Q87 / 160 · TS EAMCET 2016 · Mathematics
The equation of the pair of straight lines through the point $(1,1)$ and perpendicular to the pair of straight lines $3 x^2-8 x y+5 y^2=0$ is
- A$5 x^2+8 x y+3 y^2-14 x-18 y+16=0$
- Correct answerB$5 x^2+8 x y+3 y^2-18 x-14 y+16=0$
- C$5 x^2-8 x y+3 y^2-18 x-14 y+32=0$
- D$5 x^2-8 x y+3 y^2-14 x-18 y+32=0$
Solution
We know that, the equations of the line passing through the origin and perpendicular to the lines given by $3 x^2-8 x y+5 y^2=0$ is $5 x^2+8 x y+3 y^2=0$
Now, shifting the origin at $(1,1)$ the equations of the required lines are
$$
\begin{aligned}
& 5(x-1)^2+8(x-1)(y-1)+3(y-1)^2=0 \\
\Rightarrow & 5\left(x^2+1-2 x\right)+8(x y-x-y+1)+3\left(y^2+1-2 y\right)=0 \\
\Rightarrow & 5 x^2+5-10 x+8 x y-8 x-8 y+8+3 y^2 \\
& +3-6 y=0 \\
\therefore \quad & 5 x^2+8 x y+3 y^2-18 x-14 y+16=0
\end{aligned}
$$
Hence, the correct answer is option: 'B'.