Parabola
Equilateral triangle inscribed in parabola
nta_pyq_2023_jan
Grade 11
Question:
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
Step-by-Step Solution
Key Concept: Use Bayes' theorem to find $\lambda$, then use the standard result for equilateral triangle inscribed in $y^2 = 4ax$
By Bayes: $P(C|R) = 0.4 \Rightarrow \lambda = 6$. Parabola: $y^2 = 6x$, so $a = 3/2$. For equilateral triangle with vertex at origin: vertices at $(3t^2/2, \pm 3t)$ where $\tan 30° = 3t/(3t^2/2) \Rightarrow 1/\sqrt{3} = 2/t \Rightarrow t = 2\sqrt{3}$. Side vertex: $(18, 6\sqrt{3})$. Side length$^2 = 18^2 + (6\sqrt{3})^2 = 324 + 108 = 432$. Answer: 432
Correct Answer: 432