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Question

ax2+2hxy+by2=0 represents a pair of straight lines through origin & angle between them is given by
tanθ=2h2aba+b. If the lines are perpendicular then a+b=0 and the equation of bisectors is given by x2y2ab=xyh
The general equation of second degree given by
ax2+2hxy+by2+2gx+2fy+c=0 represent a pair of straight lines if =0 or
∣ ∣ahghbfgfc∣ ∣=0 or abc+2fghaf2bg2ch2=0
On the basis of above information answer the following question
If the angle between the lines represented by 6x2+5xy4y2+7x+13y3=0 is tan1(m) & a2+b2abab+10, then 5a+6b is equal to

A
1m
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B
m
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C
m2
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D
2m
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Solution

The correct option is D 2m
Compare given equation with ax2+2hxy+by2+2gx+2fy+c=0 we have,
a=6b=4h=52

θ=tan1(m) (given)

=tan1{2h2aba+b}=tan1⎪ ⎪⎪ ⎪2254+2464⎪ ⎪⎪ ⎪

=tan1{112} m=112

(2m=11) .............(*)

Again a2+b2abab+10

12(2a2+2b22ab2a2b+2)0

12{(ab)2+(a1)2+(b1)2}0

Which is possible only if ab=0,a1=0,b1=0 i.e. a=b,a=1,b=1

5a+6b=11=2m (Using *)

Hence choice (d) is the correct answer.

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