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Analytical Geometry

Blazing goIITian

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8 Mar 2009 19:31:09 IST

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692

Really Tough Ellipse questions

None

1.With a given point and line as focus and directrix, a series of ellipses are described.The locus of the extremities of their minor axes is

A.an ellipse

B.a hyperbola

C.a pair of straight lines

D.a parabola

2.An ellipse of major axis and minor axis 20 slides along the coordinate axes and always remains confined in 1st quadrant.The locus of the ellipse therefore describes athe arc of a circle.Find the length of this arc and also the equation of the locus of the centre of the ellipse.

Please solve with appropriate explanation....

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Comments (11)

aNdRoMeDa

Blazing goIITian

Joined: 10 Sep 2007

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8 Mar 2009 21:43:46 IST

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a pair of staright line

Rohit

Blazing goIITian

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9 Mar 2009 16:13:03 IST

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No,that isn't correct.

shishu

Hot goIITian

Joined: 25 Sep 2008

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9 Mar 2009 16:16:38 IST

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a st line as only 2 ellipse can be described.

Rohit

Blazing goIITian

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9 Mar 2009 16:29:18 IST

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No.Any other opinions please..

Rohit

Blazing goIITian

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10 Mar 2009 23:47:30 IST

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Please reply anyone..Are u afraid of the questions??

shashank

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11 Mar 2009 00:52:37 IST

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the locus is x^2*c^2+2cxy-y^2c^2-2cx^3-2y^2x^2

where is 0,0 is given point and x=c is the given line

I think the answer is Pair of straight lines coz the| abc.....| determinant turns out to be 0.

Rohit

Blazing goIITian

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11 Mar 2009 01:58:33 IST

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No,it isn't...

Mr.Nasty ® Retired from Action.

Blazing goIITian

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11 Mar 2009 05:55:24 IST

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A HYPERBOLA, for the first question innit?

Rate it dude!

Anant Kumar

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11 Mar 2009 09:16:49 IST

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1.) Let the fixed point be $(a,,0)$ and the fixed line be the $y$ -axis. Then, if $e$ be the eccentricity of the ellipse, its equation is
$(x-a)^2 + y^2 = e^2 x^2$
$Rightarrow (1-e^2)x^2-2ax+a^2+y^2=0$
$Rightarrow (1-e^2)left(x^2-2xdfrac{a}{1-e^2}+dfrac{a^2}{(1-e^2)^2}-dfrac{a^2}{(1-e^2)^2}ight)+a^2+y^2=0$
$(1-e^2)left(x-dfrac{a}{1-e^2}ight)^2+y^2=dfrac{a^2e^2}{1-e^2}$
$dfrac{left(x-dfrac{a}{1-e^2}ight)^2}{left(dfrac{ae}{1-e^2}ight)^2} + dfrac{y^2}{left(dfrac{ae}{sqrt{1-e^2}}ight)^2} =1$
This ellipse has its center at the point $left(dfrac{a}{1-e^2},,0ight)$ , the semi-major axis is $dfrac{ae}{1-e^2}$ and the semi-minor axis is $dfrac{ae}{sqrt{1-e^2}}$ .
The extremities of the ends of the minor axis, therefore, are $x=dfrac{a}{1-e^2}$ and $y=pm dfrac{ae}{sqrt{1-e^2}}$
From the first relation we get $e^2=1-dfrac{a}{x}$ .
Squaring the second one, we get $y^2=dfrac{a^2e^2}{1-e^2}=dfrac{a^2left(1-dfrac{a}{x}ight)}{dfrac{a}{x}}=a(x-a)$ .
Therefore, the locus of the extremities of the minor axis of the ellipses is $oxed{y^2=a(x-a)}$ , which is a parabola with vertex at the fixed point $(a,0)$ and the and the focus at the point $left(dfrac{5a}{4},,0ight)$ . Hence option (D).

Anant Kumar

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11 Mar 2009 09:28:42 IST

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For the second case, we need to know the initial position from where it starts sliding. However the locus of the center will be simply

x² + y² = 20² .

Rohit

Blazing goIITian

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11 Mar 2009 16:23:18 IST

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Thank you very much sir.Sir,in second question if it slides from vertical to horizontal position,what will be length of the arc described by its centre??

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