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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Ellipse question................
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popat.bp (13)

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1) The maximum distance of the centre of the ellipse         x2/9 + y2/4=1 from the chord of contact of mutually perpendicular tangents of the ellipse is....
ans:9/13
    
popat.bp (13)

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plzzz ans it ...give it a try........
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popat.bp (13)

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common goiitians.........
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Mr.IITIAN007 (2985)

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Hey dude I guess thats 9 and 4 are in denominator , otherwise thats not an ellipse thats  a hyperboloid  with centre at the origin where no tangents can be mutually perpendicular (unless presented in complex plane) , either of which you haven't specified  .
See , the tangents have been drawn by taking any point on the director circle of the ellipse .

Any point on the director circle can be taken as
(root over 13x cos theta , root over 13 x sin theta ).

Hence equation of the corresponding contact is
[ (root over 13) / 9] x cos theta x X + [ (root over 13) / 4] x sin theta x Y=1

Its distnce from the origin is
1 /  root over[ ( 13 / 81 cos sq theta + 13 / 16 sin sq theta ) ]
this is smaller than equal to (36/ root over 208) which is ( 9 / root over 13).

Is that  okay dude !!!!!!!!

Ken
From: UNITED STATES, Green Bay, Wisconsin
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