Conic Sections Exercise 03
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:
-
First, identify the center of the ellipse. The center of the ellipse is (0, 0).
-
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.
-
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:
-
The equation of an ellipse is given by the formula: (x-h)²/a² + (y-k)²/b² = 1
-
Since the major axis is on the x-axis, the equation can be written as: (x-0)²/a² + (y-0)²/b² = 1
-
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
- Solving the above equations, we get:
a² = 4 and b² = 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:
-
The standard form of an ellipse is given by: x2/a2 + y2/b2 = 1
-
Substitute the coordinates of the two points into the equation: (3)2/a2 + (2)2/b2 = 1 (1)2/a2 + (6)2/b2 = 1
-
Solve for a and b: 9/a2 + 4/b2 = 1 1/a2 + 36/b2 = 1
-
Subtract the two equations: 8/a2 - 32/b2 = 0
-
Divide both sides by 8: 1/a2 - 4/b2 = 0
-
Divide both sides by -4: -1/b2 + 1/a2 = 0
-
Multiply both sides by a2b2: a2b2 - b2a2 = 0
-
Factorise: a2(b2 - a2) = 0
-
Solve for a and b: a2 = 0 b2 = a2
-
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:
- 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.
- Since the major axis is the x-axis, a = 6 - 4 = 2 and b = 3 - 2 = 1.
- 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:
-
An equation for an ellipse can be written in the form: (x-h)^2/a^2 + (y-k)^2/b^2 = 1
-
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.
-
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.
-
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:
-
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
-
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
-
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:
-
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
-
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
-
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
-
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
