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Mohit replied | 13/12/2016

When a point moves in a plane according to some given conditions the path along which it moves is called a locus.

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Sunil replied | 13/12/2016

There is nothing called as a locus of a point because locus itself is set of points which satisfy certain condition. It can be visualized as curve. For example, locus of all points that has a fixed distance from fixed point(centre) is called a circle. Finding Locus based problems are important in JEE exams they should be solved thoroughly. Also there are sufficient video lectures...  more»
There is nothing called as a locus of a point because locus itself is set of points which satisfy certain condition. It can be visualized as curve. For example, locus of all points that has a fixed distance from fixed point(centre) is called a circle. Finding Locus based problems are important in JEE exams they should be solved thoroughly. Also there are sufficient video lectures on YouTube so go ahead and improve your understanding. If any doubt ask here. «less

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Medha replied | 21/11/2016

The Standard equation of Ellipse

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Sarvajeet replied | 03/12/2016

The special form of an ellipse: square of (x/a)+square of (y/b)=1, a>b;
The major axis is (2a) on x-axis and minor axis (2b) y-axis and the eccentricity e<1.

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Ajay replied | 23/11/2016

The direction of ellipse is based on its transverse axis. If its along x axis, its horizontal ellipse and if the axis is along y axis, its vertical ellipse.

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Sarvajeet replied | 03/12/2016

In horizontal ellipse: square of (x/a)+square of (y/b)=1, a>b;
and in vertical ellipse: square of (x/a)+square of (y/b)=1, a

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Sarvajeet replied | 02/12/2016

In an ellipse square of (x/a)+square of (y/b)=1, a>b;
The major axis is (2a) and minor axis (2b).

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Sanjeev replied | 03/12/2016

No, doubt ellipse being elliptical in shape has two diameters, center being the origin of the axes in which longest and shortest diameters are aligned.

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Sarvajeet replied | 03/12/2016

m= the slope of the tangent= dy/dx
Tangent A: y= mx + square root( square of (am) + square of b), where m=the slope of the tangent=dy/dx
or at (x1, y1), xx1/square of a + yy1/square of b=1.

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Rajesh replied | 09/12/2016

X Square Divided by A square + Y Square Divided by B square = 1.

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Ankit replied | 09/12/2016

Same as horizontal ellipse but here, B>A.

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Ajay replied | 23/11/2016

It means (square of x)/4 + ( square of y)/9 = 1. Hence, x intercept is 2 and y intercept is 3.

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First replied | 23/11/2016

9(square of x) + 4(square of y) = 36 can be written as:
(square of x)/(square of 2) + (square of y)/( square of 3) = 1
which is the equation of an ellipse.
So, its x intercepts are : (2,0) ; (-2,0).
And its y intercepts are : (3,0) ; (-3,0).

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Nazia 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

What do you mean by ellipse?

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Guddu replied | 22/11/2016

Locus of a point such that whose sum of distances from two fixed point is constant.

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Sarvajeet replied | 03/12/2016

(square of (x/a) ) +(square of (y/b) )=1 and a>b
and eccentricity e <1.
It must contain 1 major axes, 1 minor axes.

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Saddam 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

What are the properties of ellipse?

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Rohit replied | 13/12/2016

Properties 1. Tangents drawn from a common point, outside the curve are equally inclined to the focal points.
Properties 2. A circle containing the foci and a point p on the curve will intersect the minor axis at the points of intersection of the tangent and the normal to the curve from point.

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Rahul replied | 13/12/2016

Property 1:
Tangents drawn from a common point, outside the curve are equally inclined to the focal points.



PROPERTY 2

A circle containing the foci and a point p on the curve will intersect the minor axis at the points of intersection of the tangent and the normal to the curve from point p.





PROPERTY 3

The length...  more»
Property 1:
Tangents drawn from a common point, outside the curve are equally inclined to the focal points.



PROPERTY 2

A circle containing the foci and a point p on the curve will intersect the minor axis at the points of intersection of the tangent and the normal to the curve from point p.





PROPERTY 3

The length of the minor axis of an ellipse can be found by constructing a line perpendicular to the axis from the focal points, where this line intersects the auxiliary circle will give the length of the minor axis.









PROPERTY 4

Conjugate Diameters





Any diameter of the ellipse, as shown here, may be referred to as a conjugate diameter. This example shows the short conjugate diameter, the long conjugate diameter would pass through the centre and be parallel to a tangent to the curve at r and s.









Constructing the Ellipse given the Conjugate Diameter

PROCEDURE

1. Construct the conjugate diameters TU (long) and RS (short).

2. Take a line from R perpendicular to TU of length CU to find point D.

3. Join D to C.

4. Construct a circle of diameter DC.

5. Take a line from R through the centre of the circle and project it on to meet the circle at G.

6. A line from G through C will give the direction of the minor axis.

7. RE = 1/2 the minor axis.

8. A line from E through C will give the direction of the major axis.

9. RG = 1/2 the major axis.

10. You now have the major and minor axis and can construct the curve.

NOTE: The curve passes through points R, S, T and U.








PROPERTY 1

Tangents drawn from a common point, outside the curve are equally inclined to the focal points.




PROPERTY 2

A circle containing the foci and a point p on the curve will intersect the minor axis at the points of intersection of the tangent and the normal to the curve from point p.





PROPERTY 3

The length of the minor axis of an ellipse can be found by constructing a line perpendicular to the axis from the focal points, where this line intersects the auxiliary circle will give the length of the minor axis.









PROPERTY 4

Conjugate Diameters





Any diameter of the ellipse, as shown here, may be referred to as a conjugate diameter. This example shows the short conjugate diameter, the long conjugate diameter would pass through the centre and be parallel to a tangent to the curve at r and s.









Constructing the Ellipse given the Conjugate Diameter

PROCEDURE

1. Construct the conjugate diameters TU (long) and RS (short).

2. Take a line from R perpendicular to TU of length CU to find point D.

3. Join D to C.

4. Construct a circle of diameter DC.

5. Take a line from R through the centre of the circle and project it on to meet the circle at G.

6. A line from G through C will give the direction of the minor axis.

7. RE = 1/2 the minor axis.

8. A line from E through C will give the direction of the major axis.

9. RG = 1/2 the major axis.

10. You now have the major and minor axis and can construct the curve.

NOTE: The curve passes through points R, S, T and U. «less

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Aditya 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

What are the special cases of ellipse?

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Pankaj replied | 13/12/2016

If you put a=b then equation of ellipse reduces to x^2+y^2=a^2 and it becomes circle that is the special case of ellipse.

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Rajneesh replied | 13/12/2016

When a=b it must be a circle.

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Swarnalakshmi 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

What is the equation of horizontal ellipse?

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Pro replied | 21/11/2016

Hi Swarna, any ellipse is horizontal (elongated along x-axis) if its semi-major axis is greater than semi-minor axis (a>b) in canonical equation of ellipse i.e. x2/a2 + y2/b2 = 1, if b>a then it looks vertically elongated and a=b it is circle. You can try some experiments with ellipse here http://www.desmos.com/calculator...  more»
Hi Swarna, any ellipse is horizontal (elongated along x-axis) if its semi-major axis is greater than semi-minor axis (a>b) in canonical equation of ellipse i.e. x2/a2 + y2/b2 = 1, if b>a then it looks vertically elongated and a=b it is circle. You can try some experiments with ellipse here http://www.desmos.com/calculator «less

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Eti replied | 21/11/2016

(x^2)/(a^2)+(y^2)/(b^2)=1,where a>b

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Ajay 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

Define vertical ellipse?

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Sarvajeet replied | 26/11/2016

(square of (x/a) ) +(square of (y/b) )=1 and a

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Srinivas replied | 28/11/2016

In it the major axis is y-axis its length is 2b and minor axis is x-axis its length is 2a.

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Medha 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

What do you mean by eccentric angle of a point?

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Prateek replied | 19/11/2016

Is equal to tan[inverse (a*y / b*x)].

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Sagar replied | 21/11/2016

The eccentric angle of a point on an ellipse with semi major axes of length a and semi minor axes of length b is the angle t in the parametrization.
x = acost
(1)
y = bsint,
(2)
i.e., for a point (x,y),
t=tan^(-1)((ay)/(bx)).

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J

Jitender 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

Find the length of major and minor axes of the given equation: 4(square of x) + 9(square of y) = 36?

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Sri Veera Bhadraswamy replied | 10/12/2016

6,4 are the lengths of major and minor axes.

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Parveen replied | 12/12/2016

(Major axis)^2=9 so major axis is 3 and (minor axis)^2=4. So, min axis is 2.

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Himangi 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

Define the horizontal ellipse?

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Srinivas replied | 28/11/2016

In ellipse major axis is x-axis major axis length = 2a
minor axis is y- axis its length = 2b.

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Sarvajeet replied | 03/12/2016

square of (x/a) + square of (y/b)=1, a>b.

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Rema 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

What is the special form of ellipse having direction of the axes are parallel to the coordinate axes?

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S

Sudarshan Menon replied | 03/12/2016

circle.

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Sarvajeet replied | 03/12/2016

The special form of an ellipse: square of (x/a)+square of (y/b)=1, a>b;
The major axis is (2a) on x-axis and minor axis (2b) y-axis.

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Rekha 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

What is the equation of the chord with respect to ellipse?

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Vanaja replied | 29/11/2016

An ellipse is the locus of all points of the plane whose distances to two fixed points add to the same constant.

An ellipse can be obtained as a result from the intersection of a cone by a plane in a way that produces a closed curve. Circles are special cases of ellipses, obtained when the cutting plane is orthogonal to the cone's axis.
If equation of an ellipse...  more»
An ellipse is the locus of all points of the plane whose distances to two fixed points add to the same constant.

An ellipse can be obtained as a result from the intersection of a cone by a plane in a way that produces a closed curve. Circles are special cases of ellipses, obtained when the cutting plane is orthogonal to the cone's axis.
If equation of an ellipse is x2 / a2 + y2 / b2 = 1, then equation of director circle is x2 + y2 = a2 + b2. «less

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Tamil replied | 29/11/2016

X^2+y^2=a^2+b^2.

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Shyam 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

What is the equation of normal?

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S

Srinivas replied | 25/11/2016

a^2 x/x1 - b^2 y/y1 = a^2 - b^2.

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Abhilash replied | 14/12/2016

The normal to a curve is a line perpendicular to the tangent to curve through the point of contact. Point form: The equation of the normal to the ellipse x2/a2 + y2/b2 = 1 at the point (x1, y1) is a2x/x1 - b2y/y1 = a2 - b2.

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Krishna 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

What is the difference between the diameter and conjugate diameter with respect to ellipse?

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Manibhushan replied | 07/12/2016

Any chord that passes through the center of an ellipse is call its diameter. It follows that the family of parallel chords define two diameters: one in the direction to which they are all parallel and the other the locus of their midpoints. Such two diameters are called conjugate.The statement is easily proved analytically if we start with the equation x /a + y /b = 1 and the...  more»
Any chord that passes through the center of an ellipse is call its diameter. It follows that the family of parallel chords define two diameters: one in the direction to which they are all parallel and the other the locus of their midpoints. Such two diameters are called conjugate.The statement is easily proved analytically if we start with the equation x²/a² + y²/b² = 1 and the associated parameterization: x = a cos(t), y = b sin(t).
Thus ellipse is a curve defined by the radius-vector:-
r(t) = (a cos(t), b sin(t)).
For a fixed t, we are interested in two points, r(t ± v). We shall use the addition formulas for sine and cosine:
r(t + v) = (a (cos(t)cos(v) - sin(t)sin(v)), b (sin(t)cos(v) + cos(t)sin(v))),
r(t - v) = (a (cos(t)cos(v) + sin(t)sin(v)), b (sin(t)cos(v) - cos(t)sin(v))).
The slope of the difference, say, r(t + v) - r(t - v) is -b/a cot(t), independent of v, meaning that we thus produce a family of parallel chords. Their midpoints satisfy:-
(r(t + v) + r(t - v)) / 2 = cos(v) (a cos(t), b sin(t))
which is a parameterization (with parameter cos(v)) of the chord with the slope of b/a tan(t). To summarize, the midpoints of the chords parallel to the direction with the slope -b/a cot(t) lie on the line with the slope b/a tan(t). Applying the formulas of sine and cosine of the complementary angles we see that starting with the chords with the latter slope we would have found their midpoints on a line with the former slope, thus justifying the symmetric terminology. The two directions are conjugate. Observe in passing that, although the conjugate diameters correspond to the complementary values of the parameter t, the product of the two slopes is -b²/a² which is -1 only for circles, i.e. when a = b, so that in general, the conjugate diameters are not perpendicular.
Making use of the parameterization r(t) = (a cos(t), b sin(t)) I tacitly assumed that the origin of the system of coordinates has been placed at the center of the ellipse. We now evaluate the distance from the center to the end points of the conjugate diameters, i.e., P = (a cos(t), b sin(t)) and, say, Q = (-a sin(t), b cos(t)):
OP² + OQ² = (a² cos²(t) + b² sin²(t)) + (a² sin²(t) + b² cos²(t))
= (a² cos²(t) + a² sin²(t)) + (b² sin²(t) + b² cos²(t))
= a² + b²,
independent of t. This is known as the first theorem of Apollonius: For the conjugate (semi)diameters OP and OQ, OP² + OQ² = a² + b². «less

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Mayank replied | 07/12/2016

Consider family of chords which is parallel to the diameter (chord passing through the center of the ellipse) of the ellipse. Joining the mid points of these chords the line thus formed is the conjugate diameter of the ellipse.

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Ashish 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

What are the properties of conjugate diameters?

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Vanaja replied | 29/11/2016

Any chord that passes through the center of an ellipse is call its diameter. It follows that the family of parallel chords define two diameters: One in the direction to which they are all parallel and the other the locus of their midpoints. Such two diameters are called conjugate.

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Vidyadhan replied | 12/12/2016

In geometry, two diameters of a conic section are said to be conjugate if each chord parallel to one diameter is bisected by the other diameter. For example, two diameters of a circle are conjugate if and only if they are perpendicular.

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Aswini 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

Length of major axis 26, foci (5, 0) find the equation for the ellipse that satisfies the given condition?

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Shiv replied | 03/12/2016

length of major axis 2a=26;
a=13;
f=5;
(square of f)=(square of a)-(square of b);
25=169-(square of b);
(square of b)=169-25;
b=12;
the (sauare of x)/(169)-(square of y)/(144)=1

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Sarvajeet replied | 03/12/2016

The length of major axis 2a=26;
a=13;
ae=5;
So, e=5/13
square of b=(square of a)(1-(square of e))
So, b=12;
square of (x/13)-square of (y/12)=1.
(square of f)=(square of a)-(square of b);
25=169-(square of b);
(square of b)=169-25;
b=12;
the (square of x)/(169)-(square of y)/(144)=1.

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Prasenjit 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

Find the eccentricity and the length of the latus rectum of the ellipse: 16(square of x) + (square of y) = 16?

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Barsha replied | 18/11/2016

Please take a look Prasenjit.

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Sarvajeet replied | 03/12/2016

The eccentricity = (square root 15)/4;
The length of the latus rectum of the ellipse = 1/2

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Dr 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

Find the major axis on the x - axis and phases through the points (4, 3) and (6, 2)?

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Shiv replied | 19/11/2016

x+2y=10;
on x-axis is 10.
on y-axis is 5 and angle is tan[inverse(1/2)].

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Ankit replied | 09/12/2016

x+2y=10;
on x-axis is 10.
on y-axis is 5 and angle is tan[inverse(1/2)].

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Kiran 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

Find the foci on a x - axis if b = 3, c = 4?

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Eti replied | 21/11/2016

Foci is (4,0) and (-4,0).

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Neha 14/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Ellipse

Find the locus of the extremities of the latus recta of all ellipses having a given major axis 6a?

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Saurabh replied | 01/12/2016

Let LSI be the latus rectum, C be the centre of the ellipse and the coordinates of L be (x, y) then x = CS = 3 ae ...... (1)
And y = SL = b2/3a = (9a2(1 - e2))/3a = 3a (1 - e2) ...... (2)
Eliminating the variable 'e' from (1) and (2) we get the locus of L.
Hence putting the value of e from (1) and (2), we get:-
y = 3a(1-x2/9a2)
x2 = 9a2 - 3ay
x2 = 3a(3a -...  more»
Let LSI be the latus rectum, C be the centre of the ellipse and the coordinates of L be (x, y) then x = CS = 3 ae ...... (1)
And y = SL = b2/3a = (9a2(1 - e2))/3a = 3a (1 - e2) ...... (2)
Eliminating the variable 'e' from (1) and (2) we get the locus of L.
Hence putting the value of e from (1) and (2), we get:-
y = 3a(1-x2/9a2)
x2 = 9a2 - 3ay
x2 = 3a(3a - y), which is clearly a parabola. Similarly, we can show that the locus of L' is x2 = 3ay(y + 3a) which is again a parabola. «less

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Tejas replied | 02/12/2016

Co-ordinate of extreme point of latus rectum would b (c,y).You can find ‘y’ by taking ‘x’ as ‘c’ in general equation,which comes out to b2/a.So the co-ordinate becomes {(a2-b2)1/2,b2/a}.Eliminating ‘b’ you will get a2-x2=ay.

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