Question

If variable parameter $\alpha ,\beta ,\gamma \in R$ and satisfy the relations ${\alpha }^2+2{\beta }^2+3\alpha =1+{\gamma }^2$ and $2{\alpha }^2+4{\beta }^2=2{\gamma }^2+5\beta ,$ then

• A. Minimum value of ${\alpha }^2+{\beta }^2$ is $\frac{4}{61}$
• B. Minimum value of ${\alpha }^2+{\beta }^2$ is $\frac{2}{17}$
• C. Minimum value of ${\alpha }^2+{\beta }^2-2\alpha +1$ is $\frac{16}{61}$
• D. Minimum value of ${\alpha }^2+{\beta }^2-2\alpha +1$ is $\frac{1}{17}$

Answer 1

Answer: (A), (C)

The correct options are
A Minimum value of ${\alpha }^2+{\beta }^2$ is $\frac{4}{61}$
C Minimum value of ${\alpha }^2+{\beta }^2-2\alpha +1$ is $\frac{16}{61}$

${\alpha }^2+2{\beta }^2+3\alpha =1+{\gamma }^2\cdots \left(1\right)$
$2{\alpha }^2+4{\beta }^2=2{\gamma }^2+5\beta \cdots \left(2\right)$
From equation $\left(1\right)\times 2$ $-$ equation $\left(2\right),$ we get
$6\alpha +5\beta =2$

Perpendicular distance from centre $(0,0)$ to line $6x+5y=2$ is $\frac{|-2|}{\sqrt{36+25}}$
Hence, the minimum value of ${\alpha }^2+{\beta }^2$ is ${\left(\frac{|-2|}{\sqrt{36+25}}\right)}^2=\frac{4}{61}$

Perpendicular distance from centre $(1,0)$ to line $6x+5y=2$ is $\frac{|6-2|}{\sqrt{36+25}}$
Hence, the minimum value of ${\alpha }^2+{\beta }^2-2\alpha +1$ is ${\left(\frac{|6-2|}{\sqrt{36+25}}\right)}^2=\frac{16}{61}$