An ellipse having foci at (3,3) and (-4,4) and passing though the origin has eccentricity equal to
Ellipse passing through O (0, 0) and having foci P (3, 3) and Q (-4, 4) ,
The length of the major axis of the ellipse (5x - 10)2 + (5y + 15)2
is an ellipse, whose focus is (2, -3) , directrix 3x - 4y + 7 = 0 and eccentricity is 1/2.
Length offrom focus to directrix is
So length of major axis is 20/3
The eccentric angle of a point on the ellipse at a distance of 5/4 units from the focus on the positive x-axis, is
Any point on the ellipse is (2 cosθ, √3 sinθ). The focus on the positive x-axis is (1,0).
Given that (2cosθ-1)2 + 3sin2 θ = 25/16
⇒ cos θ = 3/4
From any point P lying in first quadrant on the ellipse PN is drawn perpendicular to the major axis and produced at Q so that NQ equals to PS, where S is a focus. Then the locus of Q is
Let point Q be (h,k) , where k < 0
Given that k = SP = a + ex1 , where P (x1,y1) lies on the ellipse
If a tangent of slope 2 of the ellipse is normal to the circle x2 + y2 + 4x + 1 = 0 , then the maximum value of ab is
Let be the tangent
It is passing through(-2,0)
An ellipse has the points (1, -1) and (2, -1) as its foci and x +y - 5 = 0 as one of its tangents. Then the point where this line touches the ellipse is
Let image of S'' be with respect to x +y - 5 = 0
Let P be the point of contact.
Because the line L = 0 is tangent to the ellipse, there exists a point P uniquely on the line such that PS + PS ' = 2a .
Since PS ' = 2a Hence, P should be the collinear with SS ''
Hence P is a point of intersection of SS '' (4x - 5 y = 9) , and
From point P (8, 27) , tangent PQ and PR are drawn to the ellipse Then the angle subtended by QR at origin is
Equation of QR is T = 0 (chord of contact)
⇒ 2x + 3y = 1 .......(i)
Now, equation of the pair of lines passing through origin and points Q, R is given by
(making equation of ellipse homogeneous using Eq (i)
∴ 135x2 + 432xy + 320 y2 = 0
An ellipse is sliding along the co-ordinate axes. If the foci of the ellipse are (1,1) and (3,3), then area of the director circle of the ellipse (in sq. units) is
Since x-axis and y-axis are perpendicular tangents to the ellipse, (0,0) lies on the director circle and midpoint of foci (2,2) is centre of the circle.
Hence, radius = 2√2
⇒ the area is 8π units.
The equation of the ellipse whose axes are coincident with the co-ordinates axes and which touches the straight lines 3x - 2y - 20 = 0 and x + 6y - 20 = 0 is
Let the equation of the ellipse be
We know that the general equation of the tangent to the ellipse is
Since 2x - 2y - 20 = 0
is tangent to the ellipse .
Comparing with eq (i)
is tangent to the ellipse, therefore comparing with eq. …(i)
⇒ a2 + 36b2 = 400.....(iii)
Solving eqs. (ii) and (iii) , we get a2 = 40 and b2 = 10 , Therefore , the required equation of the ellipse is
Number of point on the ellipse from which pair of perpendicular tangents are drawn to the ellipse
For the ellipse
Equation of director circle is x2 +y2 = 25. The director circle will cut the ellipse at 4 points
If maximum distance of any point on the ellipse x2 + 2y2 + 2xy = 1 from its centre be r, then r is equal to
Here centre of the ellipse is (0,0) Let P (r cos θ, r sinθ) be any point on the given ellipse then r2 cos2θ + 2r2 sin2θ + 2r2 sin θ cos θ = 1
The equation of the line passing through the centre and bisecting the chord 7x +y -1 = 0 of the ellipse
Let (h,k) be the midpoint of the chord 7 x +y -1 = 0
Represents same straight line
⇒ Equation of the line joining (0,0) and (h,k) is y - x = 0.
Eccentricity of ellipse such that the line joining the foci subtends a right angle only at two points on ellipse, is
The tangent at the point ‘a’ on the ellipse meets the auxiliary circle in two points which subtends a right angle at the centre, then the eccentricity ‘e’ of the ellipse is given by the equation :
Equation of auxillary circle is x2 + y2 = a2
Let P is (acosa,bsina)
Equatio of AB is
bcosα.x+a sinαy = ab
To get combined equation of CA and CB, homogenize equation of circle with equation (i),
b2x2 + b2y2 - (bcosα.x+a sinα.y)2 = 0
since ∠BCA = 90°
∴ coefficient of x2 + coefficient of y2 = 0
The number of values of c such that the straight line y = 4x+ c touches the curve
For given slope there exists two parallel tangents to ellipse. Hence, there are two values of c .
If tangents are drawn to the ellipse x2 + 2y2 = 2, then the locus of the midpoint of the intercept made by the tangent between the coordinate axes is
For any tangent to ellipse
Using midpoint formula, we have
An ellipse with major and minor axes, 6√3 and 6 respectively slides along the coordinate axes and always remains confined in the first quadrant. If the length of arc described by the centre of the ellipse is then the value of k is
Consider the ellipse and the parabola y2 = 2x. They intersect at P and Q in the first and fourth quadrants respectively. Tangents to the ellipse at P and Q intersect the x-axis at R and tangents to the parabola at P and Q intersect the x-axis at S. If the area of quadrilateral PQRS, is λ then
Area of quadrilateral
The area of the quadrilateral formed by the tangents at the end points of latus rectum of the ellipse is k then k/9 is
The length of the focal chord of the ellipse which is inclined to x – axis at an angle 45° is λ, then
Q is a point on the auxiliary circle corresponding to the point P of the ellipse If T is the foot of the perpendicular dropped from the focus S on to the tangent to the auxiliary circle at Q then the ΔSPT is :
Tangent at Q is x cos θ + y sin θ = a
ST = |a e cosθ - a| = a(1 - e cos θ)
Also SP = e PM
ST = SP ⇒ isosceles.
Let x and y satisfy the relation x2 + 9y2 - 4x + 6y + 4 = 0, then maximum value of the expression (4x - 9y),
Given equation is
Let x - 2 = cos θ
Then 4x - 9y = 11+ 4 cosθ - 3sinθ
A rectangle ABCD has area 200 square units. An ellipse with area 200π passes through A and C and has foci at B and D, then
A tangent to the ellipse at any point P meet the line x = 0 at a point Q. Let R be the image of Q in the line y = x, then circle whose extremities of a diameter are Q and R passes through a fixed point. The fixed point is
Equation of the tangent to the ellipse at P (5 cos θ , 4 sin θ) is
It meets the line x = 0 at Q (0, 4 cosec θ)
Image of Q in the line y = x is R (4 cosec θ , 0)
∴ Equation of the circle is
X (x – 4 cosec θ) + y(y – 4 cosec θ) = 0
i.e. x2 + y2 – 4 (x + y) cosec θ = 0
∴ each member of the family passes through the intersection of x2 + y2 = 0 and x + y = 0
i.e. the point (0, 0).
If the curve x2 + 3y2 = 9 subtends an obtuse angle at the point (2α, α), then a possible value of α2 is
The given curve is whose director circle is x2 + y2 = 12. For the required condition (2α, α) should lie inside the circle and outside the ellipse i.e.,
Let S = 0 be the equation of reflection of about the line x – y – 2 = 0. Then the locus of point of intersection of perpendicular tangents of S is
(x - 5)2 + (y - 2)2 = 16 + 9
Any ordinate MP of a ellipse meets the auxiliary circle in Q, then locus of point of intersection of normals at P and Q to the respective curves, is
Equation of normal to the ellipse at ‘P’ is
5x sec θ – 3y cosec θ = 16 …(1)
Equation of normal to the circle x2 + y2 = 25 at point Q is –
y = x tan θ
Eliminating θ from (1) & (2). We get x2 + y2 = 64.
The radius of the largest circle whose centre at (-3,0) and is inscribed in the ellipse 16x2 + 25y2 = 400 is
The line passing through the extremely A of the major axis and extremity B of the minor axis of the ellipse x2 + 9y2 = 9 meets its auxiliary circle at the point M . Then the area of the triangle with vertices at A,M and the origin O is
Equation of line AM is
x +3y -3 = 0
Perpendicular distance of line from the origin
Length of AM
Coordinate of the vertices B and C of triangle ABC are respectively (2, 0) and (8, 0). The vertex ‘A’ is varying in such a way that Then the locus of ‘A’ is an ellipse whose major axis is of length.
Let BC = a, CA = b & AB = c
∴ b + c = 2s – a = 10.
Locus of A is an ellipse with major axis of length 10