If P and Q are the length of perpendicular from the origin to the line − =
2 and + = respectively. Prove that
2 + 4
2 =
2
Answers
Answer:
The equation given lines are xcosθ−ysinθ=kcos2θ....(1)
xsecθ+ycscθ=k....(2)
The perpendicular distance (d) of a line Ax+By+C= 0 from a point(x
1
,y
1
)is given by d=
A
2
+B
2
∣Ax
1
+By
1
+C∣
On comparing equation(1) to the general equation of line i.e; Ax+By+C =0, we obtain A=cosθ,B=−sinθ, and C=−kcos2θ
It is given that p is the length of the perpendicular from(0,0) to line (1)
∴p=
A
2
+B
2
∣A(0)+B(0)+C∣
=
A
2
+B
2
∣C∣
=
cos
2
θ+sin
2
θ
∣−kcos2θ∣
=∣−kcos2θ∣......(3)
On comparing equation(2) to the general equation of line i.e; Ax+By+C=0, we obtain A=secθ,B=cscθ, and C=−k
It is given that q is the length of the perpendicular from (0,0) to line (2)
∴q=
A
2
+B
2
∣A(0)+B(0)+C∣
=
A
2
+B
2
∣C∣
=
sec
2
θ+csc
2
θ
∣−k∣
.....(4)
From (3) and (4) we have
p
2
+4q
2
=(∣−kcos2θ∣)
2
+4{
sec
2
θ+csc
2
θ
∣−k∣
}
=k
2
cos
2
2θ+
(sec
2
θ+csc
2
θ)
4k
2
=k
2
cos
2
2θ+
{
cos
2
θ
1
+
sin
2
θ
1
}
4k
2
=k
2
cos
2
2θ+
{
sin
2
θcos
2
θ
sin
2
+cos
2
θ
}
4k
2
=k
2
cos
2
2θ+4k
2
sin
2
θcos
2
θ
=k
2
cos
2
2θ+k
2
(2sinθcosθ)
2
=k
2
cos
2
2θ+k
2
sin
2
2θ
=k
2
(cos
2
2θ+sin
2
2θ)=k
2
Hence, we proved that p
2
+4q
2
=k
2
_______________________________
We know, perpendicular distance of point (x1, y1) from ax + by + c = 0 is
d = |ax1 + by1 + c|/√(a² + b²)
now, P is the perpendicular distance of origin from line xcos∅ - ysin∅ = k cos2∅
then, P = |0 - 0 - kcos2∅|/√(sin²∅ + cos²∅)
P = Kcos2∅/√1 [ because sin²∅ + cos²∅ = 1]
P = Kcos2∅ -------------(1)
similarly , q is the perpendicular distance of origin from line xsec∅ + ycosec∅ = k
q = |0 + 0 - k|/√(sec²∅ + cosec²∅)
= ksin²∅.cos²∅ -----------(2)
from equations (1)and (2),
LHS = P² + 4q²
= (Kcos2∅)² + 4(Ksin²∅.cos²∅)²
= K²{ cos²∅ - sin²∅}² +. 4sin²∅.cos²∅}
= K²{sin²∅ + cos²∅} [ use, (a + b)² = (a - b)²+4ab]
= K² = RHS