. find P and & with Q represent a perpendecular line 12x²+7xy-py²-18x+qy+6 following equation
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Answer:
Correct option is
A
p=12,q=1
Given equation is
12x
2
+7xy−py
2
−18x+qy+6=0
On comparing with
ax
2
+by
2
+2hxy+2gx+2fy+c=0, we get
a=12,b=−p,h=
2
7
,g=−9,f=
2
q
,c=6
Conditions for pair of lines and pair of perpendiculars are
abc+2fgh−af
2
−bg
2
−ch
2
=0 and a+b=0
⇒12×(−p)(6)+2×(
2
q
)(−9)(
2
7
)−12(
2
q
)
2
−(−p)(−9)
2
−6(
2
7
)
2
=0
and 12−p=0
⇒−72p−
2
63
q−3q
2
+81p−
2
147
=0
and 12−p=0
⇒2q
2
+21q−23=0 and p=12
$\Rightarrow q = 1andp = 12$$.
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