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Question

Find area of the ellipse x225+y216=1.

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Solution

We know x2a2+y2b2=1 is an ellipse,
Let us consider general point by (x,y)=(acosθ,bsinθ)
So,
x2a2+y2b2=1

x=5cosθ,y=4sinθdxdθ=5cosθ,dydθ=4cosθSlopeoftangent=dydx=dtdθdxdθ=4cosθ5sinθ=4cotθ5Slopeofnormal=1dydx=14cotθ5=54cotθ=5tanθ4
Now, normal equation
(5cosθ,4sinθ)=y4sinθ=1dydx(x5cosθ)y4sinθ=5tanθ4(x5cosθ)4y16sinθ=5xtanθ25sinθ4y16sinθ5xtanθ+25sinθ=0

Line distance from ellipse's centre
d=(5242)sinθ(1+tan2θ)12=9sinθsecθ=9sinθcosθ=9sin2θ2
sin2θ maximum value = 1
Maximum value dmax=92

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