Conic Sections Exercise 04
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 (x0)²/14²  (y0)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:

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

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).
 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).
 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).
 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:

Write the equation of a hyperbola in standard form: (xh)^2/a^2  (yk)^2/b^2 = 1

Substitute the coordinates of the foci for the h and k variables: (x±5)^2/a^2  (y0)^2/b^2 = 1

Use the fact that the transverse axis is of length 8 to substitute for the a and b variables: (x±5)^2/8^2  (y0)^2/8^2 = 1

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

Solve for y: y^2 = 64  (x±5)^2

Rewrite the equation in standard form: (xh)^2/a^2  (yk)^2/b^2 = 1

Substitute the values for h, k, a, and b: (x0)^2/8^2  (y0)^2/8^2 = 1

The equation of the hyperbola is: (x0)^2/8^2  (y0)^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:
 Foci: (±3√3, 0)
 Vertices: (±9, 0)
 Eccentricity: 3/9
 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:

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

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 semimajor and semiminor 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).

The vertices of the hyperbola can be found by using the equation: x = a ± c, where a is the length of the semimajor 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).

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 semimajor axis. In this case, c = sqrt(73) and a = 8, so e = sqrt(73)/8.

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 semimajor 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:
(x4)^2/a^2  (y0)^2/b^2 = 1
Step 2: The latus rectum is of length 12, so the equation of the hyperbola is:
(x4)^2/36  (y0)^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:
(x4)^2/36  (y0)^2/18^2 = 1
Question:
Find the equation of the hyperbola whose foci are (0,±√10) and passing through the point (2,3).
Answer:

A hyperbola with foci at (0,±√10) has the equation (x0)^2/a^2  (y√10)^2/b^2 = 1

We can determine the values of a and b by substituting the coordinates of the given point (2,3): (20)^2/a^2  (3√10)^2/b^2 = 1

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

Therefore, the equation of the hyperbola is: (x0)^2/4  (y√10)^2/1/4 = 1

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 xaxis subtends an angle of 90o at the centre . The coordinate of the points (s) where it goes on to intersects the yaxis is . A (0,±4√7/13) B (0,±2√7/31) C (0,±12√7/31) D (0,±√7/31)
Answer:

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

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

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

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

Simplify the equation

9x^2 + 4x^2 = 144

13x^2 = 144

Divide both sides by 13

x^2 = 11

Take the square root of both sides

x = ± √11

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

y = ± (√11/4)

The coordinates of the points where the chord intersects the yaxis 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√(1e^2), 0).
a=3, b=2, e=√(1(b^2/a^2))=√(1(2^2/3^2))=√(14/9)=√(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√(1e^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