IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

mehrotra.pulkit

Q1) Find the value of 'm' for which the area of the triangle included between the axes and any tangent to the curve (x^m)*y=(b^m) is constant

Q2) if the tangent at(a,b) to the curve x^3+y^3=c^3 meet the curve again in(a1,b1),then prove that (a1/a)+(b1/b)=-1.

deedee

first q is d ansewr 1??

srujana

$Differnentiating\ wrt\ x,we\ get$

$\frac{dy}{dx}=-\frac{x^2}{y^2}$

$Tangent\ at\ the\ point,(a,b)\ is\ -\frac{a^2}{b^2}$

$as\ (a_1,b_1)\ is\ also\ a\ point\ on\ the\ curve,we\ get$

${a_1}^3+{b_1}^3=c^3$

$a^3+b^3=c^3\ [as\ (a,b)\ is\ also\ a\ point\ on\ the\ curve]$

$as\ (a,b)\ and\ (a_1,b_1)\ are\ the\ points\ on\ the\ tangent,slope\ of\ the\ tangent\ is$

$-\frac{a^2}{b^2}=\frac{b_1-b}{a_1-a}$

$solving\ the\ following\ equations,$

${a_1}^3+{b_1}^3=c^3$

$-\frac{a^2}{b^2}=\frac{b_1-b}{a_1-a}$

$we\ get\ the\ required\ equation$

$\frac{a_1}{a}+\frac{b_1}{b}=-1$

ashish_banga

Q 1 m = 1
as u can see curve passes through P ( b, 1 )
also slope of the curve = - my/ x
slope at P = -m/b
the equation of tangent at P =
(y - 1) / ( x - b) = -m /b

this line meets x and y axis at ( b( 1 + m)/ m , 0 ) and ( 0 , 1 + m )

area of triangle = 1/2 b ( 1 + m )^2 / m

to get m differentiate it and put equal to 0 as it is constant

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