Question:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 36x^2+4y^2=144

Answer:

Foci: (0, ±6√2)

Vertices: (±6, 0)

Major axis: 12

Minor axis: 6√2

Eccentricity: √2/3

Latus rectum: 12√2

Question:

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (±3,0) ends of minor axis (0,±2).

Answer:

Step 1: An ellipse can be represented by the equation x^2/a^2 + y^2/b^2 = 1, where a and b are the lengths of the major and minor axes respectively.

Step 2: We know that the ends of the major axis are (±3,0), so a = 3.

Step 3: We also know that the ends of the minor axis are (0,±2), so b = 2.

Step 4: Substituting a and b into the equation, we get x^2/3^2 + y^2/2^2 = 1, which is the equation for the ellipse that satisfies the given conditions.

Question:

Find the equation for the ellipse that satisfies the given condition. Ends of major axis (0,±√5) ends of minor axis (±1,0)

Answer:

1. First, identify the center of the ellipse. The center of the ellipse is (0, 0).

2. Next, calculate the length of the major and minor axes. The length of the major axis is √5, and the length of the minor axis is 1.

3. Finally, use the formula for an ellipse to calculate the equation of the ellipse. The equation of the ellipse is: (x^2)/(√5)^2 + (y^2)/1^2 = 1

Question:

Find the equation for the ellipse that satisfies the given conditions: Major axis on the x-axis, centre is origin and passes through the points (4,3) and (6,2).

Answer:

1. The equation of an ellipse is given by the formula: (x-h)²/a² + (y-k)²/b² = 1

2. Since the major axis is on the x-axis, the equation can be written as: (x-0)²/a² + (y-0)²/b² = 1

3. Substituting the points (4,3) and (6,2) in the equation, we get:

(4-0)²/a² + (3-0)²/b² = 1 and (6-0)²/a² + (2-0)²/b² = 1

4. Solving the above equations, we get:

a² = 4 and b² = 5

5. Hence, the equation of the ellipse is: (x-0)²/4 + (y-0)²/5 = 1

Question:

The eccentricity of the ellipse x2​/36+y2​/16=1 is A 2√5/4 ​​B 2√13/6 ​​C 2√5/6 ​​D 2√13/14

Answer:

Answer: B 2√13/6

Question:

Centre at (0,0), major axis on the y−axis and passes through the points (3,2) and (1,6).Find the values of a,b.

Answer:

1. The standard form of an ellipse is given by: x2/a2 + y2/b2 = 1

2. Substitute the coordinates of the two points into the equation: (3)2/a2 + (2)2/b2 = 1 (1)2/a2 + (6)2/b2 = 1

3. Solve for a and b: 9/a2 + 4/b2 = 1 1/a2 + 36/b2 = 1

4. Subtract the two equations: 8/a2 - 32/b2 = 0

5. Divide both sides by 8: 1/a2 - 4/b2 = 0

6. Divide both sides by -4: -1/b2 + 1/a2 = 0

7. Multiply both sides by a2b2: a2b2 - b2a2 = 0

8. Factorise: a2(b2 - a2) = 0

9. Solve for a and b: a2 = 0 b2 = a2

10. Therefore, a = 0 and b = 0.

Question:

If major axis is the x-axis and passes through the points (4,3) and (6,2), then the equation for the ellipse whose centre is the origin is satisfies the given condition. A x^2​/52+y^2​/13=1 B x^2​/13+y^2​/52=1 C x^2​/52−y^2​/13=1 D x^2​/13−y^2​/52=0

Answer:

Answer: C

Explanation:

1. The equation of an ellipse whose centre is the origin is given by x^2/a^2 + y^2/b^2 = 1, where a and b represent the major and minor axes respectively.
2. Since the major axis is the x-axis, a = 6 - 4 = 2 and b = 3 - 2 = 1.
3. Substituting these values in the equation, we get x^2/2^2 + y^2/1^2 = 1, which simplifies to x^2/4 + y^2/1 = 1, which is option C.

Question:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse x^2​/49+y^2​/36=1.

Answer:

Foci: The foci of an ellipse can be found using the formula (c, 0) and (-c, 0), where c is equal to the square root of the sum of the squares of the major and minor axes.

In this case, c = sqrt(49 + 36) = sqrt(85) = 9.22

Therefore, the foci of the ellipse are (9.22, 0) and (-9.22, 0).

Vertices: The vertices of an ellipse can be found using the formula (a, 0) and (-a, 0), where a is equal to the length of the major axis.

In this case, a = sqrt(49) = 7.

Therefore, the vertices of the ellipse are (7, 0) and (-7, 0).

Major Axis: The length of the major axis can be found using the formula a = sqrt(x^2).

In this case, a = sqrt(49) = 7.

Minor Axis: The length of the minor axis can be found using the formula b = sqrt(y^2).

In this case, b = sqrt(36) = 6.

Eccentricity: The eccentricity of an ellipse can be found using the formula e = c/a, where c is equal to the square root of the sum of the squares of the major and minor axes and a is equal to the length of the major axis.

In this case, e = 9.22/7 = 1.31.

Latus Rectum: The length of the latus rectum of an ellipse can be found using the formula l = 2b^2/a, where b is equal to the length of the minor axis and a is equal to the length of the major axis.

In this case, l = 2(6^2)/7 = 8.57.

Question:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse x^2​/100+y^2​/400=1.

Answer:

Foci: (0, ±√300)

Vertices: (±10, 0)

Length of Major Axis: 20

Length of Minor Axis: 10

Eccentricity: √(100-400)/100 = 0.5

Length of Latus Rectum: 2√100 = 20

Question:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 16x^2+y^2=16

Answer:

Step 1: Find the coordinates of the foci.

The coordinates of the foci of an ellipse are given by (h ± c, k) where h and k are the coordinates of the center of the ellipse and c is the distance from the center to the foci.

In this case, the equation of the ellipse is 16x^2+y^2=16.

We can see that the center of the ellipse is at (0, 0).

Therefore, the coordinates of the foci are (±4, 0).

Step 2: Find the coordinates of the vertices.

The vertices of an ellipse are given by (h ± a, k) where h and k are the coordinates of the center of the ellipse and a is the length of the major axis.

In this case, the equation of the ellipse is 16x^2+y^2=16.

We can see that the center of the ellipse is at (0, 0).

Therefore, the coordinates of the vertices are (±4, 0).

Step 3: Find the length of the major axis.

The length of the major axis of an ellipse is given by 2a where a is the distance from the center to the vertices.

In this case, the equation of the ellipse is 16x^2+y^2=16.

We can see that the distance from the center to the vertices is 4.

Therefore, the length of the major axis is 8.

Step 4: Find the length of the minor axis.

The length of the minor axis of an ellipse is given by 2b where b is the distance from the center to the foci.

In this case, the equation of the ellipse is 16x^2+y^2=16.

We can see that the distance from the center to the foci is 4.

Therefore, the length of the minor axis is 8.

Step 5: Find the eccentricity.

The eccentricity of an ellipse is given by e = c/a where c is the distance from the center to the foci and a is the distance from the center to the vertices.

In this case, the equation of the ellipse is 16x^2+y^2=16.

We can see that the distance from the center to the foci is 4 and the distance from the center to the vertices is 4.

Therefore, the eccentricity is e = 4/4 = 1.

Step 6: Find the length of the latus rectum.

The length of the latus rectum of an ellipse is given by 2b/e where b is the distance from the center to the foci and e is the eccentricity.

In this case, the equation of the ellipse is 16x^2+y^2=16.

We can see that the distance from the center to the foci is 4 and the eccentricity is 1.

Therefore, the length of the latus rectum is 2b/e = 8/1 = 8.

Question:

Find the equation for the ellipse that satisfies the given conditions: Length of major axis 26, foci (±5,0).

Answer:

1. An equation for an ellipse can be written in the form: (x-h)^2/a^2 + (y-k)^2/b^2 = 1

2. The major axis of an ellipse is the longest diameter and it’s length is given as 26. This means that the length of the semi-major axis, a, is 13.

3. The foci of an ellipse can be found using the equation: c^2 = a^2 - b^2, where c is the distance between the center and the foci. In this case, c = 5 and a = 13. Therefore, b = sqrt(a^2 - c^2) = sqrt(169 - 25) = 12.

4. The equation for the ellipse is: (x-5)^2/13^2 + (y-0)^2/12^2 = 1

Question:

Length of minor axis 16, foci (0,±6).

Answer:

1. First, identify the equation of an ellipse with the given information. The equation of an ellipse with a minor axis of length 16 and foci at (0, ±6) is: (x^2)/16 + (y^2)/36 = 1

2. Next, calculate the length of the major axis. The length of the major axis can be calculated using the formula a^2 = c^2 - b^2, where a is the length of the major axis, b is the length of the minor axis, and c is the distance between the foci. Therefore, the length of the major axis is: a^2 = 36 - 16 = 20 a = √20 = 4.472135955

3. Finally, calculate the coordinates of the center of the ellipse. The coordinates of the center of the ellipse can be calculated using the formula (h, k) = (0, 0). Therefore, the coordinates of the center of the ellipse are: (h, k) = (0, 0)

Question:

Find the equation for the ellipse that satisfies the given conditions b=3, c=4, centre at the origin foci on x−axis

Answer:

1. The standard equation of an ellipse with the centre at origin and major and minor axis of length 2a and 2b respectively is given by: (x^2/a^2) + (y^2/b^2) = 1

2. Given that b = 3 and c = 4, the equation of the ellipse can be written as: (x^2/4^2) + (y^2/3^2) = 1

3. To shift the foci to the x-axis, we need to add the condition that the distance between the foci is equal to 2c. Therefore, the equation of the ellipse is: (x^2/4^2) + (y^2/3^2) = 1 + (x^2/c^2) = 1

4. Simplifying, the equation of the ellipse is: (x^2/4^2) + (y^2/3^2) - (x^2/16) = 1

Question:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse x^2​/25+y^2​/100=1

Answer:

A. Coordinates of the foci: (0, ±√75)

B. Coordinates of the vertices: (±5, 0)

C. Length of major axis: 10

D. Length of minor axis: 5

E. Eccentricity: √75/5

F. Length of the latus rectum: 5√75

Question:

A chord of the ellipse x^2​/16+y^2​/9=1 passing through the focus on positive x-axis subtends an angle of 90o at the centre . The co-ordinate of the points (s) where it goes on to intersects the y-axis is . A (0,±4√7​/13) B (0,±2√7​/31) C (0,±12√7​/31) D (0,±√7/31)

Answer:

Step 1: An ellipse is a type of conic section, and its equation can be written as Ax^2 + By^2 + Cxy + Dx + Ey + F = 0, where A, B, C, D, E, and F are constants.

Step 2: The equation for the ellipse given in the problem can be written as x^2/16 + y^2/9 = 1, which can be simplified to 16x^2 + 9y^2 = 144.

Step 3: The equation for the ellipse can then be written in the standard form as 16x^2 + 9y^2 - 144 = 0.

Step 4: The equation for the chord of the ellipse passing through the focus on the positive x-axis can be written as y = mx, where m is the slope of the chord.

Step 5: The angle that the chord subtends at the centre is 90°, which means that the slope of the chord is 0.

Step 6: The equation for the chord can then be written as y = 0x, or simply y = 0.

Step 7: Substituting y = 0 into the equation for the ellipse, we get 16x^2 = 144, which can be simplified to x^2 = 9.

Step 8: Taking the square root of both sides, we get x = ±3.

Step 9: Substituting x = ±3 into the equation for the ellipse, we get y = ±4√7/13, ±2√7/31, ±12√7/31, and ±√7/31.

Therefore, the coordinates of the points where the chord intersects the y-axis are (0, ±4√7/13), (0, ±2√7/31), (0, ±12√7/31), and (0, ±√7/31).

Question:

The ellipse 4x2+9y2=36 and the hyperbola 4x2−y2=4 have the same food and they intersect at right angles then the equation of the circle through the points of intersection of two conics is A x2+y2=5 B √5(x^2+y^2)−3x−4y=0 ​C √5(x^2+y^2)+3x+4y=0 D x^2+y^2=25

Answer:

Answer: B √5(x^2+y^2)−3x−4y=0

Question:

Find equation of ellipse, If foci (±3,0),a=4.

Answer:

Answer: Step 1: The equation of an ellipse with major axis of length 2a and two foci at (±c,0) is given by (x - 3)^2/a^2 + y^2/b^2 = 1

Step 2: Substituting the values of c and a, we get (x - 3)^2/4^2 + y^2/b^2 = 1

Step 3: Solving for b, we get b = √(4^2 - (x - 3)^2)

Question:

For curve x^2​/4+y^2​/25=1,, find those points where tangent is (i) Parallel to x-axis (ii) Parallel to y-axis

Answer:

(i) Parallel to x-axis

Step 1: Differentiate the given equation with respect to y to get the slope of the tangent line.

\frac{dy}{dx} = -\frac{x}{2y}

Step 2: The slope of the tangent line should be 0 for it to be parallel to x-axis.

\frac{dy}{dx} = 0

Step 3: Solve the equation for x.

-\frac{x}{2y} = 0

x = 0

Step 4: Substitute x = 0 in the given equation to get the corresponding value of y.

\frac{0^2}{4} + \frac{y^2}{25} = 1

y^2 = 25

y = \pm 5

Therefore, the points where the tangent is parallel to x-axis are (0, 5) and (0, -5).

(ii) Parallel to y-axis

Step 1: Differentiate the given equation with respect to x to get the slope of the tangent line.

\frac{dx}{dy} = -\frac{2y}{x}

Step 2: The slope of the tangent line should be infinite for it to be parallel to y-axis.

\frac{dx}{dy} = \infty

Step 3: Solve the equation for y.

-\frac{2y}{x} = \infty

y = 0

Step 4: Substitute y = 0 in the given equation to get the corresponding value of x.

\frac{x^2}{4} + \frac{0^2}{25} = 1

x^2 = 4

x = \pm 2

Therefore, the points where the tangent is parallel to y-axis are (2, 0) and (-2, 0).

Question:

The equation of the ellipse having vertices at (±5,0) and foci (±4,0) is A x^2​/25+y^2​/16=1 B x^2/9+y^2/25=225 C x^2​/9+y^2​/25=1 D x^2/4+y^2/5=20

Answer:

Answer: C x^2​/9+y^2​/25=1

Question:

Find the equation of the ellipse whose vertices are (±6,0) and foci are (±4,0)

Answer:

Answer: Step 1: Calculate the length of the major and minor axes. Major axis = 6 - 4 = 2 Minor axis = 6 + 4 = 10

Step 2: Calculate the eccentricity of the ellipse. Eccentricity = (Major axis/Minor axis) = (2/10) = 0.2

Step 3: Calculate the center of the ellipse. Center = (6 + 4)/2 = 5

Step 4: Calculate the equation of the ellipse. Equation of the ellipse = (x-5)²/2² + (y-0)²/10² = 1