If all chords of the curve $2 x^2-y^2+3 x+2 y=0$, which subtend a right angle at the origin always passing through the point $(\alpha, \beta)$, then $(\alpha, \beta)=$

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Aicharya · Accepted Answer

$2 x^2-y^2+3 x+2 y=0$...(i)
Let $y=m x+c$ be the chord $\Rightarrow \frac{y-m x}{c}=1$
Substitute in (i), $\begin{aligned} & 2 x^2-y^2+3 x(1)+2 y(1)=0 \\ & \Rightarrow 2 x^2-y^2+3 x\left(\frac{y-m x}{c}\right)+2 y\left(\frac{y-m x}{c}\right)=0 \\ & \Rightarrow(3 c-3 m) x^2+(2-c) y^2+(3-2 m) x y=0 \end{aligned}$
Chord is perpendicular to the curve So, the slope $\frac{3 c-3 m}{2-c}=-1 \Rightarrow 2 c-3 m=-2$ Comparing with equation of chord $y=m x+c$ $(x, y)=(-3,-2)$

If all chords of the curve $2 x^2-y^2+3 x+2 y=0$, which subtend a right angle at the origin always passing…

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