IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

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let A ( p^2, -p) B (q^2 , q) and C ( r^2 , -r) be the vertices of traingle ABC.

a parallelogram AFDE is drawn with D , E , F on the line segment BC, CA, AB respectively. using calculas show that the maximum area of such parallelogram is (p+q)(q+r)(p-r)/4...

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ramyadiamond

Given

A( p²,-p), B(q²,q) C(r²,-r)

Now take one side of the paralellogram AF as y and AE as x. Drop a perpendicular to AF from E so that it meets AF at R.

Now area of parallelogram= S = AF*ER

= y* (x.sinA)

Now in triangle ABC,

ED is parallel to AB, Angle A = Angle CED, angle B= angle CDE

Hence, triangle ECD is similar to triangle ACB (by ASA prop)

therefore, EC/AC = ED/AB

(AC-x)/AC = y/AB

1-x/b = y/c and b and c are the sides AC and AB

x = b(1-y/c)

Substituting back in the eqn, we get area as

S = by(1-y/c).sinA

Differenciating w.r.t y

dS/dy = (b -2yb/c).sinA

on equating it to zeo, u get y =c/2

again differenciating,

d²S/dy² = -2sinA/c <0

negative sign implies that there's a maxima at y=c/2

Now putting y=c/2, we get area as

S= bc.sinA/4

But area of triangle ABC= 1/2 bc.sinA

So S= 1/2 (area of triangle ABC)

Area of triangle ABC can be found be determinant method since u have been given all three coordinates.

on solving u get

S = (p+q)(q+r)(p-r)/4

Hence, proved.

Since i haven't posted any diagram, so plz make the diagram first else it can be confusing!!

Hope its clear!!!

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