Find the envelope of the straight line y-x=2
Answers
Answer:
Solution.
The equation for the given family of curves can be written as
x
2
a
2
+
y
2
1
−
a
2
=
1
,
where the semi-axis \(a\) is a parameter and \(0 \lt a \lt 1.\) Differentiating this equation with respect to the parameter \(a,\) we obtain the second equation:
a
is a parameter and
0
<
a
<
1.
Differentiating this equation with respect to the parameter
a
,
we obtain the second equation:
−
2
x
2
a
3
−
y
2
⋅
(
−
2
a
)
(
1
−
a
2
)
2
=
0
,
⇒
−
2
x
2
a
3
+
2
y
2
a
(
1
−
a
2
)
2
=
0
,
⇒
y
2
a
(
1
−
a
2
)
2
=
x
2
a
3
,
⇒
y
2
a
4
=
x
2
(
1
−
a
2
)
2
.
Take the square root of both sides of then equation:
y
2
a
4
=
x
2
(
1
−
a
2
)
2
,
⇒
|
y
|
a
2
=
|
x
|
(
1
−
a
2
)
.
Express \({{a^2}}\) from the last equation:
a
2
from the last equation:
|
y
|
a
2
=
|
x
|
−
|
x
|
a
2
,
⇒
(
|
x
|
+
|
y
|
)
a
2
=
|
x
|
,
⇒
a
2
=
|
x
|
|
x
|
+
|
y
|
.
Substituting this into the first equation and assuming that the family of curves has no singular points, we find the envelope:
x
2
|
x
|
|
x
|
+
|
y
|
+
y
2
1
−
|
x
|
|
x
|
+
|
y
|
=
1
,
⇒
x
2
(
|
x
|
+
|
y
|
)
|
x
|
+
y
2
(
|
x
|
+
|
y
|
)
|
x
|
+
|
y
|
−
|
x
|
=
1
,
⇒
|
x
|
2
+
|
x
|
|
y
|
+
|
x
|
|
y
|
+
|
y
|
2
=
1
,
⇒
(
|
x
|
+
|
y
|
)
2
=
1
,
⇒
|
x
|
+
|
y
|
=
±
1.
Obviously, the negative root here does not make sense. Therefore, the final equation of the envelope is given by
|
x
|
+
|
y
|
=
1.
This equation describes a square (Figure
4
).
Envelope of a family of ellipses