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Classical Topics of Mathematical Conic Curve
Let A(0, b),

B(0,-b),

M(m,-b),

P(p,q)

P is on the ellipse, p 2/a 2+q 2/b 2 = 1.

( 1)

k(AM)=k(AP)

= & gt

-2b/m=(q-b)/p

(2)

MO⊥PB

= & gt

k(MO)*k(PB)=- 1

= & gt

-b/m*(q+b)/p=- 1

(3)

(2)/(3) Available

2p/(q+b)=-(q-b)/p, available after finishing.

2p^2+q^2=b^2

(4)

(4)-( 1) * b 2, available

2p^2-p^2*b^2/a^2=0

That is 2 = b 2/a 2.

It can be seen that for ellipses, there are

B>a, that is, the long axis is on the y axis.

∴e=c/b, then b 2 = a 2+c 2.

2=b^2/a^2=b^2/(b^2-c^2)= 1/( 1-e^2)

solve

e= 1/√2=√2/2

∴ Eccentricity of ellipse is e=√2/2.