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
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(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)
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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
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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.
