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JEE MATHEMATICS MCQ: PAIR OF STRAIGHT LINES

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A point moves in such a way that the distance between the feet of the perpendiculars drawn from it to the lines ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0 is a constant 2λ2 \lambda. Then, the locus of the point is
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Correct option: B
(x2+y2)(h2−ab)=λ2[(a−b)2+4h2]\left(x^2+y^2\right)\left(h^2-a b\right)=\lambda^2\left[(a-b)^2+4 h^2\right]
Option B ((x2+y2)(h2−ab)=λ2[(a−b)2+4h2]\left(x^2+y^2\right)\left(h^2-a b\right)=\lambda^2\left[(a-b)^2+4 h^2\right]) is correct because it represents the locus of the point derived from the given condition about the distance between the feet of perpendiculars.

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If ax2+2hxy+by2+2gx+2fy+c=0a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 represents parallel straight lines, then Image of the point (−1,3)(-1,3) with respect to the line y=2xy=2 x is If the pair of lines represented by ax2+2hxy+by2+2gx+2fy+c=0a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 intersect on yy-axis, then If the sum of the slopes of the lines given by 4x2+2λxy−7y2=04 x^2+2 \lambda x y-7 y^2=0 is equal to the product of the slopes, then λ\lambda is equal to The equation 3x2+2hxy+3y2=03 x^2+2 h x y+3 y^2=0 represents a pair of straight lines passing through the origin. The two lines are The value of hh for which 3x2−3hxy+4y2=03 x^2-3 h x y+4 y^2=0 represents a pair of coincident lines are
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