If the area(in sq. units) bounded by the parabola y2 = 4λx and y= λx, λ

Engineering

Mathematics

Introduction to Area

Question

If the area(in sq. units) bounded by the parabola y² = 4λx and y= λx, λ > 0, is $\frac{1}{9}$ then λ is equal to

$4 \sqrt{3}$

$2 \sqrt{6}$

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

y² = 4λx & y = λx

λ²x² = 4λx

$x = 0 & x = \frac{4}{λ}$

$Area = \int_{0}^{4 / λ} (\sqrt{4 λx} - λx) dx = \frac{1}{9}$

$\Rightarrow 2 \sqrt{λ} \times {(\frac{x^{3 / 2}}{\frac{3}{2}})}^{4 / λ} - λ {(\frac{x^{2}}{2})}_{0}^{4 / λ} = \frac{1}{9}$

$\frac{4}{3} \sqrt{λ} \times \frac{{(2^{2})}^{3 / 2} x}{λ^{3 / 2}} - \frac{λ}{2} \times \frac{16}{λ^{2}} = \frac{1}{9}$

$\Rightarrow \frac{32}{3 λ} - \frac{8}{λ} = \frac{1}{9}$

$\Rightarrow \frac{32}{3 λ} = \frac{1}{9}$ ⇒ = 24