### Parabola Question 2

#### Question 2 - 24 January - Shift 2

The urns A, B and C contain 4 red, 6 black; 5 red, 5 black and $\lambda$ red, 4 black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn $C$ is 0.4 then the square of the length of the side of the largest equilateral triangle, inscribed in the parabola $y^{2}=\lambda x$ with one vertex at the vertex of the parabola is ________

## Show Answer

#### Answer: 432

#### Solution:

#### Formula: Slope of the tangent

$m^{2}+2 m-3=0$

$\Rightarrow m=1,-3$

So tangent which touches in first quadrant at $T$ is

$T \equiv(\frac{a}{m^{2}}, \frac{2 a}{m})$

$\equiv(\frac{3}{2}, 3)$

$\Rightarrow CT=\sqrt{4+4}=2 \sqrt{2}$

Urn | Red | Black |
---|---|---|

A | 4 | 6 |

B | 5 | 5 |

C | $\lambda$ | 4 |

$P(\frac{C}{R})=\frac{P(C) P(\frac{R}{C})}{P(A) P(\frac{R}{A})+P(B) P(\frac{R}{B})+P(C) P(\frac{R}{C})}$

$0.4=\frac{\frac{1}{3} \times \frac{\lambda}{(\lambda+4)}}{\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}+\frac{1}{3} \frac{\lambda}{(\lambda+4)}}$

$\Rightarrow \lambda=6$

$\tan 30^{\circ}=3 t=\frac{3}{2} t^{2}$

$\frac{1}{\sqrt{3}}=\frac{2}{t}$

$t=2 \sqrt{3}$ $\stackrel{Q^{2}}{ }{ }^{2}=18^{2}+(6 . \sqrt{3})^{2}$