Transposing one term in each to the second member and dividing, we get
x
2
a
2
=
y
2
b
2
{\displaystyle {\frac {x^{2}}{a^{2}}}={\frac {y^{2}}{b^{2}}}}
Therefore, from (A ) ,
x
2
a
2
=
1
2
{\displaystyle {\frac {x^{2}}{a^{2}}}={\frac {1}{2}}}
y
2
b
2
=
1
2
{\displaystyle {\frac {y^{2}}{b^{2}}}={\frac {1}{2}}}
giving
a
=
±
x
2
{\displaystyle a=\pm x{\sqrt {2}}}
b
=
±
y
2
{\displaystyle b=\pm y{\sqrt {2}}}
Substituting these values in (B ) , we get the envelope
x
y
=
±
k
2
π
{\displaystyle xy=\pm {\frac {k}{2\pi }}}
a pair of conjugate rectangular hyperbolas (see last figure).
1. Find the envelope of the family of straight lines
y
=
2
m
x
+
m
4
{\displaystyle y=2mx+m^{4}}
m
{\displaystyle m}
Ans.
x
=
−
2
m
3
{\displaystyle x=-2m^{3}}
,
y
=
−
3
m
4
{\displaystyle y=-3m^{4}}
; or
16
y
3
+
27
x
4
=
0
{\displaystyle 16y^{3}+27x^{4}=0}
.
[1]
2. Find the envelope of the family of parabolas
y
2
=
a
(
x
−
a
)
{\displaystyle y^{2}=a(x-a)}
a
{\displaystyle a}
Ans.
x
=
2
a
{\displaystyle x=2a}
,
y
=
±
a
{\displaystyle y=\pm a}
; or
y
=
±
1
2
x
{\displaystyle y=\pm {1 \over 2}x}
.
3. Find the envelope of the family of circles
x
2
+
(
y
−
β
)
2
=
r
2
{\displaystyle x^{2}+(y-\beta )^{2}=r^{2}}
β
{\displaystyle \beta }
Ans.
x
=
±
r
{\displaystyle x=\pm r}
4. Find the equation of the curve having as tangents the family of straight lines
y
=
m
x
±
a
2
m
2
+
b
2
{\displaystyle y=mx\pm {\sqrt {a^{2}m^{2}+b^{2}}}}
m
{\displaystyle m}
Ans. The ellipse
b
2
x
2
+
a
2
y
2
=
a
2
b
2
{\displaystyle b^{2}x^{2}+a^{2}y^{2}=a^{2}b^{2}}
.
5. Find the envelope of the family of circles whose diameters are double ordinates of the parabola
y
2
=
4
p
x
{\displaystyle y^{2}=4px}
Ans. The parabola
y
2
=
4
p
(
p
+
x
)
{\displaystyle y^{2}=4p(p+x)}
.
6. Find the envelope of the family of circles whose diameters are double ordinates of the ellipse
b
2
x
2
+
a
2
y
2
=
a
2
b
2
{\displaystyle b^{2}x^{2}+a^{2}y^{2}=a^{2}b^{2}}
Ans. The ellipse
x
2
a
2
+
b
2
+
y
2
b
2
=
1
{\displaystyle {\frac {x^{2}}{a^{2}+b^{2}}}+{\frac {y^{2}}{b^{2}}}=1}
.
7. A circle moves with its center on the parabola
y
2
=
4
a
x
{\displaystyle y^{2}=4ax}
Ans. The [
cissoid ]
y
2
(
x
+
2
a
)
+
x
3
=
0
{\displaystyle y^{2}(x+2a)+x^{3}=0}
.
8. Find the curve whose tangents are
y
=
l
x
a
l
2
+
b
l
+
c
{\displaystyle y=lx{\sqrt {al^{2}+bl+c}}}
l
{\displaystyle l}
Ans.
4
(
a
y
2
+
b
x
y
+
c
x
2
)
=
4
a
c
−
b
2
{\displaystyle 4(ay^{2}+bxy+cx^{2})=4ac-b^{2}}
.
9 Find the evolute of the ellipse
b
2
x
2
+
a
2
y
2
=
a
2
b
2
{\displaystyle b^{2}x^{2}+a^{2}y^{2}=a^{2}b^{2}}
b
y
=
a
x
tan
ϕ
−
(
a
2
−
b
2
)
sin
ϕ
{\displaystyle by=ax\tan \phi -(a^{2}-b^{2})\sin \phi }
ϕ
{\displaystyle \phi }
Ans.
x
=
a
2
−
b
2
a
cos
3
ϕ
{\displaystyle x={\frac {a^{2}-b^{2}}{a}}\cos ^{3}\phi }
,
y
=
b
2
−
a
2
a
sin
3
ϕ
{\displaystyle y={\frac {b^{2}-a^{2}}{a}}\sin ^{3}\phi }
; or
(
a
x
)
2
3
+
(
b
y
)
2
3
=
(
a
2
−
b
2
)
2
3
{\displaystyle (ax)^{2 \over 3}+(by)^{2 \over 3}=(a^{2}-b^{2})^{2 \over 3}}
.
10. Find the evolute of the hypocycloid
x
2
3
+
y
2
3
=
a
2
3
{\displaystyle x^{2 \over 3}+y^{2 \over 3}=a^{2 \over 3}}
y
cos
τ
−
x
sin
τ
=
a
cos
2
τ
{\displaystyle y\,\cos \tau -x\,\sin \tau =a\,\cos 2\tau }
τ
{\displaystyle \tau }
Ans.
(
x
+
y
)
2
3
+
(
x
−
y
)
2
3
=
2
a
2
3
{\displaystyle (x+y)^{2 \over 3}+(x-y)^{2 \over 3}=2a^{2 \over 3}}
.
↑ When two answers are given, the first is in parametric form and the second in rectangular form.
