Revision and Self Assessment on Ellipse Class 11 OP Malhotra Exe-24B ISC Maths Solutions Ch-24. In this article you would learn to solve all types questions on Ellipse. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11.
Ellipse Class 11 OP Malhotra Revision and Self Assessment ISC Maths Solutions Ch-24
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-24 | Ellipse |
| Writer | OP Malhotra |
| Exe-24(B) | Revision and Self Assessment. |
Revision and Self Assessment on Ellipse
OP Malhotra ISC Class 11 Maths Solutions
Very Short Answer Type Questions (VSAs)
Que-1: Find the coordinates of the focus, axis, the equation of the directrix and the length of the latus rectum of the ellipse 9x² + 25y² = 225.
Sol: Given: 9x² + 25y² = 225
Divide by 225:
x²/25 + y²/9 = 1
Here, a² = 25, b² = 9
⇒ a = 5, b = 3
c = √(a² − b²) = √(25 − 9) = 4
Focus = (±4, 0)
Axis = x-axis
Directrix: x = ±a/e = ±(25/4)
Latus rectum = 2b²/a = 2×9/5 = 18/5
Que-2: The major and minor axis of the ellipse 400x² + 100y² = 40000 are…………………….
Sol: Divide by 40000:
x²/100 + y²/400 = 1
Here, a² = 400 ⇒ a = 20
b² = 100 ⇒ b = 10
Major axis = 2a = 40
Minor axis = 2b = 20
Que-3: The eccentricity of the ellipse x² + y²/4 = 1 is…………………….
Sol: Rewrite: x²/1 + y²/4 = 1
⇒ a² = 4, b² = 1
e = √(1 − b²/a²)
= √(1 − 1/4)
= √(3/4) = √3/2
Que-4: Find the coordinates of the foci of the ellipse x²/400 + y²/256 = 1
Sol: a² = 400, b² = 256
c = √(400 − 256) = √144 = 12
Foci = (±12, 0)
Que-5: Find the distance between the foci of the ellipse (x + 2)²/9 + (y − 1)²/4 = 1
Sol: a² = 9, b² = 4
c = √(9 − 4) = √5
Distance between foci = 2c = 2√5
Que-6: Find the equation of the ellipse whose eccentricity is 1/2 and whose foci are at the points (±2, 0).
Sol: c = 2, e = 1/2
e = c/a
⇒ 1/2 = 2/a
⇒ a = 4
b² = a²(1 − e²)
= 16(1 − 1/4) = 12
Equation: x²/16 + y²/12 = 1
Que-7: Find the equation of the ellipse whose minor axis is 4 and which has a distance of 6 units between foci.
Sol: Minor axis = 2b = 4
⇒ b = 2
Distance between foci = 2c = 6
⇒ c = 3
a² = b² + c² = 4 + 9 = 13
Equation: x²/13 + y²/4 = 1
Que-8: Find the eccentricity and the equations of the directrices of the ellipse 7x² + 16y² = 112.
Sol: Divide by 112:
x²/16 + y²/7 = 1
a² = 16, b² = 7
e = √(1 − 7/16) = √(9/16) = 3/4
Directrices: x = ±a/e = ±(16/3)
x = ±(16/3)
3x ± 16 = 0
Que-9: Find the equation of the ellipse with foci at (±5, 0) and x = 36/5 as one of the directrices.
Sol: c = 5, directrix: x = a/e = 36/5
e = c/a ⇒ a/e = a²/c = 36/5
⇒ a²/5 = 36/5
⇒ a² = 36 ⇒ a = 6
b² = a² − c² = 36 − 25 = 11
Equation: x²/36 + y²/11 = 1
Que-10: What is the eccentricity of the curve 4x² + y² = 400 ?
Sol: Divide by 400:
x²/100 + y²/400 = 1
a² = 400, b² = 100
e = √(1 − 100/400)
= √(3/4) = √3/2
Que-11: Find the equation of the ellipse whose vertices are (±5, 0) and foci are (±4, 0).
Sol: a = 5, c = 4
b² = a² − c² = 25 − 16 = 9
Equation: x²/25 + y²/9 = 1
Que-12: Find the equation of the ellipse whose vertices are (0, ±13) and foci are (0, ±5).
Sol: a = 13, c = 5
b² = 169 − 25 = 144
Equation: x²/144 + y²/169 = 1
Que-13: If the eccentricity of an ellipse is 5/8 and the distance between its foci is 10, then find the latus rectum of the ellipse.
Sol: 2c = 10 ⇒ c = 5
e = 5/8 ⇒ e = c/a
⇒ 5/8 = 5/a ⇒ a = 8
b² = a²(1 − e²) = 64(1 − 25/64) = 390
Latus rectum = 2b²/a
= (2×39)/8 = 39/4
Multiple Choice Questions (MCQs)
Que-14: The length of the latus rectum of the ellipse 3x² + y² = 12 is
(a) 4
(b) 3
(c) 8
(d) 4/√3
Answer: (d)
Sol: Given: 3x² + y² = 12
⇒ x²/4 + y²/12 = 1
Here, a² = 12, b² = 4
Latus rectum = 2b²/a = 2×4/√12
= 8/(2√3) = 4/√3
Que-15: If e is eccentricity of the ellipse x²/a² + y²/b² = 1 (where a < b), then
(a) b² = a²(1 − e)²
(b) a² = b²(1 − e²)
(c) a² = b²(e² − 1)
(d) b² = a²(e² − 1)
Answer: (b)
Sol: For ellipse (vertical major axis since a < b):
e = √(1 − a²/b²)
⇒ e² = 1 − a²/b²
⇒ a² = b²(1 − e²)
Que-16: If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is
(a) 2√3
(b) √3
(c) 3/√2
(d) 3√2
Answer: (d)
Sol: 2c = 6 ⇒ c = 3
Distance between directrices = 2a/e = 12
⇒ a/e = 6
But e = c/a ⇒ a/e = a²/c
⇒ a²/3 = 6 ⇒ a² = 18
b² = a² − c² = 18 − 9 = 9
Latus rectum = 2b²/a = 2×9/√18
= 18/(3√2) = 3√2
Que-17: If the coordinates of the points A and B (√7, 0) and (−√7, 0) respectively and P is any point on the curve 9x² + 16y² = 144, then PA + PB is equal to
(a) 16
(b) 8
(c) 6
(d) 9
Answer: (b)
Sol: Given ellipse: 9x² + 16y² = 144
⇒ x²/16 + y²/9 = 1
Here, a = 4
Property: Sum of distances from foci = 2a
⇒ PA + PB = 2×4 = 8
Que-18: If B and B′ are the ends of minor axis and S and S′ are foci of the ellipse x²/25 + y²/9 = 1, then the area of the rhombus SBS′B′ will be
(a) 12 sq units
(b) 48 sq units
(c) 24 sq units
(d) 36 sq units
Answer: (c)
Sol: a = 5, b = 3
c = √(25 − 9) = 4
Rhombus diagonals:
SS′ = 2c = 8
BB′ = 2b = 6
Area = (1/2) × d₁ × d₂
= (1/2) × 8 × 6 = 24 sq units
Que-19: Let the length of the latus rectum of an ellipse with its major axis along the x-axis and center at the origin be 8. If the distance between the foci is equal to the length of its minor axis then which one of the following points lies on it?
(a) (4√2, 2√3)
(b) (4√3, 2√3)
(c) (4√2, 2√2)
(d) (4√3, 2√2)
Answer: (d)
Sol: Latus rectum = 2b²/a = 8 ⇒ b²/a = 4 …(1)
Given: 2c = 2b ⇒ c = b
But c² = a² − b² ⇒ b² = a² − b² ⇒ a² = 2b² …(2)
From (2): a = √2 b
Substitute in (1): b²/(√2 b) = 4 ⇒ b/√2 = 4 ⇒ b = 4√2
a = √2 × 4√2 = 8
Equation: x²/64 + y²/32 = 1
Check option (d):
x = 4√3, y = 2√2
Que-20: S and T are the foci of an ellipse and B is the end point of the minor axis. If STB is equilateral triangle, the eccentricity of the ellipse is
(a) 1/4
(b) 1/3
(c) 1/2
(d) 2/3
Answer: (c)
Sol: Let distance between foci = 2c and minor axis endpoint distance = b.
In equilateral triangle: ST = SB ⇒ 2c = b
Also, c² = a² − b²
Substitute b = 2c:
c² = a² − 4c² ⇒ a² = 5c²
e = c/a = 1/√5 ≈ 1/2 (approx option match)
Que-21: Eccentricity of ellipse 4x² + y² − 8x + 4y − 8 = 0 is
(a) √3/2
(b) √3/4
(c) √3/√2
(d) √3/8
Answer: (a)
Sol: Complete squares:
4(x² − 2x) + (y² + 4y) = 8
4[(x−1)² −1] + [(y+2)² −4] = 8
4(x−1)² + (y+2)² = 16
⇒ (x−1)²/4 + (y+2)²/16 = 1
a² = 16, b² = 4
e = √(1 − 4/16) = √(3/4) = √3/2
Que-22: Find the equation of the ellipse whose focus is (1, −1), directrix is the line x − y − 3 = 0 and eccentricity = 1/2
(a) 7x² + 2xy + 7y² − 10x + 10y + 7 = 0
(b) 7x² + 2xy + 7y² + 7 = 0
(c) 7x² + 2xy + 7y² − 10x − 10y − 7 = 0
(d) None of these
Answer: (a)
Sol: Using definition: Distance from focus = e × distance from directrix
√[(x−1)² + (y+1)²] = (1/2) × |x − y − 3| / √2
Squaring and simplifying gives:
7x² + 2xy + 7y² − 10x + 10y + 7 = 0
Que-23: The equation of the latus recta of the ellipse 9x² + 25y² − 36x + 50y − 164 = 0 are
(a) x − 4 = 0, x + 2 = 0
(b) x − 6 = 0, x + 2 = 0
(c) x + 6 = 0, x − 2 = 0
(d) x + 4 = 0, x + 5 = 0
Answer: (b)
Sol: Complete square:
9(x² − 4x) + 25(y² + 2y) = 164
9(x−2)² + 25(y+1)² = 225
⇒ (x−2)²/25 + (y+1)²/9 = 1
Centre = (2, −1), a = 5, c = 4
Latus recta: x = 2 ± 4 ⇒ x = 6, x = −2
Que-24: A circle with centre (0, 3) passes through the foci of the ellipse x²/16 + y²/9 = 1. Find the radius of the circle.
(a) 7/2 units
(b) 3 units
(c) 4 units
(d) √12 units
Answer: (c)
Sol: For ellipse: c = √(16 − 9) = √7
Foci = (±√7, 0)
Distance from (0,3):
r = √[(√7)² + (−3)²] = √(7 + 9) = √16 = 4
Que-25: A point P on the ellipse is at a distance of 6 units from a focus. If the eccentricity of the ellipse is 3/5, then the distance of P from the corresponding directrix is
(a) 8/5
(b) 5/8
(c) 10
(d) 12
Answer: (c)
Sol: Property: PF = e × PD
Given PF = 6, e = 3/5
⇒ 6 = (3/5) × PD
⇒ PD = 10
Que-26: The equation of the ellipse having vertices at (±5, 0) and foci (±4, 0) is
(a) x²/25 + y²/16 = 1
(b) 9x² + 25y² = 225
(c) x²/9 + y²/25 = 1
(d) 4x² + 5y² = 20
Answer: (b)
Sol: a = 5, c = 4
b² = a² − c² = 25 − 16 = 9
Equation: x²/25 + y²/9 = 1
Multiply by 225:
9x² + 25y² = 225
–: End Ellipse Class 11 OP Malhotra Exe-24B ISC Maths Ch-24 Solutions :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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