prove that the straight line to origin to the point of intersection of lin x÷a+y÷b=1 and the curve x^2+y^2=c^2 are at right angle if 1÷a^2+1÷b^2=2÷c^2
Answers
Answer:
Step-by-step explanation:
(x−h)
2
+(y−k)
2
=c
2
kx+hy=2hk
The equation of pair straight line joining the origin is find by method of homogenisation. We make the coefficient of constant term of straight line 1 and put it the equation of the curve so that degree of each term of the curve become 2
1=
2hk
kx+hy
=
2h
x
+
2k
y
x
2
−2hx+h
2
+y
2
−2ky+k
2
−c
2
=0
x
2
+y
2
−2hx(1)−2ky(1)+(h
2
+k
2
−c
2
)(1)
2
=0
x
2
+y
2
−2hx(
2h
x
+
2k
y
)−2ky(
2h
x
+
2k
y
)+(h
2
+k
2
−c
2
)(
2h
x
+
2k
y
)
2
=0
(1−1+
4h
2
h
2
+k
2
−c
2
)x
2
+(1−1+
4k
2
h
2
+k
2
−c
2
)y
2
−(
k
h
+
h
k
−
2kh
h
2
+k
2
−c
2
)xy=0
(
4h
2
h
2
+k
2
−c
2
)x
2
+(
4k
2
h
2
+k
2
−c
2
)−(
k
h
+
h
k
−
2kh
h
2
+k
2
−c
2
)xy=0
Lines are prerpendicular if a+b=0
⇒
4h
2
h
2
+k
2
−c
2
+
4k
2
h
2
+k
2
−c
2
=0
(h
2
+k
2
−c
2
)(k
2
+h
2
)=0
⇒h
2
+k
2
−c
2
=0
h
2
+k
2
=c
2
Hence proved