The line x+2y=k meets the curve 2x^2-2 xy+3y^2+2x-y-1=0 at two points A and B. Let O be the origin. If the line segments OA and OB are perpendicular to each other, then k=

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Correct answer: A. (a) Given line x+2y=k meets the curve 2x^2-2 xy+ 3y^2+2x-y-1=0 Take, x+2yk=1.......(i) From curve 2x^2-2 xy+3y^2+2x-y-1=0. (2x^2-2 xy+3y^2)+ (2x-y ) (1 )- (1 )^2=0(2x^2-2 xy+3y^2)+ (2x-y ) (x+2y )k- (x+2y )^2k^2=0k^2(2x^2-2 xy+3y^2)+(2x^2+4 yx- xy-2y^2)k-(x^2+)(4y^2+4 xy)=02k^2x^2-2k^2 xy+3k^2y^2+2kx^2+3 xyk-2ky^2-x^2-4y^2-4 xy=0x^2(2k^2+2k-1)- xy(2k^2-3k+4)+y^2(3k^2-2k-4)=0 Here lines from intersected points A and B are perpendicular. So, coefficient of (x^2 )+ coefficient of (y^2 )=0 2k^2+2k-1+3k^2-2k-4=0 ⇒k= 1

$The line x + 2 y = k meets the curve 2 x^{2} - 2 xy + 3 y^{2} + 2 x - y - 1 = 0 at two points A and B . Let O be the origin. If the line segments OA and OB are perpendicular to each other, then k =$

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