Conic sections:
Conic sections are formed when a right cone is intercepted by a plane.The shape so formed depends on the angle at which the conic is cut.
Consider the below figure of right cone intercepted by plane and we can see the different shapes formed.
Conic sections:
Conic sections are formed when a right cone is intercepted by a plane.The shape so formed depends on the angle at which the conic is cut.
Consider the below figure of right cone intercepted by plane and we can see the different shapes formed.
The first picture formed is known as parabola
Second one is ellipse
The third is hyperbola.
Eccentricity of 1 :-
Eccentricity is the measure of the deviation of a conic from a circular path .The parabola is the conic haing eccentricity as 1.
Definition of parabola :-
Parabola is locus of points whose distance from fixed line, called directrix, and a from a fixed point ,called focus,is equal. Vertex is a point where parabola changes its direction .The distance from focus to vertex and vertex to directrix is equal.
The main compenents of Parabola are :-
i)Vertex
ii) axis
iii) focus
iv) directrix
Consider the parabola given below
Description of paraboala :-
i)Point F is called the focus of parabola
ii)Point "O" is called as the vertex of parabola.
iii)C ,D are the points on the parabola whoce distance fron focus and directrix are equal .
iv) Line AB is called directric fo parabola
v) Line ox is called the axis of parabola.
i) Y2 = 4aX
This parabola opens towards right side ie + ve x axis having vertex at (0,0). and focal distance "a" and focal point (a,0) . This curve transforms to (Y-k)2 = 4a(X-h) when vertex shifted to a point (h,k). with focus shifting to (a+h ,k)
ii) Y2= -4aX
This parabola opens towards left side ie -ve x axis having vertex at (0,0). and focal distance "a" and focal point (-a,0).This curve transforms to (Y-k)2 = - 4a(X-h) when vertex shifted to a point (h,k) and focus shifts to (h-a ,k)
iii) X2= 4aY
This parabola opens towards top side ie +ve y axis having vertex at (0,0). and focal distance "a"and focal point (0,a).This curve transforms to (X-h)2 = 4a(Y-k) when vertex shifted to a point (h,k) and focus shifts to (h,k+a).
iv) X2=-4aY
This parabola opens towards down side ie -ve y axis having vertex at (0,0). and focal distance "a" and focal point (0,-a).This curve transforms to (X-h)2 = - 4a(Y-k) when vertex shifted to a point (h,k-a).
Ex 1:
Given the equation of parabola Y2 = 8X find the focus of parabola
solution:
Comparing with standard form of equation
Y2 = 4aX
we get 4a =8
=> a=2
so focus of parabola is (a,0) = (2,0)
Ex 2:
find focus of parabola y = 1/4 x2
solution :
y = 1/4 x2
=> x2= 4y
=> a= 1 on comparing with standard form
so focus of parabola is (0,1) as axis of symmetry is +ve y axis
My forthcoming post is on The Domain of the Function and samacheer kalvi textbooks online will give you more understanding about Algebra
Ex 3:
Given a parabolic equation Y2= -16X-32
Find i) Vertex of parabola
ii) Focus of parabola
Solution:
Given Y2= -16X-32
=> Y2= -16(X+2)
=> Comparing with standard equation (Y-k)2 = - 4a(X-h)
(h, k) =(-2,0)
a=4
i) vertex of parabola is (h,k) =(-2,0)
ii)focus of parabola is (-6,0) as parabola is opening towards left side.
Conic sections are formed when a right cone is intercepted by a plane.The shape so formed depends on the angle at which the conic is cut.
Consider the below figure of right cone intercepted by plane and we can see the different shapes formed.
Conic sections:
Conic sections are formed when a right cone is intercepted by a plane.The shape so formed depends on the angle at which the conic is cut.
Consider the below figure of right cone intercepted by plane and we can see the different shapes formed.
The first picture formed is known as parabola
Second one is ellipse
The third is hyperbola.
explanation to eccentricity of 1
Eccentricity of 1 :-
Eccentricity is the measure of the deviation of a conic from a circular path .The parabola is the conic haing eccentricity as 1.
Definition of parabola :-
Parabola is locus of points whose distance from fixed line, called directrix, and a from a fixed point ,called focus,is equal. Vertex is a point where parabola changes its direction .The distance from focus to vertex and vertex to directrix is equal.
The main compenents of Parabola are :-
i)Vertex
ii) axis
iii) focus
iv) directrix
Consider the parabola given below
Description of paraboala :-
i)Point F is called the focus of parabola
ii)Point "O" is called as the vertex of parabola.
iii)C ,D are the points on the parabola whoce distance fron focus and directrix are equal .
iv) Line AB is called directric fo parabola
v) Line ox is called the axis of parabola.
i) Y2 = 4aX
This parabola opens towards right side ie + ve x axis having vertex at (0,0). and focal distance "a" and focal point (a,0) . This curve transforms to (Y-k)2 = 4a(X-h) when vertex shifted to a point (h,k). with focus shifting to (a+h ,k)
ii) Y2= -4aX
This parabola opens towards left side ie -ve x axis having vertex at (0,0). and focal distance "a" and focal point (-a,0).This curve transforms to (Y-k)2 = - 4a(X-h) when vertex shifted to a point (h,k) and focus shifts to (h-a ,k)
iii) X2= 4aY
This parabola opens towards top side ie +ve y axis having vertex at (0,0). and focal distance "a"and focal point (0,a).This curve transforms to (X-h)2 = 4a(Y-k) when vertex shifted to a point (h,k) and focus shifts to (h,k+a).
iv) X2=-4aY
This parabola opens towards down side ie -ve y axis having vertex at (0,0). and focal distance "a" and focal point (0,-a).This curve transforms to (X-h)2 = - 4a(Y-k) when vertex shifted to a point (h,k-a).
examples to eccentricity of 1
Ex 1:
Given the equation of parabola Y2 = 8X find the focus of parabola
solution:
Comparing with standard form of equation
Y2 = 4aX
we get 4a =8
=> a=2
so focus of parabola is (a,0) = (2,0)
Ex 2:
find focus of parabola y = 1/4 x2
solution :
y = 1/4 x2
=> x2= 4y
=> a= 1 on comparing with standard form
so focus of parabola is (0,1) as axis of symmetry is +ve y axis
My forthcoming post is on The Domain of the Function and samacheer kalvi textbooks online will give you more understanding about Algebra
Ex 3:
Given a parabolic equation Y2= -16X-32
Find i) Vertex of parabola
ii) Focus of parabola
Solution:
Given Y2= -16X-32
=> Y2= -16(X+2)
=> Comparing with standard equation (Y-k)2 = - 4a(X-h)
(h, k) =(-2,0)
a=4
i) vertex of parabola is (h,k) =(-2,0)
ii)focus of parabola is (-6,0) as parabola is opening towards left side.
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