The combined equation of the diagonals of the | vedclass

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(D) The given equation is $(7 x^2-4 x y+8 y^2)^2+(4 x-8 y-32)(7 x^2-4 x y+8 y^2)=0$.Factoring out $(7 x^2-4 x y+8 y^2)$,we get $(7 x^2-4 x y+8 y^2)(7

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) The given equation is $(7 x^2-4 x y+8 y^2)^2+(4 x-8 y-32)(7 x^2-4 x y+8 y^2)=0$.Factoring out $(7 x^2-4 x y+8 y^2)$,we get $(7 x^2-4 x y+8 y^2)(7 x^2-4 x y+8 y^2+4 x-8 y-32)=0$.This implies either $7 x^2-4 x y+8 y^2=0$ or $7 x^2-4 x y+8 y^2+4 x-8 y-32=0$.The first part $7 x^2-4 x y+8 y^2=0$ represents a pair of lines passing through the origin. Since the discriminant $B^2-4AC = (-4)^2 - 4(7)(8) = 16 - 224 = -208 The second part $7 x^2-4 x y+8 y^2+4 x-8 y-32=0$ also represents imaginary lines. Since the lines forming the parallelogram are not real,the question as stated does not form a real parallelogram. Therefore,none of the given options are correct.

The combined equation of the diagonals of the parallelogram formed by the lines $(7 x^2-4 x y+8 y^2)^2+(4 x-8 y-32)(7 x^2-4 x y+8 y^2)=0$ is

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