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Analytical Geometry

Cool goIITian

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18 Mar 2009 23:51:43 IST

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366

try this

None

find the area of quadrilateral formed by the tangents at the ends of the latus rectum of the ellipse 9x²+ 16y² -144 = 0.

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Viraj Shah

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Joined: 6 Mar 2009

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19 Mar 2009 00:08:23 IST

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Is the answer 8 /root7??

Ankit Rana

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19 Mar 2009 00:12:34 IST

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arrange in normal form

i.e. x²/a² + y²/b² = 1

then find e.......e = root(7) / 4

now eq. of tangent at (ae, b²/a²) ....end pt of LT is xe / a + y /a² = 1

Find area is first quad. and multiply by four coz four such area exist (ellipse is symm. about x and y axis)

hence the area must be 4 * 0.5 * (a/e) * a²

here a = root(144/9) = 12 /3 = 4 and e = root(7) /4

hence req. area = 512 / root(7)

abhilash

Cool goIITian

Joined: 24 Oct 2008

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19 Mar 2009 10:43:00 IST

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i got some other ans 128/........................the steps are correct i think their is mistake in calculation either mine or urs.......................

can u plz send me the eqⁿ of tangent and normal found by u so that i can verify it with mine...............................

Mirka

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21 Mar 2009 20:29:09 IST

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PLS READ ENTIRE THING CAREFULLY ND UNDERSTAND

Eqⁿ of ellipse is x²/16 + y²/9 = 1

a = 4, b = 3, so e = root(7) / 4

foci = ( plus or minus ae, 0 ) = ( +- rt(7) , 0 )

Refer the figure :

Consider only 1^st & 4^th Quadrant,

U get coordinates of end-pts of latus rectum as

( rt (7) , 9/4 ) and ( rt (7) , -9/4 )

For ellipse, v know eqn of tangent at point ( x₁, y₁ ) is given by

x . x₁ / a² + y. y₁ / b² = 1

with this formula, u get tangent at A is

rt(7) . x / 16 + y / 4 = 1

take it in intercept form

u get

x intercept = 16 / rt(7)

y intercept = 4

Similarly u find tht tangents at all other points ( B, C and D )

Also have same intercepts on each axes (this is obvious due 2 symmetry!!)

Area of reqd region is 4 times d area of wat I have coloured in d diagram

This is 4 * ( ½ * 16/rt(7) * 4 )

= 128 / root(7)

Answer confirmed ….

U see dat reqd region is actually a RHOMBUS (diagonals perpendicular)

wid diagonals of length 32/ rt(7) and 8 units....

so area = ½ * d1 * d2

= ½ * 32 / rt(7) * 8

= 128 / rt(7)

This way also u can doooo

Hope its clear !

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