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.

Download pdf of NCERT solutions class 11 chapter 11-Conic Section

pdf-NCERT solutions class 11 chapter 11-Conic Section

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

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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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