Q82 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics
Let $A(2,1)$ be a point and equation of the straight line $L$ be $x-y=0$. Let $a$ and $b$ respectively represent the distances from a variable point $P(\alpha, \beta)$ to $A$ and to the line $L$. If $C$ is distance of the point $A$ from origin such that $a=b c$, then locus of $P$ is
- A$3 x^2+3 y^2+10 x y+8 x+4 y+10=0$
- Correct answerB$3 x^2+3 y^2-10 x y+8 x+4 y-10=0$
- C$3 x^2+2 y^2-10 x y+8 x+4 y+10=0$
- D$2 x^2+3 y^2-10 x y-8 x-4 y-10=0$
Solution
Given equation of line $x-y=0$
Distance from $P(\alpha, \beta)$ to line $x-y=0$ is
$b=\left|\frac{\alpha-\beta}{\sqrt{2}}\right| \Rightarrow(\alpha-\beta)^2=2 b^2$ $\ldots(i)$
Distance between $P(\alpha, \beta)$ and $A(2,1)$ is
$a^2=(\alpha-2)^2+(\beta-1)^2$ $\ldots(ii)$
Distance from origin to $A(2,1)$ is $C^2=5$
Also $\quad a=b c$
$\therefore \quad a^2=5 b^2$ $\ldots(iii)$
From Eqs. (i), (ii) and (iii), we get
$(\alpha-2)^2+(\beta-1)^2=\frac{5(\alpha-\beta)^2}{2}$
$2\left(\alpha^2-4 \alpha+4+\beta^2-2 \beta+1\right]=5\left[\alpha^2+\beta^2-2 \alpha \beta\right]$
$\Rightarrow 3\left(\alpha^2+\beta^2\right)-10 \alpha \beta+8 \alpha+4 \beta-10=0$
$\therefore$ Locus of $P$ is
$3\left(x^2+y^2\right)-10 x y+8 x+4 y-10=0$
Hence, the correct answer is option: 'B'.