All the chords of the curve 3x² − y² − 2x + 4y = 0 which subtend a right angle at the origin are concurrent. Does this result also hold for the curve, 3x² + 3y² − 2x + 4y = 0 ? If yes, what is the point of concurrence.
Text Solution
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(1, − 2), yes (1/3, − 2/3)
Sol. Let equation of chords hx + ky = 1
By homogenisation
3x 2 – y 2 – 2x (hx + ky) + 4y (hx + ky) = 0
∴ it makes 90º. Hence
coeff. x 2 + coeff. y 2 = 0
3 – 2h – 1 + 4k = 0 ⇒ h – 2k = 1
Hence all chords are concurrent at (1, – 2)
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