find the center of curvature of hyperbola x^2/a^2 - y^2/b^2=1
Answers
Given hyperbola:x2−y2=1--------------------------(3)
Given hyperbola:x2−y2=1--------------------------(3)Here a=1,b=1
Given hyperbola:x2−y2=1--------------------------(3)Here a=1,b=1⊥ar drawn from C meets the tangent at M and curve at N.
Given hyperbola:x2−y2=1--------------------------(3)Here a=1,b=1⊥ar drawn from C meets the tangent at M and curve at N.Let the point P on the hyperbola be
Given hyperbola:x2−y2=1--------------------------(3)Here a=1,b=1⊥ar drawn from C meets the tangent at M and curve at N.Let the point P on the hyperbola beP(secθ,tanθ)
Given hyperbola:x2−y2=1--------------------------(3)Here a=1,b=1⊥ar drawn from C meets the tangent at M and curve at N.Let the point P on the hyperbola beP(secθ,tanθ)equation of tangent from P is
Given hyperbola:x2−y2=1--------------------------(3)Here a=1,b=1⊥ar drawn from C meets the tangent at M and curve at N.Let the point P on the hyperbola beP(secθ,tanθ)equation of tangent from P is T:1xsecθ−1ytanθ=1-------------------------(1)
Given hyperbola:x2−y2=1--------------------------(3)Here a=1,b=1⊥ar drawn from C meets the tangent at M and curve at N.Let the point P on the hyperbola beP(secθ,tanθ)equation of tangent from P is T:1xsecθ−1ytanθ=1-------------------------(1)Now, CM=⊥ar distance from (0,0) to T
Given hyperbola:x2−y2=1--------------------------(3)Here a=1,b=1⊥ar drawn from C meets the tangent at M and curve at N.Let the point P on the hyperbola beP(secθ,tanθ)equation of tangent from P is T:1xsecθ−1ytanθ=1-------------------------(1)Now, CM=⊥ar distance from (0,0) to TCMˉ=sec2θ+tan2θ∣0−0−1∣=sec2θ+tan2θ1
Given hyperbola:x2−y2=1--------------------------(3)Here a=1,b=1⊥ar drawn from C meets the tangent at M and curve at N.Let the point P on the hyperbola beP(secθ,tanθ)equation of tangent from P is T:1xsecθ−1ytanθ=1-------------------------(1)Now, CM=⊥ar distance from (0,0) to TCMˉ=sec2θ+tan2θ∣0−0−1∣=sec2θ+tan2θ1Foot of ⊥ar from (0,0)at T: 1h−0=1k−0=(sec2θ+tan2θ)2−1
Given hyperbola:x2−y2=1--------------------------(3)Here a=1,b=1⊥ar drawn from C meets the tangent at M and curve at N.Let the point P on the hyperbola beP(secθ,tanθ)equation of tangent from P is T:1xsecθ−1ytanθ=1-------------------------(1)Now, CM=⊥ar distance from (0,0) to TCMˉ=sec2θ+tan2θ∣0−0−1∣=sec2θ+tan2θ1Foot of ⊥ar from (0,0)at T: 1h−0=1k−0=(sec2θ+tan2θ)2−1M:((sec2θ+tan2
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Step-by-step explanation:
Given hyperbola:x
2
−y
2
=1--------------------------(3)
Here a=1,b=1
⊥
ar
drawn from C meets the tangent at M and curve at N.
Let the point P on the hyperbola be
P(secθ,tanθ)
equation of tangent from P is
T:
1
xsecθ
−
1
ytanθ
=1-------------------------(1)
Now, CM=⊥
ar
distance from (0,0) to T
CM
ˉ
=
sec
2
θ+tan
2
θ
∣0−0−1∣
=
sec
2
θ+tan
2
θ
1
Foot of ⊥
ar
from (0,0)at T:
1
h−0
=
1
k−0
=
(
sec
2
θ+tan
2
θ
)
2
−1
M:(
(
sec
2
θ+tan
2
θ
)
2
−1
,
(
sec
2
θ+tan
2
θ
)
2
−1
):
sec
2
θ+tan
2
θ
−1
,
sec
2
θ+tan
2
θ
−1
Line passing through(0,0) and M ⇒(y−0)=
secθ
−tanθ
(x−0)[Slope=
SlopeofT
−1
] (as this line is perpendicular to T)
⇒secθy=−tanθx
----------------------------(2)
Coordinates of N (intersecting equation (1) and (3)⇒x
2
−
sec
2
θ
tan
2
θ
x
2
=1
⇒x
2
(
sec
2
θ
1
)=1
x=±secθ&y=±tanθ).
CN=
sec
2
θ+tan
2
θ
---------------(5)
Now CM.CN=
sec
2
θ+tan
2
θ
1
.
sec
2
θ+tan
2
θ
(from equation (4) and (5)
CM.CN=1.