**An ellipse is inscribed in a semi circle touches the circular arc at two distinct points and also touches the bounding diameter. its major axis is parallel to the bounding diameter, When the ellipse has maximum possible area, its eccentricity is?**

How do you solve this question, this came in the **KVPY** exam that was held on 2nd november, and i couldnt solve this problem,
though i did guess it by simply finding which **e** gave the largest area,

Can any one give the actual solution

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## Comments

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TopNewestMy \(e\) is coming out to be \(\sqrt{\frac{2}{3}}\). Kindly tell me whether it is correct or not. If I am correct I will post the solution.

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will you please upload Your Solution @Ronak Agarwal

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Here's the solution.

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Let the point of intersection as shown in the figure be \((acos(t),b(1+sin(t)))\)

With respect to circle the point be \(A=(rcos(\theta),rsin(\theta))\)

Since the points coincide hence :

\(rcos(\theta)=acos(t)\) (i)

\(rsin(\theta)=b(sin(t)+1)\) (ii)

Dividing them we get :

\(tan(\theta)=\frac{b(1+sin(t))}{acos(t)}\) (iii)

We know that the tangents of the circle and ellipse at that point coincide.

Hence their slopes are equal.

\({m}_{ellipse}={m}_{circle}\)

\(\Rightarrow \frac{-bcot(t)}{a}=-cot(\theta)\) (iv)

Multiplying (iii) and (iv)

\(1=\frac{{b}^{2}(1+sin(t))}{{a}^{2}sin(t)}\) (v)

Squaring and adding (i) and (ii) :

\({r}^{2}={a}^{2}{cos}^{2}(t)+{b}^{2}{(1+sin(t))}^{2}\)

Using (v) we get :

\({a}^{2}{cos}^{2}(t)+{a}^{2}sin(t)(1+sin(t))={r}^{2}\)

\(\Rightarrow {a}^{2}=\frac{{r}^{2}}{1+sin(t)}\)

Hence \({b}^{2}=\frac{{r}^{2}sin(t)}{{1+sin(t)}^{2}}\)

So area of ellipse =\(A=\pi ab = {r}^{2} \sqrt{\frac{sin(t)}{{(1+sin(t)}^{3}}}\)

Maximising this we get the maximum at \(sin(t)=\frac{1}{2}\)

Hence we get :

\(\frac{{b}^{2}}{{a}^{2}}=\frac{1}{3}\) (Using (v))

\(\Rightarrow e=\sqrt{\frac{2}{3}}\)

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@Ronak Agarwal You did excellent work ! I used Co-ordinate geometry which is really too bad method in front of you :)

oh greatI used \(C:\quad { x }^{ 2 }\quad +\quad { (y+b) }^{ 2 }=\quad { R }^{ 2 }\\ \quad \\ E:\quad \cfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\cfrac { { \quad y }^{ 2 } }{ { b }^{ 2 } } \quad =\quad 1\).

And further which needs at-least 2 pages which is useless in front of Yours :)

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@Ronak Agarwal But Ronak shoudn't it is b(sin(t)) instead of b(1+ sin(t) ) Plz Explain me ! Thanks

Great !!! Thanks a lot !Log in to reply

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Okay uploading it.

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How much marks are you getting @Mvs Saketh . I'm just asking.

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also if a question turns out to be wrong, then will everyone be given marks? ( i am asking since you are already a kvpy scholar)

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Hey! which question are you talking about??

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Really bad, should have done level 2 chem instead of math,,, infact its shamefully 58, It was an assymetric distribution with physics way too easy and maths part 2 hard for me,

what about you? @Ronak Agarwal

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I am already a KVPY scholar hence I haven't given KVPY this year.

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And if u want to spend the rest of your life wondering,, i think its awesome ,,,

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Comment deleted Nov 04, 2014

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@Mvs Saketh !! what is procedure to get into IISC if a person never given the K.V.P.Y ?? Does JEE advance Helps in it ? If It is Then What is restrictions in ranks in it for getting an IISC ??

HeyAnd Also in this above Question Is Radius is fixed or not ? I can't understand what does this question want to convey will you please clarify it to me ?

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@DEEPANSHU GUPTA yes JEE advance does help,, minimum is above 60 percent,, but thats just the minimum, i heard that it is recommended to get a within 1000 rank,, and as far as i know olympiads help in it, which is why i am highly interested in NSEP,,

Radius is fixed,, its value has not been given, andyeah question says of all ellipses that can be inscribed in a semi circle , the ellipse has maximum area for what value of e (eccentricity)

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national standard examination of physicsmy bad i assumed you were,Log in to reply

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