. The tangent and normal cut the y-axis at T and G repsectively. Show that the circle on GT as diameter passes through the foci S and S'.

GARRRRRRrR i keep getting pages and miles of algebra involving many tans and secs and squares and shitty and i need help
URGENTLY!!!!!!!!
 

edd91

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Find the tangent and normal at P, then find their y intercept (sub x=0), then use distance formula to get distance between those two lines, then use that distance as a diameter of a circle, x^2 + y^2 = r^2, then sub in the focii points (+-ae,0) and show they satisfy the circle equation
 

fashionista

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i did but i got the distance as being some insanely long equation with many tan^2 and sec^2 and other squares in it all in fraction form...oh the pain of it all. i got the distance between T and G (diameter) and then halved it (radius) and got huge equations. and subbing the focii in just made it longer
 

edd91

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you shoudlnt have any secs in your working
I have done this one so heres some stuff to do:
Find where G is, find where T is, (draw a diagram for clarity)
find GT, which is just G-T because theres no x values (on the y axis)
If the circle lies on GT, and has diameter GT, it meens the centre is at the midpoint of GT, and the radius is GT/2

:)
 

CM_Tutor

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If the circle has GT as diameter, then it must pass through both foci S and S', or neither, by symmetry - so, it is sufficient to prove that it passes through S. The circle will pass through S if, and only if, SG and ST are perpendicular. So, I'd just show these are perpendicular, rather than trying to put (ae, 0) into the equation of the circle.
 

edd91

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How would you show that SG and ST are perpindicular CMtutor? ALso could you goto the extracirricular thread and look at my directors circle thread, is that what you'd use in this example?
 

CM_Tutor

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Tangent at P(asec A, btan A) to (x<sup>2</sup> / a<sup>2</sup>) - (y<sup>2</sup> / b<sup>2</sup>) = 1 is (x / a)sec A - (y / b)tan A = 1.
So, T (at x = 0) is (0, -bcot A).

Normal at P is axcos A + bycot A = a<sup>2</sup> + b<sup>2</sup>. So, G is at (0, tan A(a<sup>2</sup> + b<sup>2</sup>) / b).

Noting S is (ae, 0), m<sub>ST</sub> = -(bcot A) / ae
Similarly, m<sub>SG</sub> = tan A(a<sup>2</sup> + b<sup>2</sup>) / abe

So, m<sub>ST</sub>m<sub>SG</sub> = -[bcot A * tanA(a<sup>2</sup> + b<sup>2</sup>)] / a<sup>2</sup>be<sup>2</sup>
= -(a<sup>2</sup> + b<sup>2</sup>) / a<sup>2</sup>e<sup>2</sup>, noting cot A * tan A = 1, and cancelling the 'b' on the numerator and denominator
= -[a<sup>2</sup> + a<sup>2</sup>(e<sup>2</sup> - 1)] / a<sup>2</sup>e<sup>2</sup>, noting b<sup>2</sup> = a<sup>2</sup>(e<sup>2</sup> - 1) for a hyperbola
= -a<sup>2</sup>e<sup>2</sup> / a<sup>2</sup>e<sup>2</sup>
= -1

So, SG _|_ ST, and so (as noted above) S and S' lie on the circle with GT as diameter, using the converse if the angle in a semicircle theorem.
 
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CM_Tutor

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As a general rule, if a conics question is turning into an algebra bash, STOP. Draw a diagram (big) that you can draw all over. There are very few conics questions that can't be made relatively simple (algebraically) by applying geometric simplifications. So, mark all the geometric relationships that you can find - similar and congruent triangles, circles and related circle geometry theorems, intercept theorems, right angles and Pythagoras' theorem, simple trigonometry, etc, etc. Remember that many conics problems have a 10 page, a 2 page and a half a page solution - in this way, an exam question can penalise you time for doing it the wrong way, even if you get all the marks. Always try to find the half page approach.
Originally posted by edd91
ALso could you goto the extracirricular thread and look at my directors circle thread, is that what you'd use in this example?
I'll see what I can do later today or tonight. :)
 

fashionista

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its ok i suck..i guess i confirmed that for myself after todays assessment :( i hate maths.
 

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