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for a>0 the line a(a^2+1)y = a - x forms a triangle with co ordinate axes then greatest area of this triangle is

for a>0 the line a(a^2+1)y = a - x forms a triangle with co ordinate axes then greatest area of this triangle is

Grade:11

1 Answers

Riddhish Bhalodia
askIITians Faculty 434 Points
5 years ago
we can modify the equation of the line as
\frac{y}{\frac{1}{1+a^2}} + \frac{x}{a} = 1
as a>0 the x and y intercepts are a and 1/(a2 +1)
hence the area of the triangle formed is
\Delta = \frac{a}{2(a^2+1)}
we need to maximize the area wrt a
so
\frac{d\Delta}{da} = \frac{(1+a^2) - a(2a)}{(1+a^2)^2} = 0
and hence solving that we get a = 1 as a possible solution and on checking the double derivative at a=1 is -ve and hence its a maxima
so the greates area is
\Delta = 1/4

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