All chords of a curve $3 x^{2} - y^{2} - 2 x + 4 y = 0$ which subtend a right angle at origin, pass through a fixed point which is

(1, 2)

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(2, 1)

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(1, - 2)

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(- 2, 1)

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Solution

The correct option is D $(1, - 2)$
Let, $y = m x + c$ is a equation of chord.
$∵ \frac{y - m x}{c} = 1$
By homogenization,
$3 x^{2} - y^{2} - 2 x (\frac{y - m x}{c}) + 4 y (\frac{y - m x}{c}) = 0$
$3 c x^{2} - c y^{2} - 2 x y + 2 m x^{2} + 4 y^{2} - 4 m x y = 0$
$(3 c + 2 m) x^{2} - (c - 4) y^{2} - (2 + 4 m) x y = 0$
So, angle between these lines is given by
$t a n θ = ∣ ∣ ∣ \frac{2 \sqrt{(} 1 + 2 m)^{2} - (3 c + 2 m) (4 - c)}{(2 c + 4 + 2 m)} ∣ ∣ ∣$
Given. $θ = 90^{\circ},$
$∴ 2 c + 2 m + 4 = 0$
$- 2 = m + c$
$y = m x + c$
$y = - 2$ & $x = 1$

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