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  • In a square matrix A of order <img alt="\mathrm{3, \mathrm{a}_{\mathrm{ii}}=\mathrm{m}_{\mathrm{i}}+\mathrm{i}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B3%2C%20%5Cmathrm%7B

In a square matrix A of order \mathrm{3, \mathrm{a}_{\mathrm{ii}}=\mathrm{m}_{\mathrm{i}}+\mathrm{i}} where \mathrm{\mathrm{i}=1,2,3} and \mathrm{\mathrm{m}_{\mathrm{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 \mathrm{y^2=4 x. } Rest all other entries of the matrix are one. The value of det. (A) is equal to

Option: 1

37


Option: 2

-6


Option: 3

-4


Option: 4

-9


Answers (1)

best_answer

equation of normal to \mathrm{y^2=4 x \quad(a=1)}
\mathrm{ \mathrm{y}=\mathrm{mx}-2 \mathrm{~m}-\mathrm{m}^3 }
passes through \mathrm{(9,-6)}

\mathrm{\begin{aligned} & -6=9 m-2 m-m^3 \\ & \mathrm{~m}^3-7 \mathrm{~m}-6=0 \\ & (\mathrm{~m}+1)(\mathrm{m}+2)(\mathrm{m}-3)=0 \\ & \mathrm{~m}=-1 \text { or }-2,3 \\ & \therefore \quad \mathrm{m}_1=-1 ; \mathrm{m}_2=-2 ; \mathrm{m}_3=3 \\ & \therefore \quad \mathrm{a}_{11}=1+\mathrm{m}_1=0 \\ & \mathrm{a}_{22}=2+\mathrm{m}_2=0 \\ & \mathrm{a}_{33}=3+\mathrm{m}_3=6 \\ & \therefore \quad \operatorname{det}(\mathrm{A})=\left|\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 6 \end{array}\right|=-4 \text { Ans } \\ & \end{aligned}}

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Irshad Anwar

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