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show that all the chords of the curve 3x²-y²-2x+4y=0 which subtend a right angle at the origin pass through a fixed point
Let (x,y) be a common point of chord y=mx+c, y=kx & 3x²−y²−2x+4y=0First two give x & y which can be used in third to give (3−k²)c + (4k−2)(k−m) = 0As a quadratic in k this is k²(4−c)−2k(2m+1)+(2m+3c) = 0If the two values of k arising from this give perp lines thro O then (2m+3c)/(4−c) = −1→ 2m+3c = −4+c → c=−m−2Chord is thus y = mx–m−2 → y+2 = m(x−1) which always contains (1,−2)
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