In a square matrix $A$ of order $3,a_{ii} = m_{i} + i$ where $i = 1,2,3$ and $m_{i}$ 's are the slop

Question

In a square matrix $A$ of order $3,a_{ii} = m_{i} + i$ where $i = 1,2,3$ and $m_{i}$ 's are the slopes (in increasing order of their absolute value) of the 3 normals concurrent at the point $(9, - 6)$ to the parabola $y^{2} = 4x$. Rest all other entries of the matrix are one. The value of det. A is equal to :

Accepted Answer

Sol. [Key Concept: The three concurrent normal' slopes satisfy ${am}^{3} + (2a - h)m + k = 0$; Vieta gives $\sum\ m = 0$ and the symmetric sums — build the matrix and expand the determinant.]Equation of normal at $\left( t^{2},2t ight)$ is$$y = - tx + 2t + t^{3}$$$$\begin{matrix} \Rightarrow t^{3} - 7t + 6 = 0 \Rightarrow (t - 1)(t - 2)(t + 3) = 0 \ \Rightarrow t = 1,2 - 3 \ a_{11} = m_{1} + 1 = - 1 + 1 = 0 \ a_{22} = m_{2} + 2 = - 2 + 2 = 0 \ a_{33} = m_{3} + 3 = 3 + 3 = 6 \ |A| = \left| \begin{matrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 6 \end{matrix} ight| = - 4 \end{matrix}$$

Question

Options

Solution

More Parabola Questions