L, 1) N-1, 2) 4) (2, -1)
14. The first degree terms of
ax² + 2hxy +by? +2gx + 2fy +c= 0 are
removed by shifting origin to (a,b). The
new equation is
1) ax² + 2hxy + by2 + 2 ya +2bB+c = 0
2) ax² + 2hxy + by2 + ga + f B+c=0
3) ax² + 2hxy + by2 + ha + bB+c=0
4) ax +2hxy – by – ha – bt – c = 0
Answers
Step-by-step explanation:
1)Given: ax
2
+2hxy+by
2
+2gx+2fy+c=0
2ax+2hy+2hx
dx
dy
+2by
dx
dy
+2g+2f
dx
dy
=0
2hx+2by+2f
−(2ax+2hy+2g)
=
dx
dy
2)ax
2
+2hxy+by
2
+2gx+2fy+c=0
let the twe lines be.
L
1
;y=m
1
x+c
1
L
2
:y=m
2
x+c
2
(m
1
x+c
1
−y)(m
2
x+c
2
−y)=0− (ii)
on comparing (i) and(ii) , we got
on comparing
(i) and (ii), we got
m
1
c
2
+m
2
c
1
=
b
2g
& c
1
+c
2
=−
b
2f
m
2
m
2
=
b
a
& c
1
c
2
=
b
c
Now,
1+m
2
2
∣c
1
∣
=
1+m
2
2
∣c
21
Squaring both sides and cross-multiply
⇒c
1
2
+c
1
2
m
2
2
=c
2
2
+c
2
2
m
1
2
⇒(e
1
2
−c
2
2
)=(c
2
2
m
2
2
−c
1
2
m
2
2
)
⇒(c
1
+c
2
)(c
1
−c
2
)=(c
2
m
1
+c
1
m
2
)(c
2
m
1
−c
1
m
2
)
⇒(c
1
+c
2
)
(c
1
+c
2
)
2
−4c
1
c
2
=(c
2
m
1
+c
1
m
2
)
(E
2
m
1
+C
1
m
2
)
2
−4m
1
m
2
C
1
C
2
⇒−
b
2f
b
2
4f
2
−
b
4c
=
b
2q
b
2
4g
2
−
b
4a
⇒g
2
(
b
2
g
2
−ac
)=f
2
(
b
2
f
2
−bc
)
⇒f
4
−g
4
=c(bf
2
−ag
2
3) If two lines ax
2
+2hxy+by
2
=0 make angles α and β with X−axis, then the (α+β)=
a−b
2h
Explanation:
m
1
=tanα, m
2
=tanβ
∴tan(α+β)=
1−tanαtanβ
tanα+tanβ
=
1−m
1
m
2
m
1
+m
2
=
1−(a/b)
−2h/b
=
a−b
2h