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Gruma

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another question from our friends at cambridge.
can someone show me how to set out this question.


exercise 3.3

note: .0 is theta

Q5) P(asec.0, btan.0) lies on the hyperbola
x^2/a^2 - y^2/b^2 = 1

The tangent at P cuts the x-axis at X and the y-axis at Y. Show that PX/PY = sin^2.0 and deduce that if P is an extremity of a latus rectum, then PX/PY = (e^2 - 1)/e^2.
 

Affinity

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approach (Omit):
deduce equation of tangent then find it's X and Y intercepts, use ratios for similar triangles to find PX/PY.

Then use the fact that the latus rectum has x coordinate ae to deduce the required equality


+++++++++++++++++++++++++++++++++++
(Derive equation of tangent in parametric form)
-My teacher insist that we can't quote the equation-

The equation of the tangent at P (a cos(t), b sin(t)) is

[x- asec(t)]*[dy/dx]=[y-btan(t)]
[x- asec(t)]*[(dy/dp)/(dx/dp)]=[y-btan(t)] (insert evaluation sign after the derivatives (p=t)

[x- asec(t)]*[bsec(t)/-atan(t)]=[y-btan(t)]

simplify etc (refer to book)
++++++++++++++++++++++++++++++++++++

xsec(t)/a - ytan(t)/b=1 (1)

Solve (1) simulatenously with y=0 to find X
y=0
x=a/sec(t)
X=(a/sec(t), 0)

Solve (1) simulatenously with x=0 to find Y
x=0
y= -b/tan(t)
Y=(0, -b/tan(t))

construct line L_1 through Y parallel to the x-axis, construct line through P perpendicular to the x-axis L_2, L_2 meets x axis at A and L_1 at B

PXA is similar to PYB
-you might want to justify this, although I don't think it is required-

PX/PY = PA/PB = | btan(t)/[btan(t)-(-b/tan(t))] |
=| tan(t) / [ ( tan^2(t) +1 ) / tan(t) ] |
=| tan^2(t) / sec^2(t) |
=sin^2(t)

first part done
++++++++++++++++++++++++++++++++++++

Now the foci are (ae,0) and (-ae,0) and the latus rectum is parallel to the Y axis

Therefore if P is an extremity of the latus rectum,
P=(+/-ae, btan(t))
asec(t) = +/- ae because P=(asec(t), btan(t))
sect(t) = +/- e

(e^2 - 1)/e^2
= 1 - 1/e^2
= 1 - 1/sec^2(t)
= 1- cos^2(t)
= sin^2(t)

(e^2 - 1)/e^2= PX/PY by above and result of part 1.
 
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Gruma

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hey thanks ive been trying to get a grasp on these types of questions all weekend, this helps.
 

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