NCERT solutions of class 11 maths exercise 11.2 -Conic sections
All the NCERT solutions of class 11 maths exercise 11.2 chapter 11-Conic sections are created here are the solutions of unsolved questions of the chapter 11-conic sections which is one of the important chapter.The concept of the chapter conic section of NCERT text book is utilised in the industry and science and technology as an example planetry motion,in designing telescope,antennas,reflector in flashlights and automobile headlights.Conic sections are actually the surfaces generated by the intersection of a plane on a double napped right circular cone.
All the NCERT solutions are explained by an expert of maths,each NCERT solutions are explained here by a step by step method,so every students can understand it properly,it will help all the students of class 11 studens in boosting their preparation of fortcomming CBSE board exams.
Click here for exercise 11.1 class 11 maths NCERT solutions
NCERT Solutions of class 11 maths exercise 11.1 chapter 11-Conic sections
Q1.Find the coordinates of the focus, axis of parabola, the equation of directrix and the length of the lactus rectum for y² = 12x
Ans. The given equation is y² = 12x
Here, coefficient of x is positive. Hence, the parabola opens toward the right.
On comparing the equation with y² = 4ax, we obtain
4a = 12
a = 12/4
a = 3
Coordinates of the focus = (a,0) = (3,0)
Since the given equation involves y², the axis of the parabola is the x-axis.
Equation of directrix, x = -a
Length of lactus rectum = 4a = 4 × 3 = 12
Q2.Find the coordinates of the focus, axis of parabola, the equation of directrix and the length of the lactus rectum for x² = 6y
Ans. The given equation is x² = 6y
Here, coefficient of y is positive. Hence, the parabola opens upwards.
On comparing the equation with x² = 4ay, we obtain
6y = 4ay
4a = 6
a = 6/4
a = 3/2
Coordinates of the focus = (0,a) = (0,3/2)
Since the given equation involves x², the axis of the parabola is the y-axis.
Equation of directrix, y = -a
Length of lactus rectum = 4a = 6
Q3.Find the coordinates of the focus, axis of parabola, the equation of directrix and the length of the lactus rectum for y² = -8x
Ans. The given equation is y² = -8x
Here, coefficient of x is negative. Hence, the parabola opens towards the left.
On comparing the equation with y² = -4ax, we obtain
-4a = -8
a = 8/4
a = 2
Coordinates of the focus = (-a,o) = (-2,0)
Since the given equation involves y², the axis of the parabola is the x-axis.
Equation of directrix, x = a
Length of lactus rectum = 4a = 8
Q4.Find the coordinates of the focus, axis of parabola, the equation of directrix and the length of the lactus rectum for x² = -16y
Ans. The given equation is x² = -16y
Here, coefficient of y is negative. Hence, the parabola opens downwards.
On comparing the equation with x² = -4ay, we obtain
-4a = -16
a = 16/4
a = 4
Coordinates of the focus = (o,-a) = (0,-4)
Since the given equation involves x², the axis of the parabola is the y-axis.
Equation of directrix, y = a
Length of lactus rectum = 4a = 16
Q5.Find the coordinates of the focus, axis of parabola, the equation of directrix and the length of the lactus rectum for y² = 10x
Ans. The given equation is y² = 10x
Here, coefficient of x is positive. Hence, the parabola opens towards the right.
On comparing the equation with y² = 4ax, we obtain
4a = 10
a = 10/4
a = 5/2
Coordinates of the focus = (a,o) = (5/2,0)
Since the given equation involves y², the axis of the parabola is the x-axis.
Equation of directrix, x = -a⇒ x = -5/2
Length of lactus rectum = 4a = 4×5/2 =10
Q6.Find the coordinates of the focus ,axis of parabola ,the equation of directrix and length of latus rectum for x² = -9y.
Ans. The given equation of the parabola is x² = -9y
Comparing the given equation with the standard equation of parabola
x² = 4ay
4a = -9
a = -9/4
Coordinates of the focus = (a,0) =(-9/4, 0)
Since the given equation(x² =-9y) is symmetric about y-axis so the axis of parabola is y-axis
The equation of directrix= x = -a⇒x = -(-9/4) ⇒x =9/4
And length of latus rectum =4a = 4×9/4 =9
In each of the section of the exercises 7 to 12, find the equation of the parabola that satisfies the given conditions.
Q7. Focus (6,0), directrix x = -6
Ans. Focus of the parabola is : (a,0) and directrix x = -a
We are given here focus : (6,0)=(a,0)⇒ a = 6
directrix is x =-6 indicates that the parabola is symmetric about x-axis and negative sign shows that it opens at left side of y-axis
So,the equation of parabola is y² = 4ax
Putting a = 6,we get
y² = 4×6x⇒ y² = 24x
y² = 24x
Q8.Focus (0,-3): direcrix, y = 3
Ans. Focus of the parabola is : (0,a) and directrix y= -a
We are given here focus : (0,a)=(0,-3)⇒ a = -3 and directrix y =3
directrix is y =3 indicates that the parabola is symmetric about y-axis and positive sign shows that it opens downwards of x -axis
So,the equation of parabola is x² = 4ay
Putting a = -3,we get
x² = 4×-3y⇒ x² = -12y
x² = -12y
Q9.Vertex (0,0): focus (3,0)
Ans. Focus of the parabola is : (a,0) and the coordinates of the vertex are (0,0)
We are given here focus : (a,0)=(3,0)⇒ a = 3 and directrix will be x= -a ⇒x = -3
directrix is x = -3 indicates that the parabola is symmetric about x-axis and negative sign shows that it opens right of y -axis
So,the equation of parabola is y² = 4ax
Putting a = 3,we get
y² = 4×3x⇒ y² = 12x
y² = 12x
Q10.Vertex (0,0) : focus (-2,0)
Ans. Focus of the parabola is : (a,0) and the coordinates of the vertex are (0,0)
We are given here focus : (a,0)=(-2,0)⇒ a = -2 and directrix will be x= -a ⇒x = -(-2) = 2⇒ x = 2
directrix is x = 2 indicates that the parabola is symmetric about x-axis and positive sign shows that it opens left of y -axis
Equation of parabola is y² = 4ax
Putting a = -2,we get
y² = 4×-2x⇒ y² = -8x
y² = -8x
Q11. Vertex (0,0) passing through (2,3) and axis is along x-axis.
Ans. Since axis of parabola is on x-axis therefore equation of parabola is
y² =4ax, it is given to us that parabola is passing through (2,3)
Putting x =2,y =3 in the equation of parabola
3² = 4×a×2⇒a =9/8
The focus of parabola is : (a,0)=(9/8,0)⇒ a = 9/8 and directrix will be x= -a ⇒x = -9/8
directrix is x = -9/8 indicates that the parabola is symmetric about x-axis and negative sign shows that it opens right of y -axis
So,the equation of parabola is
y² =4ax, putting a =9/8 in the equation
y² = 4×(9/8)x
2y² =9x
Q12.Vertex (0,0) passing through (5,2) and symmetric with respect to y-axis.
Ans. Since axis of parabola is symmetric with respect to y-axis. therefore equation of parabola is
x² =4ay, it is given to us that parabola is passing through (5,2)
Putting x =5,y =2 in the equation of parabola
5² = 4×a×2 ⇒a =25/8
The focus of parabola is : (0,a)=(0,25/8)⇒ a = 25/8 and directrix will be y= -a⇒x = -25/8
directrix is x = -25/8 indicates that the parabola is symmetric about x-axis and negative sign shows that it opens right of y -axis
So,the equation of parabola is
x² =4ay, putting a =25/8 in the equation
x² = 4×(25/8)y
x² =(25/2)x ⇒2x² =25y
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