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In this article, we will provide detailed solutions to Exercise 25.1 from Chapter 2 of the Class 11 RD Sharma textbook. This exercise is part of the Set 1 and focuses on the fundamental concepts of the parabolas including their standard forms and properties.
Understanding the parabola is essential for the mastering various topics in the coordinate geometry and calculus making this chapter a critical part of the Class 11 mathematics syllabus.
A parabola is a plane curve that is mirror-symmetrical and U-shaped. It can be defined as the set of the all points in a plane that are equidistant from the fixed point called the focus and fixed straight line called the directrix. The parabola has several important properties including its vertex, axis of symmetry and focal length in which are crucial in solving the problems related to its geometry.
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Solution:
Given that focus is (3, 0) and the directrix is 3x + 4y = 1
Let us assume that P (x, y) be any point on the parabola.
So, first we draw PM perpendicular to 3x + 4y = 1.
Then,
SP = PM
SP2 = PM2
(x - 3)2 + (y - 0)2 =
(x - 3)2 + y2 =
25 (x2 + 9 - 6x + y2) = (3x + 4y - 1)2
25x2 - 150x + 25y2 + 225 = 9x2+ 16y2 + 1 + 24xy - 8y - 6x
=> 16x2 + 9y2 - 24xy - 144x + 8y + 224 = 0
Hence, the equation of the parabola is 16x2 + 9y2 - 24xy - 144x + 8y + 224 = 0
Solution:
Given that, focus is (1, 1) and the directrix is x + y + 1 = 0
Let us assume P (x, y) be any point on the parabola.
So, first we draw PM perpendicular to x + y + 1 = 0.
Then,
SP = PM
SP2 = PM2
(x - 1)2 + (y - 1)2 =
(x - 1)2 + (y - 1)2 =
2 (x2 + 1 - 2x + y2 + 1 - 2y) = x2 + y2 + 1 + 2xy + 2y + 2x
(2x2 + 2 - 4x + 2 y2 + 2 - 4y) = x2 + y2 + 1 + 2xy + 2y + 2x
=> x2 + y2 - 2xy - 6x - 6y + 3 = 0
Hence, the equation of the parabola is x2 + y2 - 2xy - 6x - 6y + 3 = 0
Solution:
Given that focus is (0, 0) and the directrix 2x − y − 1 = 0
Let us assume P (x, y) be any point on the parabola.
So, the first we draw PM perpendicular to 2x − y − 1 = 0.
Then,
SP = PM
SP2 = PM2
(x - 0)2 + (y - 0)2 =
x2 + y2 =
5x2 + 5 y2 = 4x2 + y2 + 1 - 4xy + 2y - 4x
=> x2 + 4y2 + 4xy - 2y + 4x - 1 = 0
Hence, the equation of the parabola is x2 + 4y2 + 4xy - 2y + 4x - 1 = 0
Solution:
Given that focus is (2, 3) and the directrix x − 4y + 3 = 0
Let us assume P (x, y) be any point on the parabola.
So, the first we draw PM perpendicular to x − 4y + 3 = 0.
Then,
SP = PM
SP2 = PM2
(x - 2)2 + (y - 3)2 =
(x - 2)2 + (y - 3)2 =
17 (x2 + 4 - 4x + y2 - 6y + 9) = x2 + 16y2 + 9 - 8xy - 24y + 6x
17x2 - 68x - 102y + 17y2 + 13 (17) = x2 + 16y2 + 9 - 8xy - 24y + 6x
=> 16x2 + y2 + 8xy - 74x - 78y + 212 = 0
Hence, the equation of the parabola is 16x2 + y2 + 8xy - 74x - 78y + 212 = 0
Solution:
Given that, focus is (2, 3) and directrix is the line x − 4y + 3 = 0.
Let us assume P (x, y) be any point on the parabola.
So, first we draw PM perpendicular to x − 4y + 3 = 0.
Then,
SP = PM
SP2 = PM2
(x - 2)2 + (y - 3)2 =
(x - 2)2 + (y - 3)2 =
17 (x2 + 4 - 4x + y2 - 6y + 9) = x2 + 16y2 + 9 - 8xy - 24y + 6x
17x2 - 68x + 17y2 - 102y + 9 (17) = x2 + 16y2 + 9 - 8xy - 24y + 6x
16x2 + y2 + 8xy - 74x - 78y + 212 = 0
Length of the latus rectum = 2 (Length of the perpendicular from the focus on the directrix)
=
=
=
= 4/√17
Hence, the length of its latus-rectum is 4/√17
Solution:
Give that, the focus is at (−6, −6) and the vertex is at (−2, 2)
Let us assume (x1, y1) be the coordinates of the point of intersection of the axis and directrix.
So, the slope of the axis of the parabola = = -8/-4 = 2
And the slope of the directrix = -1/2
Let us assume the directrix intersects the axis at point K (r, s), so
(r - 6)/2 = -2, (s - 6)/2 = 2
=> r = 2, s = 10
The equation of the directrix is,
y - 10 = -1/2 (x - 2)
=> 2y + x - 22 = 0
Now, let us assume P (x, y) be any point on the parabola
focus is S (−6, −6).
And the directrix is 2y + x - 22 = 0.
So, first we draw PM perpendicular to 2x + y + 22 = 0.
Then,
SP = PM
SP2 = PM2
(x + 6)2 + (y + 6)2 =
5 (x2 + 12x + 36 + y2 + 12y + 36) = 4y2 + x2 + 484 + 4xy - 88y - 44x
4x2 + y2 - 4xy + 104x + 148y - 124 = 0
=> (2x - y)2 - 4 (26x + 37y - 31) = 0
Hence, the equation of the parabola is (2x - y)2 - 4 (26x + 37y - 31) = 0
Solution:
Given that the focus is at (0, −3) and the vertex is at (0, 0)
Let us assume (x1, y1) be the coordinates of the point of intersection of the axis and directrix.
Thus, the slope of the axis of the parabola cannot be defined.
So, the slope of the directrix = 0
Let us assume the directrix intersect the axis at K (r, s).
(r + 0)/2 = 0, (s - 3)/2 = 0
=> r = 0, s = 3
So, the equation of directrix is y = 3
Let us assume P (x, y) be any point on the parabola
With focus is S (0, −3) and the directrix is y = 3.
So, first we draw PM perpendicular to y = 3.
Then,
SP = PM
SP2 = PM2
(x - 0)2 + (y + 3)2 =
x2 + y2 + 6y + 9 = y2 - 6y + 9
=> x2 = -12y
Hence, the equation of the parabola is x2 = -12y
Solution:
Given that, focus is at (0, −3) and the vertex is at (−1, −3)
Let us assume (x1, y1) be the coordinates of the point of intersection of the axis and directrix.
So, the slope of the axis of the parabola is zero and, the slope of the directrix is
(r + 0)/2 = -1, (s - 3)/2 = - 3
=> r = - 2, s = - 3
The equation of the directrix is x + 2 = 0.
Let us assume P (x, y) be any point on the parabola whose focus is S (0, −3) and the directrix is x + 2 = 0.
So, first we draw PM perpendicular to x + 2 = 0
Then,
SP = PM
SP2 = PM2
(x - 0)2 + (y + 3)2 =
x2 + y2 + 6y + 9 = x2 + 4x + 4
=> y2 + 6y - 4x + 5 = 0
Hence, the equation of the parabola is y2 + 6y - 4x + 5 = 0
Solution:
Given that focus is at (a, 0) and the vertex is at (a', 0)
Let us assume (x1, y1) be the coordinates of the point of intersection of the axis and directrix.
So, the slope of the axis of the parabola is zero and, the slope of the directrix cannot be defined.
Let us assume the directrix intersect the axis at point K (r, s).
So, (r + a)/2 = a', (s + 0)/2 = 0
=> r = 2a' - a, s = 0
The equation of the directrix is x - 2a' + a = 0
Let us assume P (x, y) be any point on the parabola
with focus is S (a, 0), and the directrix is x - 2a' + a = 0.
So, first we draw PM perpendicular to x - 2a' + a = 0
Then,
SP = PM
SP2 = PM2
(x - a)2 + (y - 0)2 =
y2 = (x - 2a' + a)2 - (x - a)2
y2 = x2 + 4a'2 + a2 - 4a'x - 4aa' + 2ax - x2 - a2 + 2ax
y2 = 4a'2 - 4a'x - 4aa' + 4ax
=> y2 = -4 (a' - a) (x - a')
Hence, the equation of the parabola is y2 = -4 (a' - a) (x - a')
Solution:
Given that focus is at (0, 0) and vertex is at the intersection of the lines x + y = 1 and x − y = 3.
Let us assume (x1, y1) be the coordinates of the point of intersection of the axis and directrix.
So, the slope of the axis of the parabola = = -8/-4 = 2
And the slope of the directrix = -1/2
Let us assume the directrix intersect the axis at point K (r, s).
so, (r - 6)/2 = -2, (s - 6)/2 = 2
=> r = 2, s = 10
The equation of the directrix is,
y - 10 = -1/2 (x - 2)
=> 2y + x - 22 = 0
Now, let us assume P (x, y) be any point on the parabola
With focus is S (−6, −6) and the directrix is 2y + x - 22 = 0.
So, first we draw PM perpendicular to 2x + y + 22 = 0.
Then,
SP = PM
SP2 = PM2
(x + 6)2 + (y + 6)2 =
5 (x2 + 12x + 36 + y2 + 12y + 36) = 4y2 + x2 + 484 + 4xy - 88y - 44x
4x2 + y2 - 4xy + 104x + 148y - 124 = 0
=> (2x - y)2 - 4 (26x + 37y - 31) = 0
Hence, the equation of the parabola is (2x - y)2 - 4 (26x + 37y - 31) = 0
Solution:
Given that, y2 = 8x
Now, on comparing the given equation with y2 = 4ax, we get
=> 4a = 8
=> a = 2
So, the vertex = (0, 0)
Focus = (a, 0) = (2, 0)
The equation of the directrix is
x = −a
=> x = −2
Axis = y = 0
Length of the latus rectum = 4a = 4(2) = 8 units
Solution:
Given that,
4x2 + y = 0
=> -y/4 = x2
Now, on comparing the given equation with x2 = -4ay, we get
=> 4a = 1/4
=> a = 1/16
So, the vertex = (0, 0)
Focus = (0, −a) = (0, -1/16)
Equation of the directrix, y = a
=> y = 1/16
Axis = x = 0
Length of the latus rectum = 4a = 4 (1/16) = 1/4 units
Solution:
Given that,
y2 − 4y − 3x + 1 = 0
(y - 2)2 - 4 - 3x + 1 = 0
(y - 2)2 = 3 (x + 1)
(y - 2)2 = 3 (x - (-1))
Let us assume Y = y - 2 and X = x + 1.
Then,
Y2 = 3X
Now, on comparing the given equation with Y2 = 4aX, we get
=> 4a = 3
=> a = 3/4
So, the vertex = (X = 0, Y = 0) = (x = -1, y = 2)
Focus = (X = a, Y = 0) = (x + 1 = 3/4, y - 2 = 0) = (x = -1/4, y = 2)
Equation of the directrix: X = −a
i.e. x + 1 = -3/4
=> x = -7/4
Axis = Y = 0
i.e. y - 2 = 0
=> y = 2
Length of the latus rectum = 4a = 4 (3/4) = 3 units
Solution:
Given that,
y2 − 4y + 4x = 0
(y - 2)2 - 4 + 4x = 0
(y - 2)2 = -4 (x - 1)
Let us assume Y = y - 2 and X = x - 1.
Then,
=> Y2 = - 4X
Now, on comparing the given equation with Y2 = -4aX, we get
=> 4a = 4
=> a = 1
So, the vertex = (X = 0, Y = 0) = (x = 1, y = 2)
Focus = (X = −a, Y = 0) = (x - 1 = - 1, y - 2 = 0) = (x = 0, y = 2)
Equation of the directrix: X = a
i.e. x - 1 = 1
=> x = 2
Axis = Y = 0
i.e. y - 2 = 0
=> y = 2
Length of the latus rectum = 4a = 4 (1) = 4 units
Solution:
Given that
y2 + 4y + 4x −3 = 0
(y + 2)2 - 4 + 4x - 3 = 0
(y + 2)2 = - 4 (x - 7/4)
Let us assume Y = y + 2 and X = x - 7/4
Then, Y2 = - 4X.
Now, on comparing the given equation with Y2 = -4aX
=> 4a = 4
=> a = 1
So, the Vertex = (X = 0, Y = 0) = (x = 7/4, y = - 2)
Focus = (X = −a, Y = 0) = (x - 7/4 = - 1, y + 2 = 0) = (x = 3/4, y = - 2)
Equation of the directrix: X = a
i.e. x - 7/4 = 1
=> x = 11/4
Axis = Y = 0
i.e. y + 2 = 0
=> y = - 2
Length of the latus rectum = 4a = 4 (1) = 4 units
Solution:
Given that
y2 = 8x + 8y
(y - 4)2 = 8 (x + 2)
On putting Y = y - 4 and X = x + 2, we get
Y2 = 8X
On comparing the given equation with Y2 = 4aX, we get
=> 4a = 8
=> a = 2
So, the vertex = (X = 0, Y = 0) = (x = - 2, y = 4)
Focus = (X = a, Y = 0) = (x + 2 = 2, y - 4 = 0) = (x = 0, y = 4)
Equation of the directrix: X = −a
i.e. x + 2 = - 2
=> x + 4 = 0
Axis = Y = 0
i.e. y - 4 = 0
=> y = 4
Length of the latus rectum = 4a = 4 (2) = 8 units
Solution:
Given that
4(y − 1)2 = − 7 (x − 3)
(y - 1)2 = -7/4 (x - 3)
Let Y = y - 1 and X = x - 3.
Then,
Y2 = -7/4 X
On comparing the given equation with Y2 = - 4aX, we get
=> 4a = 7/4
=> a = 7/16
So, the vertex = (X = 0, Y = 0) = (x = 3, y = 1)
Focus = (X = −a, Y = 0) = (x - 3 = -7/16, y - 1 = 0) = (x = 41/16, y = 1)
Equation of the directrix: X = a
i.e. x - 3 = 7/16
=> x = 55/16
Axis = Y = 0
i.e. y - 1 = 0
=> y = 1
Length of the latus rectum = 4a = 4 (7/16) = 7/4 units
Solution:
Given that,
y2 = 5x − 4y − 9
y2 + 4y = 5x - 9
(y + 2)2 = 5x - 5
(y + 2)2 = 5 (x - 1)
On putting Y = y + 2 and X = x - 1, we get
Y2 = 5X
On comparing the given equation with Y2 = 4aX, we get
=> 4a = 5
=> a = 5/4
So the vertex = (X = 0, Y = 0) = (x = 1, y = -2)
Focus = (X = a, Y = 0) = (x - 1 = 5/4, y + 2 =0) = (x = 9/4, y = -2)
Equation of the directrix: X = −a
i.e. x - 1 = -5/4
=> x = -1/4
Axis = Y = 0
i.e. y + 2 = 0
=> y = - 2
Length of the latus rectum = 4a = 4 (5/4) = 5 units
Solution:
Given that
x2 = 6x − y − 14
(x - 3)2 = -y - 14 + 9
(x - 3)2 = -y - 5
(x - 3)2 = - (y + 5)
Let us assume Y = y + 5 and X = x - 3.
Then,
X2 = - Y
On comparing the given equation with X2 = - 4aY, we get
=> 4a = 1
=> a = 1/4
So, the vertex = (X = 0, Y = 0) = (x = 3, y = - 5)
Focus = (X = 0, Y = −a) = (x - 3 = 0, y + 5 = -1/4) = (x = 3, y = -21/4)
Equation of the directrix: Y = a
i.e. y + 5 = 1/4
=> y = -19/4
Axis = X = 0
i.e. x - 3 = 0
=> x = 3
Length of the latus rectum = 4a = 4 (1/4) = 1 units
Solution:
Given that the equation of the parabola is y2 = 4px.
Let us assume PQ be the double coordinate of length 8p of the parabola and let A be the vertex of parabola.
y2 = 4px
Then,
PR = RQ = 4p
Let AR = x1
So, the coordinates of P and Q are (x1 , 4p) and (x1, -4p)
Now, P lies on y2 = 4px.
So, (4p)2 = 4p x1
=> x1 = 4p
So, the coordinates of P is (4p, 4p) and Q is (4p, -4p)
The coordinates of A are (0, 0).
So, m1 = Slope of AP = = 4p/4p = 1
And, m2 = Slope of AQ = = -4p/4p = -1
Now,
m1 m2 = - 1
Thus, AP is perpendicular to AQ.
Hence proved.
Solution:
Given that the equation of the parabola is x2 = 12y.
Now on comparing the given equation with x2 = 4ay, we get
=> 4a = 12
=> a = 3
So, the coordinates of focus are (0, 3).
Now two points lie on the parabola.
=> x2 = 12 (3)
=> x2 = 36
=> x = ±6
Therefore, the points are P (6, 3) and Q (-6, 3).
Distance PQ =
=
=
= 12
Area = (PQ) (3) (1/2)
= (12) (3) (1/2)
= 18 sq. units
Therefore, the area of the triangle formed by the lines joining the vertex of the given parabola to the ends of its latus rectum is 18 sq. units.
Solution:
Given that the equation of the directrix is 3x − 4y = 2.
So, the slope of the directrix = -3/-4 = 3/4
Axis is perpendicular to the directrix.
so, the lope of the axis = -4/3
Now, the focus lies on the axis of the parabola.
So, the equation of the axis is
=> y - 3 = (-4/3) (x - 3)
=> 3y - 9 = -4x + 12
=> 3y + 4x - 21 = 0
On solving equations (1) and (2), we get
x = 18/5, y = 11/5
So, the intersection point of the axis and directrix is (18/5, 11/5).
Ad the length of the latus rectum = 2 (Length of the perpendicular from the focus on the directrix)
=
=
= 2
Solution:
We have,
=> x2 = 9y
Now by putting x = 3y in the given equation of the parabola, we get
=> 9y2 = 9y
=> 9y (y - 1) = 0
=> y = 0, 1
At y = 0, x = 0.
And also at y = 1, x = 3.
Therefore, at (1, 3), the abscissa is three times that of the coordinate.
Solution:
Let us consider the equation of the required parabola be,
=> y2 = 4ax
Since it passes through (2, 3), we get,
=> 9 = 4a (2)
=> 8a = 9
=> a = 9/8
So, the equation is y2 =
i.e. 2y2 = 9x
Let us assume the equation of the required parabola be y2 = -4ax.
Since the above equation passes through (2, 3), we get
=> 9 = - 4a (2)
=> 8a = -9
=> a = -9/8
Hence, the equation is,
=> y2 =
=> 2y2 = 9x
Hence, in either case, the required equation of the parabola is 2y2 = 9x.
Solution:
Given that, vertex is (0, 2) and the directrix is y = 2.
Let us assume (x1, y1) be the coordinates of the point of intersection of the axis and directrix.
=> (x1, y1) = (0, 2)
Let us assume the focus be (x2, y2).
Now,
(x2 + 0)/2 = 0 and (y2 + 2)/2 = 0
=> x2 = 0 and y2 = -2
The coordinates of focus are (0, -2).
First we draw PM perpendicular to y = 2.
Then, we have:
SP = PM
SP2 = PM2
(x - 0)2 + (y + 2)2 =
(x - 0)2 + (y + 2)2 = (y - 2)2
x2 + y2 + 4 + 4y = y2 + 4 - 4y
x2 = -4y - 4y
=> x2 = -8y
Hence, the equation of parabola is x2 = -8y.
Chapter 25 of RD Sharma's Class 11 mathematics textbook focuses on the parabola, a fundamental conic section. This chapter covers the definition, standard forms, and various properties of parabolas. Exercise 25.1 specifically deals with identifying and analyzing parabolas given their equations, finding key elements such as the vertex, focus, directrix, and latus rectum, and solving problems related to these concepts
