Math, asked by bommusashi, 10 days ago

# The locus of the mid point of the portion of the linex cos a + y sin a = p between the coordinate axes is (p is a nonzero constant)(A) x2 + y2 = 1(B) +4TELEy2(C) x2 + y2 = 4p?(D) 3 +4​

0

\bf{\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}}

x

2

1

+

y

2

1

=

p

2

4

Step-by-step explanation:

Let P(h,k) be the midpoint of the portion of the line

x\:cos\alpha+y\:sin\alpha=pxcosα+ysinα=p .....(1)

put y=0 in (1), we get

x\:cos\alpha=pxcosα=p

x=\frac{p}{cos\alpha}x=

cosα

p

put x=0 in(1), we get

y\:sin\alpha=pysinα=p

y=\frac{p}{sin\alpha}y=

sinα

p

Therefore, the line (1) meets the coordinate axes at

A(\frac{p}{cos\alpha},0

cosα

p

,0 ) and B(0,\frac{p}{sin\alpha}

sinα

p

)

Clearly, the midpoint of the Portion AB= P

(\frac{\frac{p}{cos\alpha}+0}{2},\frac{0+\frac{p}{sin\alpha}}{2})=(h,k)(

2

cosα

p

+0

,

2

0+

sinα

p

)=(h,k)

(\frac{p}{2cos\alpha},\frac{p}{2sin\alpha})=(h,k)(

2cosα

p

,

2sinα

p

)=(h,k)

\implies\:h=\frac{p}{2cos\alpha},\:\:\:\:k=\frac{p}{2sinalpha}⟹h=

2cosα

p

,k=

2sinalpha

p

\implies\:cos\alpha=\frac{p}{2h},\:\:\:\:sin\alpha=\frac{p}{2k}⟹cosα=

2h

p

,sinα=

2k

p

squaring and adding these equations, we get

cos^2\alpha+sin^2\alpha=\frac{p^2}{4h^2}+\frac{p^2}{4k^2}cos

2

α+sin

2

α=

4h

2

p

2

+

4k

2

p

2

\implies\:1=\frac{p^2}{4h^2}+\frac{p^2}{4k^2}⟹1=

4h

2

p

2

+

4k

2

p

2

\implies\:1=\frac{p^2}{4}(\frac{1}{h^2}+\frac{1}{k^2})⟹1=

4

p

2

(

h

2

1

+

k

2

1

)

\implies\:\frac{4}{p^2}=\frac{1}{h^2}+\frac{1}{k^2}⟹

p

2

4

=

h

2

1

+

k

2

1

\therefore\text{The locus of P is}∴The locus of P is

\boxed{\bf{\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}}}

x

2

1

+

y

2

1

=

p

2

4

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