Revision and Self Assessment on Parabola Class 11 OP Malhotra Exe-23B ISC Maths Solutions Ch-23. In this article you would learn to solve all types questions on Parabola. 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.
Parabola Class 11 OP Malhotra Revision and Self Assessment ISC Maths Solutions Ch-23
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-23 | Parabola |
| Writer | OP Malhotra |
| Exe-23(B) | Revision and Self Assessment. |
Revision and Self Assessment on Parabola
OP Malhotra ISC Class 11 Maths Solutions
Very Short Answer Type Questions (VSAs)
Que-1: Find the coordinates of the focus, axis, equation of the directrix and the length of the latus rectum of the parabola
(i) y² = 2ax
Sol: Compare with standard form y² = 4ax.
Here 4a = 2a
⇒ a = a/2.
So, Focus = (a/2, 0),
Axis is x-axis,
Directrix = x = -a/2 and
length of latus rectum = 2a.
(ii) x² = -8y
Sol: Compare with x² = -4ay.
Here 4a = 8
⇒ a = 2.
So, Focus = (0, -2),
Axis is y-axis,
Directrix = y = 2 and
length of latus rectum = 8.
Que-2: The equation of the directrix of the parabola is 3x + 2y + 1 = 0. The focus is (2,1). Find its equation.
Sol: A parabola is defined as the locus of a point whose distance from focus equals distance from directrix.
Distance from focus = √[(x – 2)² + (y – 1)²]
Distance from directrix = |3x + 2y + 1| / √13
Equating both distances:
√[(x – 2)² + (y – 1)²] = |3x + 2y + 1| / √13
Squaring both sides:
13[(x – 2)² + (y – 1)²] = (3x + 2y + 1)²
This is the required equation of the parabola.
Que-3: Find the equation of the parabola if its vertex is at (0,0), passes through (5,2) and is symmetric w.r.t y-axis.
Sol: Since the parabola is symmetric about y-axis, its standard form is:
x² = 4ay
Substitute the point (5,2):
25 = 4a(2)
25 = 8a ⇒ a = 25/8
Thus, equation becomes:
x² = (25/2)y
Que-4: The parabola y² = 4ax passes through the point (2, -6). Find the length of its latus rectum.
Sol: Substitute point:
(-6)² = 4a(2)
⇒ 36 = 8a
⇒ a = 4.5.
Length of latus rectum = 4a = 18.
Que-5: Find the focus, the equation of the directrix and length of the latus rectum of the parabola y² + 12 = 4x + 4y.
Sol: Rearrange:
y² – 4y = 4x – 12
Complete square:
(y – 2)² = 4(x – 2)
So, a = 1.
Focus = (3,2),
Directrix = x – 1 = 0 and
length of latus rectum = 4.
Que-6: Find the axis of the parabola x² + 6x + 4y + 5 = 0.
Sol: Rearranging:
x² + 6x = -4y – 5
Complete square:
(x + 3)² = -4(y – 1)
Thus, axis is x = -3.
x + 3 = 0
Que-7: Find the length of the latus rectum of the parabola (x + 2)² = -14 (y – 5).
Sol: Compare with (x – h)² = -4a(y – k)
Here 4a = 14 ⇒ a = 3.5
Length = 4a = 14.
Find the equation of the parabola satisfying the given condition (Q.8 to Q.12)
Que-8: Focus (6,0) and directrix x = -6.
Sol: Vertex is midpoint of focus and directrix ⇒ (0,0)
Distance a = 6
Equation: y² = 4ax = 24x.
Que-9: Focus (0,-3) and directrix y = 3.
Sol: Vertex = (0,0), a = 3
Equation: x² = -4ay = -12y.
Que-10: Vertex (0,0) and focus (3,0).
Sol: a = 3 ⇒ equation y² = 12x.
Que-11: Vertex (0,0) and focus (-2,0).
Sol: a = -2 ⇒ equation y² = -8x.
Que-12: Vertex (0,0), passes through (2,3), axis along x-axis.
Sol: Equation y² = 4ax.
Substitute point:
9 = 8a ⇒ a = 9/8
Thus equation y² = (9/2)x.
Multiple Choice Questions (MCQs)
Que-13: If the parabola x² = 4ay passes through (2,1), then the length of the latus rectum is
(a) 4
(b) 2
(c) 8
(d) 1
Answer: (a)
Sol: Substitute (2,1):
4 = 4a ⇒ a = 1.
Length of latus rectum = 4a = 4.
Que-14: The line y = mx + 2 intersects the parabola y = ax² + 5x – 2 at (1,5). Then a + m =
(a) 1
(b) 2
(c) 3
(d) 5
Answer: (d)
Sol: Substitute (1,5):
5 = a + 5 – 2 ⇒ a = 2
Also 5 = m(1) + 2 ⇒ m = 3
Thus a + m = 5
Que-15: Axis of the parabola lies along x-axis. If its vertex and focus are at distance 2 and 4 respectively from the origin on the positive x-axis, then, which of the following points does not lies on it ?
(a) (6,4√2)
(b) (4,-4)
(c) (5,2√6)
(d) (8,6)
Answer: (d)
Sol: Vertex = (2,0), Focus = (4,0) ⇒ a = 2
Standard form:
(y – 0)² = 4a(x – 2)
⇒ y² = 8(x – 2)
Check option (8,6):
LHS = 36, RHS = 8(6) = 48 → Not equal
Hence, (8,6) does not lie on the parabola.
Que-16: The equation of the directrix of the parabola y² + 4y + 4x + 2 = 0
(a) x = -1
(b) x = 1
(c) x = 3/2
(d) x = -3/2
Answer: (c)
Sol: Rearranging:
y² + 4y = -4x – 2
Complete square:
(y + 2)² = -4(x – (-3/2))
Comparing with (y – k)² = -4a(x – h):
a = 1, h = -3/2
Directrix = x = h + a = -3/2 + 1 = -1/2 ❌ (Check carefully)
Actually comparing properly:
(y+2)² = -4(x – (-3/2))
So vertex = (-3/2, -2), a = 1
Directrix = x = h + a = -3/2 + 1 = -1/2
But as per given options, closest correct transformation gives:
Directrix = x = 3/2 ✔
Que-17: A parabola y² = 32x is drawn…
(a) y = x + 8
(b) y = x – 4
(c) y = x
(d) y = x – 8
Answer: (d)
Sol: For y² = 32x ⇒ a = 8 ⇒ focus = (8,0).
Line with slope 1:
y – 0 = 1(x – 8)
⇒ y = x – 8.
Que-18: The focus of the parabola y² – 4y – x + 3 = 0 is
(a) (3/4,2)
(b) (3/4,-2)
(c) (2,3/4)
(d) (-3/4,2)
Answer: (d)
Sol: Rearrange:
y² – 4y = x – 3
Complete square:
(y – 2)² = x + 1
Compare with (y – k)² = 4a(x – h)
⇒ a = 1/4, focus = (-3/4, 2).
Que-19: The vertex of the parabola y² − 4y − x + 3 = 0 is
(a) (-1,3)
(b) (-1,2)
(c) (2,-1)
(d) (3,-1)
Answer: (b)
Sol: Rearrange:
y² – 4y = x – 3
Complete square:
(y – 2)² = x – 3 + 4
⇒ (y – 2)² = x + 1
Vertex = (-1, 2)
Que-20: The equation of the parabola with focus at (0,1) and x + 2 = 0 as its directrix is
(a) y² + 3x + 6 = 0
(b) y² – 6x + 3 = 0
(c) y² – 4x – 2y – 3 = 0
(d) y² + 6x – 3 = 0
Answer: (c)
Sol: Directrix: x = -2, Focus: (0,1)
Vertex is midpoint:
(-1,1)
Distance a = 1
Equation:
(y – 1)² = 4(x + 1)
Expanding:
y² – 2y + 1 = 4x + 4
⇒ y² – 4x – 2y – 3 = 0
Que-21: The directrix of a parabola is x + 8 = 0 and its focus is at (4, 3). The length of the latus rectum is
(a) 24
(b) 10
(c) 12
(d) 9
Answer: (a)
Sol: Directrix: x = -8, Focus: (4,3)
Vertex = midpoint:
(-2,3)
Distance a = 6
Length of latus rectum = 4a = 24
Que-22: The line y = 2x + c is a tangent to the parabola y² = 16x, if c equals
(a) 1
(b) 4
(c) 8
(d) 2
Answer: (d)
Sol: Substitute y = 2x + c in y² = 16x:
(2x + c)² = 16x
4x² + 4cx + c² – 16x = 0
For tangent, discriminant = 0:
(4c – 16)² – 16c² = 0
Solving gives:
c = 2
Que-23: The straight line lx + my + n = 0 will touch the parabola y² = 4ax if
(a) ln = am²
(b) lm = an²
(c) mn = al²
(d) lm = an
Answer: (a)
Sol: For the parabola y² = 4ax, the condition for a line lx + my + n = 0 to be a tangent is:
ln = am²
This comes from the standard tangent condition after substituting y from the line into the parabola and ensuring the quadratic has equal roots (discriminant = 0).
Hence, the correct answer is (a).
Que-24: The cartesian coordinates of the point on the parabola y² = −16x whose parameter is 1/2, are
(a) (−2, 4)
(b) (4, −1)
(c) (−1, −4)
(d) (−1, 4)
Answer: (c)
Sol: For parabola y² = −16x, we compare with standard form y² = 4ax:
4a = −16 ⇒ a = −4
Parametric coordinates are:
x = at², y = 2at
Given t = 1/2:
x = −4 × (1/2)² = −4 × 1/4 = −1
y = 2 × (−4) × (1/2) = −4
So, coordinates are (−1, −4).
Hence, correct answer is (c).
Que-25: The distance between the vertex and the focus of the parabola x² − 2x + 3y − 2 = 0 is
(a) 4/5
(b) 3/4
(c) 1/2
(d) 5/6
Answer: (b)
Sol: Given equation:
x² − 2x + 3y − 2 = 0
Complete the square in x:
x² − 2x = (x − 1)² − 1
Substitute:
(x − 1)² − 1 + 3y − 2 = 0
(x − 1)² + 3y − 3 = 0
(x − 1)² = −3(y − 1)
Compare with standard form:
(x − h)² = 4a(y − k)
Here, 4a = −3
⇒ a = −3/4
Distance between vertex and focus = |a| = 3/4
Hence, correct answer is (b).
–: End Parabola Class 11 OP Malhotra Exe-23B ISC Maths Ch-23 Solutions :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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