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Question

Two of the lines represented by x3−6x2y+3xy2+dy3=0 are perpendicular for

A
all real values of d
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B
two real values of d
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C
three real values of d
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D
no real value of d
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Solution

The correct option is D two real values of d
Let m1,m2,m3 be the slopes of the three lines represnted by the given equation such that m1m2=1

We have m1m2m3=1d so that

m3=1d

Since y=m3xx=dy satisfies the given equation, we get

d36d2+3d+d=0d(d26d+4)=0

If d=0, the given equation represents the line x=0 and x26xy+3y2=0 which are not perpendicular

d0 and d26d+4=0d=6±36162=3±5

which gives two real values of d

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