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Question

If r1 and r2 be the lengths of radii vectores of the parabola which are drawn at right angles to one another from the vertex, prove that r431r432=16a2(r231+r232).

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Solution


Let r1 is inclined at an angle θ with the axis then coordinates of point are P(r1cosθ,r1sinθ)

Then r2 is inclined at 90+θ with the axis, then coordinates of Q are (r2sinθ,r2cosθ)

Now P lies on the parabola

(r1sinθ)2=4ar1cosθr1sin2θ=4acosθ......(i)

Q also lies on the parabola

(r2cosθ)2=4ar2sinθr2cos2θ=4asinθ.......(ii)

Dividing (i) by (ii)

tan3θ=r2r1tanθ=(r2r1)13

cosθ=11+tan2θcosθ=r131r231+r232sinθ=tanθcosθsinθ=r132r231+r232

Substituting sinθ and cosθ in (i)

r1⎜ ⎜ ⎜ ⎜r132r231+r232⎟ ⎟ ⎟ ⎟2=4ar131r231+r232r1r232r231+r232=4ar131r231+r232r231r232=4ar231+r232

Squaring both sides

r431r432=16a2(r231+r232)


698305_641477_ans_e54b89be84f64b488796439c78194374.png

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