Question

Let $\alpha ,\beta$ are real numbers for which the system of linear equations
$2\alpha x+y+z=0\alpha x+y+2z=1{\beta }^{2}x+y-3\alpha z=-2$
never posses unique solution then point $P(\alpha ,\beta )$ lies on a conic whose

• A. center is $(-\frac{1}{2},0)$
• B. eccentricity is less than 3
• C. one vertex is origin
• D. (–1, 0) lies on it

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A

center is $(-\frac{1}{2},0)$

B

eccentricity is less than 3

C

one vertex is origin

D

(–1, 0) lies on it

$\left|\begin{array}{ccc}2\alpha & 1& 1\\ \alpha & 1& 2\\ {\beta }^2& 1& -3\alpha \end{array}\right|=0\Rightarrow 2\alpha (-3\alpha -2)-\alpha (-3\alpha -1)+{\beta }^2=0\Rightarrow 3({\alpha }^2+\alpha )-{\beta }^2=0\Rightarrow (\alpha +\frac{1}{2}{)}^2-\frac{{\beta }^2}{1}=\frac{3}{4}$
So conic is $(x+\frac{1}{2}{)}^2-\frac{y^2}{1}=\frac{3}{4}$
Whose center is $(-\frac{1}{2},0)$
$e^2-1=\frac{1}{1}$
$\Rightarrow e=\sqrt{2}$