Elipse in a semi circle

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

#Algebra #Geometry #KVPY

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

My $e$ 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.

Ronak Agarwal - 6 years, 7 months ago

will you please upload Your Solution @Ronak Agarwal

Karan Shekhawat - 6 years, 7 months ago

Here's the solution.

Image

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))

$\Rightarrow e=\sqrt{\frac{2}{3}}$

Ronak Agarwal - 6 years, 7 months ago

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

Karan Shekhawat - 6 years, 7 months ago

@Karan Shekhawat – Actually with respect to origin the ellipse has it's centre at (0,b) hence I have shifted it's co-ordinates accordingly.

Ronak Agarwal - 6 years, 7 months ago

@Ronak Agarwal – Thanks ! Now I got it completely ! I really appreciate your solution very much !!

Karan Shekhawat - 6 years, 7 months ago

@Karan Shekhawat – Thanks

Ronak Agarwal - 6 years, 7 months ago

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

Deepanshu Gupta - 6 years, 7 months ago

@Ronak Agarwal – That was awesome.... simply awesome... thanks for uploading...

Mvs Saketh - 6 years, 7 months ago

Okay uploading it.

Ronak Agarwal - 6 years, 7 months ago

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

Ronak Agarwal - 6 years, 7 months ago

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

Mvs Saketh - 6 years, 7 months ago

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

Ronak Agarwal - 6 years, 7 months ago

@Ronak Agarwal – Oh, ok so bro can you please tell me what are the chances of getting through 2nd level for me(based on score)? and is there hope for me to get into IISC,, if not through this,, will qualifying NSEP (if i am able to) help?

Mvs Saketh - 6 years, 7 months ago

@Mvs Saketh – Your are sure to go to 2nd stage but you have to perform a lot better in the interview, also can you please tell me what are the benifits of going into IISC as I am also interested in taking admission into this institution, but I am very doubtful about the oppurtunities after my graduation. I am very much interested in going into research field.

Ronak Agarwal - 6 years, 7 months ago

@Ronak Agarwal – Well IISC is perhaps the best place for BSc MSc Phd in india,, ofcourse CMI, DAE are no less, but IISC has international recognition as far as i know,, and i think oppurtunities depend on how much u are willing to put in,,

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

Mvs Saketh - 6 years, 7 months ago

@Mvs Saketh – I am very confused whether to go for BTech or BSc MSc.

Ronak Agarwal - 6 years, 7 months ago

@Ronak Agarwal – I think u will excel in whatever u do!

Mvs Saketh - 6 years, 7 months ago

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)

Mvs Saketh - 6 years, 7 months ago

Hey! which question are you talking about??

Ayush Garg - 6 years, 7 months ago

@Ayush Garg – Maths Part-2, you can see this question there.

Ronak Agarwal - 6 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$