Math, asked by devchauhanhd, 4 months ago

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

Answered by mahipoorna143
1

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

=k

2

(cos

2

2θ+sin

2

2θ)=k

2

Hence, we proved that p

2

+4q

2

=k

2

Answered by SajanJeevika
0

_______________________________

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

*****************************************************

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