MCQ
If the area (in sq. units) bounded by the parabola $y^2 =4\lambda x$ and the line $y = \lambda x$, $\lambda > 0$, is $\frac{1}{9}$, then $\lambda $ is equal to
- A$48$
- B$4\sqrt 3$
- C$2\sqrt 6$
- ✓$24$
✓
Answer
Correct option: D.
$24$
d
$y^{2}=4 \lambda x$ and $y=\lambda x$
$y^{2}=4 \lambda x$ and $y=\lambda x$
$\lambda^{2} x^{2}=4 \lambda x$
$x=0$ and $x=\frac{4}{\lambda}$
Area $ = \int\limits_0^{4\lambda } {(\sqrt {4\lambda x} - \lambda x)} dx = \frac{1}{9}$
$ \Rightarrow 2\sqrt \lambda \times \left( {\frac{{{x^{3/2}}}}{{3/2}}} \right)_0^{4/\lambda } - \lambda \left( {\frac{{{x^2}}}{2}} \right)_0^{4/\lambda } = \frac{1}{9}$
$\frac{4}{3} \sqrt{\lambda} \times\left(2^{2}\right)^{3 / 4} \frac{x}{\lambda^{3 / 2}}-\frac{x}{2} \times \frac{16}{\lambda}=\frac{1}{9}$
$\Rightarrow \frac{32}{3 \lambda}-\frac{8}{\lambda}=\frac{1}{9}$
$\Rightarrow \frac{8}{3 \lambda}=\frac{1}{9}$
$\lambda=24$
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