Question

If the area (in sq. units) bounded by the parabola y² = 4λx and the line y = λx, λ > 0, is 1/9, then λ is equal to :
(A) 48 (B) 4√3
(C) 2√6 (D) 24

Answer 1

Answer:

I am using " a " instead of lambda for ease .

First find intersection point

${y}^{2} = 4ax \\ \\ put \: y = ax \\ \\ (ax) {}^{2} = 4ax \\ (ax) {}^{2} - 4ax = 0 \\ ax(ax - 4) = 0 \\ \\ x = 0 \: \: and \: \: \frac{4}{a} \: \: \: \: \: \: \: \: a \neq0$

Now area under curve

$= \pink{ \int_{0} ^{ \frac{4}{a} } ( \sqrt{ 4ax} - ax) \: dx }\\ \\ = \frac{2}{3} x \sqrt{x} ( \sqrt{4a}) - a .\frac{ {x}^{2} }{2} \\ \\ put \: limits \\ \\ = \frac{2}{3} \times \frac{4}{a} \sqrt{4a \times \frac{4}{a} } - a \times \frac{16}{2 {a}^{2} } \\ \\ = \frac{32}{3a} - \frac{8}{a} \\ \\ = \frac{8}{3a} = \frac{1}{9} \: \: \: \: (given) \\ \\ \implies \huge \red{a = 24}$