Question:

Find the equation of the hyperbola satisfying the given conditions: Foci (±3√5,0) the latus rectum is of length 8

Answer:

Answer:

Step 1: The equation of a hyperbola centered at the origin with foci (±3√5,0) is given by:

x^2/a^2 - y^2/b^2 = 1

Step 2: The latus rectum of a hyperbola is given by 2b^2/a.

Step 3: Substitute the given value of the latus rectum (8) in the equation to get:

2b^2/a = 8

Step 4: Solve for b:

b^2 = 4a

Step 5: Substitute the value of b in the equation of the hyperbola:

x^2/a^2 - (4a/a^2) = 1

Step 6: Simplify the equation to get:

x^2 - 4a = a^2

Step 7: Solve for a:

a^2 - x^2 = 4a

Step 8: Divide both sides by 4 to get:

a^2/4 - x^2/4 = a

Step 9: Move the terms to the left side and solve for a:

(a^2 - 4x^2)/4 = a

Step 10: Substitute the value of a in the equation of the hyperbola to get:

x^2/((a^2 - 4x^2)/4) - y^2/a^2 = 1

Step 11: Simplify the equation to get:

x^2/(a^2 - 4x^2) - 4y^2/a^2 = 1

Step 12: The equation of the hyperbola satisfying the given conditions is:

x^2/(a^2 - 4x^2) - 4y^2/a^2 = 1

Question:

Find the equation of the hyperbola satisfying the give conditions: Vertices (±7,0) e=4/3

Answer:

Answer: Step 1: Find the center of the hyperbola: The center of the hyperbola is (0, 0).

Step 2: Find the length of the transverse axis: The length of the transverse axis is 14.

Step 3: Find the length of the conjugate axis: The length of the conjugate axis is 9.

Step 4: Find the equation of the hyperbola: The equation of the hyperbola is (x-0)²/14² - (y-0)²/9² = 1

Question:

Find the coordinates of the foci. the vertices the eccentricity and the length of the latus rectum of the hyperbola 49y^2−16x^2=784

Answer:

1. Identify the equation of the hyperbola: 49y^2−16x^2=784

2. Find the vertices: The vertices of the hyperbola can be found by solving for y when x=0 and x when y=0.

When x=0: 49y^2=784 y= ±√(784/49) y= ±8

When y=0: 16x^2=784 x= ±√(784/16) x= ±14

The vertices are (14, 8) and (-14, -8).

3. Find the foci: The foci of the hyperbola can be found by using the equation c^2=a^2+b^2, where a and b are the distances from the center to the vertices and c is the distance from the center to the foci.

c^2=(14)^2+(8)^2 c= √(196+64) c= √260

The foci are (√260, 0) and (-√260, 0).

4. Find the eccentricity: The eccentricity of the hyperbola can be found using the equation e=c/a, where a and c are the distances from the center to the vertices and the foci, respectively.

e= (√260)/14 e= √(260/196) e= √(13/14)

The eccentricity is √(13/14).

5. Find the length of the latus rectum: The length of the latus rectum can be found using the equation l=2b^2/a, where a and b are the distances from the center to the vertices.

l= 2(8)^2/14 l= 128/14 l= 9.14

The length of the latus rectum is 9.14.

Question:

Find the equation of the hyperbola whose foci are (± 5, 0), the transverse axis is of length 8.

Answer:

1. Write the equation of a hyperbola in standard form: (x-h)^2/a^2 - (y-k)^2/b^2 = 1

2. Substitute the coordinates of the foci for the h and k variables: (x±5)^2/a^2 - (y-0)^2/b^2 = 1

3. Use the fact that the transverse axis is of length 8 to substitute for the a and b variables: (x±5)^2/8^2 - (y-0)^2/8^2 = 1

4. Simplify the equation: (x±5)^2/64 - y^2/64 = 1

5. Solve for y: y^2 = 64 - (x±5)^2

6. Rewrite the equation in standard form: (x-h)^2/a^2 - (y-k)^2/b^2 = 1

7. Substitute the values for h, k, a, and b: (x-0)^2/8^2 - (y-0)^2/8^2 = 1

8. The equation of the hyperbola is: (x-0)^2/8^2 - (y-0)^2/8^2 = 1

Question:

Find the coordinates of the foci and the vertices the eccentricity and the length of the latus of rectum of the hyperbola y^2​/9−x^2​/27=1

Answer:

Answer:

1. Foci: (±3√3, 0)
2. Vertices: (±9, 0)
3. Eccentricity: 3/9
4. Length of latus rectum: 6

Question:

Find the coordinates of the foci, the vertices the eccentricity and the length of latus rectum of the hyperbola 16x^2−9y^2=576

Answer:

1. The equation of the hyperbola is 16x^2−9y^2=576

2. The foci of the hyperbola can be found by using the equation: c^2 = a^2 + b^2, where c is the distance between the foci and a and b are the lengths of the semi-major and semi-minor axes respectively. In this case, a = 8 and b = 3, so c = sqrt(64 + 9) = sqrt(73). Therefore, the coordinates of the foci are (8 ± sqrt(73), 0).

3. The vertices of the hyperbola can be found by using the equation: x = a ± c, where a is the length of the semi-major axis and c is the distance between the foci. In this case, a = 8 and c = sqrt(73), so the vertices are (8 ± sqrt(73), 0).

4. The eccentricity of the hyperbola can be found by using the equation: e = c/a, where c is the distance between the foci and a is the length of the semi-major axis. In this case, c = sqrt(73) and a = 8, so e = sqrt(73)/8.

5. The length of the latus rectum of the hyperbola can be found by using the equation: l = 4a^2/c, where a is the length of the semi-major axis and c is the distance between the foci. In this case, a = 8 and c = sqrt(73), so l = 4(8^2)/sqrt(73) = 32/sqrt(73).

Question:

The vertices of a hyperbola are (0,±3) and its foci are (0,±5). The equation of the hyperbola is A x^2​/16−y^2​/9=1 B y^2​/9−x^2/16​=1 C x^2​/9−y^2​/25=1 D none of these

Answer:

Answer: B y^2​/9−x^2/16​=1

Question:

Find the equation of the hyperbola satisfying the give conditions: Foci (0,±13) the conjugate axis is of length 24

Answer:

Answer:

The equation of a hyperbola with foci (0, ±13) and a conjugate axis of length 24 is given by:

(x^2/12^2) - (y^2/13^2) = 1

Question:

Find the equation of the hyperbola satisfying the give conditions: Foci (±4,0) the latus rectum is of length 12

Answer:

Answer:

Step 1: The equation of a hyperbola with foci (±4, 0) is given by:

(x-4)^2/a^2 - (y-0)^2/b^2 = 1

Step 2: The latus rectum is of length 12, so the equation of the hyperbola is:

(x-4)^2/36 - (y-0)^2/b^2 = 1

Step 3: To find the value of b, we can use the formula:

b^2 = a*c

where a is the length of the major axis and c is the length of the latus rectum.

Therefore, b = sqrt(36*12) = 18

Step 4: The equation of the hyperbola is:

(x-4)^2/36 - (y-0)^2/18^2 = 1

Question:

Find the equation of the hyperbola whose foci are (0,±√10) and passing through the point (2,3).

Answer:

1. A hyperbola with foci at (0,±√10) has the equation (x-0)^2/a^2 - (y-√10)^2/b^2 = 1

2. We can determine the values of a and b by substituting the coordinates of the given point (2,3): (2-0)^2/a^2 - (3-√10)^2/b^2 = 1

3. Solving for a and b: a^2 = 4 and b^2 = (3-√10)^2 = 1/4

4. Therefore, the equation of the hyperbola is: (x-0)^2/4 - (y-√10)^2/1/4 = 1

5. Simplifying, the equation of the hyperbola is: 4x^2 - (y-√10)^2 = 4

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:

1. Given equation is x^2/16 + y^2/9 = 1

2. Since the chord passes through the focus on positive x-axis and subtends an angle of 90o at the centre, the equation of the chord can be written as y = ± (x/4)

3. Substitute y = ± (x/4) in the given equation

4. x^2/16 + (x/4)^2/9 = 1

5. Simplify the equation

6. 9x^2 + 4x^2 = 144

7. 13x^2 = 144

8. Divide both sides by 13

9. x^2 = 11

10. Take the square root of both sides

11. x = ± √11

12. Substitute x = ± √11 in the equation y = ± (x/4)

13. y = ± (√11/4)

14. The coordinates of the points where the chord intersects the y-axis is (0, ±√7/31).

Question:

If the equation of the hyperbola is 9y^2−4x^2=36, then vertices and length of latus rectum is A (0,+2),0 B (0,−2),0 C (0,±2),9 D Noneofthese

Answer:

Answer: C (0,±2),9

Question:

Find the equation of the hyperbola satisfying the give conditions: Vertices (0,±5) foci (0,±8)

Answer:

Answer: The equation of the hyperbola satisfying the given conditions is:

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1

Where, a^2 = \frac{8^2 - 5^2}{2} = 21 b^2 = 5^2

Therefore, the equation of the hyperbola is: \frac{x^2}{21}-\frac{y^2}{25}=1

Question:

Find the coordinates of the foci, the vertices the eccentricity and the length of the latus rectum of the hyperbola 5y^2−9x^2=36

Answer:

Answer: The equation of the hyperbola is in the form of (x^2/a^2)-(y^2/b^2)=1

Therefore, the coordinates of the foci are (±a√(1-e^2), 0).

a=3, b=2, e=√(1-(b^2/a^2))=√(1-(2^2/3^2))=√(1-4/9)=√(5/9)

Therefore, the coordinates of the foci are (±3√(5/9), 0).

The vertices of the hyperbola are (±3, 0).

The length of the latus rectum is 4a√(1-e^2)=4(3)√(5/9)=4√(15/9).

Question:

Find the equation of the hyperbola satisfying the given conditions: Vertices (±2,0), foci (±3,0)

Answer:

Answer: Step 1: Calculate the distance between the foci and the vertices.

Distance = |3 - 2| = 1

Step 2: Calculate the length of the transverse axis.

Transverse axis = 2 * Distance = 2 * 1 = 2

Step 3: Calculate the length of the conjugate axis.

Conjugate axis = 2 * Distance = 2 * 1 = 2

Step 4: Calculate the coordinates of the center of the hyperbola.

Center = (2 + 3) / 2 = 2.5, 0

Step 5: Substitute the values into the standard equation of a hyperbola.

Equation of the hyperbola: (x - 2.5)^2 / 2^2 - (y - 0)^2 / 2^2 = 1