# Doubt from ellipse

Question 75 i could only found till maximum area which i got 6pi then i got confused

Its a decimal integer type question(upto 2 decimal place)

If you found maximum area then you have found a relation in a and b

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I got 9a^2+b^2=36 also and ab<6

0.95 approx

Damn how do you figure out the max area under ellipse?

Edit:Nvm got it

Use AM>= GM

Parameterise the ellipse as x=a \cos \theta, y =b\sin \theta.

To find intersection with the line we substitute for x,y to get b\sin \theta = 3 a\sin \theta +6

or b \sin \theta - 3 a \sin \theta = 6 \Rightarrow \dfrac{b}{\sqrt{9a^2+b^2}} \sin \theta -\dfrac{3a}{\sqrt{9a^2+b^2}} \cos \theta = \dfrac{6}{\sqrt{9a^2+b^2}}

So if this has a unique solution we need RHS =1 i.e. 9a^2+b^2=36 \Rightarrow 36 \ge 6ab by AM-GM or ab \le 6 with maximum occurring when b=3a

The area of the ellipse =\pi ab is maximised when b=3a so that e = \dfrac{2\sqrt2}{3}

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Thank you sir