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Polar coordinate knowledge points in senior two mathematics.
Solution:

Taking the origin of rectangular coordinates as the pole and the positive direction of X axis as the polar axis, it is relevant when establishing polar coordinate system.

x =ρ* cosθ; y=ρ*sinθ,

The elliptic equation is transformed into:

(ρ*cosθ)? /a? + (ρ*sinθ)? /b? = 1

Let the coordinates of point A be a (ρ 1, θ1); The coordinate of point b is b (ρ 2, θ 2);

* oa⊥ob

∴θ2 =θ 1π/2 = = & gt; Because? θ2 = sin? θ 1; Sin? θ2 = cos? θ 1;

A and b are points on the ellipse, so there are:

(ρ 1*cosθ 1)? /a? + (ρ 1*sinθ 1)? /b? = 1

= = & gtρ 1? = a? b? /(a? Sin? θ 1 + b? Because? θ 1) - ( 1)

(ρ2*cosθ2)? /a? + (ρ2*sinθ2)? /b? = 1

= = & gtρ2? = a? b? /(a? Sin? θ2 + b? Because? θ2)

= a? b? /(a? Because? θ + b? Sin? θ 1) - (2)

sδAOB = 1/2 * | OA | * | OB |

= 1/2 *ρ 1*ρ2

= 1/2 *√[a? b? /(a? Sin? θ 1 + b? Because? θ 1)] * √[a? b? /(a? Sin? θ 1 + b? Because? θ 1)]

= a? b? /√[4a? b? +sin? (2*θ 1)*(a? -B? )? ]

As a crime? When (2*θ 1) = 1, the triangle area has a minimum value.

Minimum value = a? b? /√[4a? b? + 1/4*(a? -B? )? ] = a? b? /(a? + b? )

As a crime? When (2*θ 1) = 0, the triangle area has a maximum value.

Maximum value = a? b? /√(4a? b? )= ab/2