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Analytical Geometry

Blazing goIITian

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4 Apr 2009 20:45:01 IST

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If r1, r2 be the lengths of the perpendicular chords of the parabola y^2=4ax drawn through the verte

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If r1, r2 be the lengths of the perpendicular chords of the parabola y^2=4ax drawn through the vertex, then show that(r1.r2)^4/3= 16a^2(r1^2/3+r2^2/3).

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anchal singhal

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4 Apr 2009 21:32:42 IST

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plz answer...

Prakhar Banga

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4 Apr 2009 21:49:24 IST

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Let the two chords be joining O and (at^2, 2at); O and (am^2, 2am)

As these are perpendicular, m = - 4 / t

The length r1^2 = (at^2)^2 + (2at)^2 = a^2(t^4+4t^2)..............1

Similarily, r2^2 = a^2(m^4+4m^2) = a^2((- 4 / t)^4+4(- 4 / t)^2) =a^2(256/t^4 + 64/t^2) = 64a^2 / t^6(4t^2 + t^4).............2

Let (4t^2 + t^4) = x

r1^2 = a^2 x and r2^2 = 64a^2 x / t^6.

r1^4/3 = a^4/3 x^2/3 and r2^4/3 = 16 a^4/3 x^2/3 / t^4

r1^2/3 = a^2/3 x^1/3 and r2^2/3 = 4 a^2/3 x^1/3 / t^2

r1^2/3 + r2^2/3 = 4 a^2/3 x^1/3 / t^2 + a^2/3 x^1/3 and r1^4/3 r2^4/3 = 16 a^4/3 x^2/3 / t^4 a^4/3 x^2/3

Now I seem to be getting nowhere.

Vikram Saxena

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5 Apr 2009 14:49:36 IST

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let the any arbiratary point of the parabola be

(at²,2at)

let the two points be::

A = (at₁² , 2at₁ ) and B =(at₂² , 2at₂)

AO = r₁ BO = r₂

AO and BO are perpendicular to each other........

2 / t₁ . 2 / t₂ = -1

now find r₁ and r₂................by applying the formula

hence ,u can prove it

Prakhar Banga

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6 Apr 2009 10:12:44 IST

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...........Continued from my last reply

r1^2 = a^2 x

r2^2 = 64 a^2 x / t^6

(r1.r2)^4/3 = (a^2 x*64 a^2 x / t^6)^2/3 = 16 a^(8/3) x^(4/3) / t^4 = 16 a^(8/3) x^(1/3) / t^4 * x = 16 a^(8/3) x^(1/3) / t^4 * (4t^2 + t^4)

= 16 a^(8/3) x^(1/3) * (4 / t^2 + 1) = 16 a^(2) * (4 a^(2/3) x^(1/3) / t^2 + a^(2/3) x^(1/3) ) = 16 a^(2) * (4 a^(2/3) x^(1/3) / t^2 + a^(2/3) x^(1/3) ) = 16 a^(2) * (r1^2/3 + r2^2/3 )

Hence Proved.

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