OP Malhotra Class 11 Maths Solutions Chapter 23 Parabola Chapter Test

Students can cross-reference their work with S Chand ISC Maths Class 11 Solutions Chapter 23 Parabola Chapter Test to ensure accuracy.

S Chand Class 11 ICSE Maths Solutions Chapter 23 Parabola Chapter Test

Question 1.
The equation of the directrix of the parabola is 3x + 2y + 1 = 0. The focus is (2, 1). Find the equation of the parabola.
Solution:
Given focus of parabola be F (2,1) and eqn. of directrix be 3x + 2y + 1 = 0.
Let P (x, y) be any point in the plane of cLiiectrix and focus.Let PM be the ⊥ drawn from P on given directrix.

On squaring both sides, we have
(x – 2)² + (y – 1)² = \(\frac{(3 x+2 y+1)^2}{13}\)
⇒ 13 [(x- 2)² + (y – 1)²] = (3x + 2y + 1)²
⇒ 13 (x² – 4x + y² – 2y + 5]
= 9x² + 4y² + 1 + 12xy + 4y + 6x
⇒ 4x² + 9y² – 12xy – 58x – 30y + 64 = 0
which is the required of parabola.

Question 2.
The points (0, 4) and (0, 2) are the vertex and focus of a parabola. Find the equation of the parabola.
Solution:
Given (0, 4) be the vertex and (0, 2) be focus of the parabola.
Since x-coordinate of both points be same. Hence axis of the parabola is y-axis. Further vertex (0, 4) lies above the focus (0, 2).

Thus the parabola be a downward parabola, length of latus rectum = 4a = 4 | VF | =4 × 2 = 8
Thus the required eqn. of parabola be
(x – 0)² = – 4a(y – 4)
⇒ x² = – 8 (y – 4)
⇒ x² + 8y – 32 = 0
which is required eqn. of parabola.

Question 3.
Find the equation of the parabola with latus rectum joining points (4, 6) and (4,-2).
Solution:
Given the ends of latus rectum are L (4, 6) and L’ (4, – 2).
Since the x-coordinates of both points L and L’ are equal.
∴ Latus-rectum is || to y-axis and hence axis of the parabola is parallel to x-axis.
[since axis ⊥ to latus rectum] Thus, the eqns. of two possible parabolas taken as, (y – k)² = ± 4a (x – h) …(1)
where (h, k) be the vertex of the parabolas.

length of latus-rectum = |LL’|
= \(\sqrt{(4-4)^2+(6+2)^2}\) = 8 = 4 a
∴ eqn. (1) reduces to,
(y – k)² = ± 8(x – h) …(2)
Since the point L (4, 6) lies on eqn. (2); we have
(6 – k)² = 8 (4 -h) …(3)
(6 – k)2 = – 8 (4 – h) …(4)
Also, the point L’ (4, – 2) lies on eqn. (2); we have

On dividing (3) and (5); we have
\(\frac{(6-k)^2}{(2+k)^2}\) = 1 ⇒ (6 – k)² = (2 + k)²
⇒ 36 + k² – 12k = k² + 4k + 4
⇒ 16k = 32 ⇒ k = 2
∴ from (5) ; (- 2 – 2)² = 8 (4 -h)
⇒ 16 = 8 (4 – h) ⇒ 2 = 4 – h ⇒ h = 2
Hence vertex of parabola be A (2, 2).
Thus eqn. of parabola be
(y – 2)² = 8(x – 2)
⇒ y² – 4y – 8x + 20 = 0
On dividing eqn. (4) and (6); we have
k = 2 ∴ from (4); 16 = -8(4 – h)
⇒ – 2 = 4 – h ⇒ h = 6
Thus vertex of required parabola be A’ (6, 2)
∴ eqn. of required parabola be given by (y – 2)² = – 8(x – 6)
⇒ y² – 4y + 8x – 44 = 0

Question 4.
Find the equation of the parabola whose focus is (- 1, – 2) and the equation of the directrix is given 4x – 3y + 2 = 0. Also find the equation of the axis.
Solution:
Given focus of parabola be F (- 1, – 2) and eqn. of directrix be

Let P (x, y) be any point in the plane of directrix and focus. Let PM be the 1 drawn from P (x, y) on given directrix. Then P (x, y) lies on parabola iff PF = PM

⇒ 25 [x² + 2x + y² + 4y + 5] = 16x² + 9y² + 4 – 24xy – 12y + 16x
⇒ 9x² + 16y² + 24xy + 34x + 112y + 121 = 0
which is the required eqn. of parabola. Since axis be the line ⊥ to directrix and passing through the focus of parabola, eqn. of line ⊥ to eqn. (1) be
3x + 4y + k = 0 …(2)
and eqn. (2) passes through the focus F (- 1, – 2)
∴ – 3 – 8 + k = 0
⇒ k = 11
∴ from (2); 3x + 4y + 11 = 0 be the required eqn. of axis of parabola.

Question 5.
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.
Solution:
We know that a curve F (x, y) = 0 be symmetric w.r.t. y-axis if
F (-x, y) = F (x, y)
Let the eqn. of parabola having vertex (0, 0) and symmetric w.r.t. y be (x – 0)² = 4a(y – 0)
i.e. x² = 4ay …(1)
it passes through the point (5, 2); we have
25 = 8a ⇒ a = \(\frac { 25 }{ 8 }\)
Thus eqn. (1) reduces to ;
x² = 4 × \(\frac { 25 }{ 8 }\)y = \(\frac { 25 }{ 2 }\)y
which is the required eqn. of parabola.

Question 6.
The parabola y² = 4ax passes through the point (2, – 6). Find the length of its latus rectum.
Solution:
Given eqn. of parabola be
y² = 4ax …(1)
Now eqn. (1) passes through the point (2, – 6).
∴ 36 = 4a × 2 ⇒ a = \(\frac { 36 }{ 8 }\) = \(\frac { 9 }{ 2}\)
∴ length of latus rectum = 4a = 4 × \(\frac { 9 }{ 2}\) = 18.

Question 7.
Find the coordinates of the vertex and the focus of the parabola y² = 4(x + y).
Solution:
Equation of given parabola be
y² = 4 (x + y)
⇒ y² – 4y = 4x
⇒ y² – 4y + 4 = 4x + 4
⇒ (y – 2)² = 4 (x + 1) …(1)
shifting the origin to point (- 1, 2),
putting x + 1 =X and y – 2 = Y in eqn. (1); we have
Y² = 4X …(2)
Here 4a – 4
⇒ a = 1
which represents a right handed parabola with axis x-axis.
Vertex of parabola (2) be given by
X = 0; Y = 0
x + 1 = 0 and y – 2 = 0
i.e. x = – 1 and y = 2
Thus (-1, 2) be the vertex of parabola (1). Further, focus of parabola (2) be given by
X = a and Y = 0
⇒ x + 1 = 1 and y – 2 = 0
⇒ x = 0 and y = 2
Thus (0, 2) be the focus of parabola (1).

Question 8.
Find the focus, the equation of the directrix and the length of the latus rectum of the parabola y² + 12 = 4x+ 4y.
Solution:
Given eqn. of parabola be
y² – 4x – 4y + 12 = 0
⇒ y² – 4y = 4x – 12
⇒ y² – 4y + 4 = 4x – 8
⇒ (y – 2)² = 4 (x – 2) …(1)
Shifting the origin to point (2, 2) and putting
x – 2 = X and y – 2 = Y in eqn. (1);
we get Y² = 4X …(2)
which represents a right handed parabola.
On comparing with Y² = 4aX
∴ length of latus rectum = 4a = 4 units
Focus of eqn. (2) be given by
X = a, Y = 0
⇒ x – 2 = 1 and y – 2 = 0
i.e. x = 3 and y = 2
Thus (3, 2) are the coordinates of focus of eqn. (1).
eqn. of directrix of parabola (2) be given by X = – a ⇒ x – 2 = – 1 ⇒ x = 1
which is the eqn. of directrix.