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  4. Let curve f(x, y) is locus of point (x, y) which satisfy |2 x 2 y 6 α 2 β 3 λ -1 2 1|+2| x y 1 4 λ 2 α 3 β 2 -1 2|+|11 -2 λ 72 2α|-81 β =0 ∀ α, β, γ ∈ R. If minimum distance between curve f(x, y) and line 4 x-3 y+12=0 is d, then the value of d is

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Q. Let curve $f(x, y)$ is locus of point $(x, y)$ which satisfy
$\begin{vmatrix}2 x & 2 y & 6 \\ \alpha & 2 \beta & 3 \lambda \\ -1 & 2 & 1\end{vmatrix}+2\begin{vmatrix} x & y & 1 \\ 4 \lambda & 2 \alpha & 3 \beta \\ 2 & -1 & 2\end{vmatrix}+\begin{vmatrix}11 & -2 \lambda \\ 72 & 2\alpha\end{vmatrix}-81 \beta$
$=0 \forall \alpha, \beta, \gamma \in R$. If minimum distance between curve $f(x, y)$ and line $4 x-3 y+12=0$ is $d$, then the value of $d$ is

Determinants

Solution:

$2 x(2 \beta-6 \lambda)-2 y(\alpha+3 \lambda)+6(2 \alpha+2 \beta)+2 x(4 \alpha+3 \beta)$
$-2 y(8 \lambda-6 \beta)+2(-4 \lambda-4 \alpha)+22 \alpha+144 \lambda-81 \beta=0$
$\therefore \alpha(8 x-2 y+26)+\beta(10 x+12 y-69)$
$+\lambda(-12 x-22 y+136)=0$
$\left.\begin{array}{l}x-y+13=0 \\ \therefore \quad 10 x+12 y-69=0 \\ \text { and } 6 x+11 y-68=0\end{array}\right\} \quad x=-\frac{3}{2}, y=7$
$\therefore f(x, y)$ is point $\left(-\frac{3}{2}, 7\right)$
$d=\left|\frac{4\left(\frac{-3}{2}\right)-3(7)+12}{5}\right|=3$