The locus of the mid point of the portion of the line
x cos a + y sin a = p between the coordinate axes is (p is a non
zero constant)
(A) x2 + y2 = 1
(B) +
4
TELE
y2
(C) x2 + y2 = 4p?
(D) 3 +
4
Answers
Answer:
\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