11. Conic Sections Mathematics class 11 exercise Miscellaneous Exercise on Chapter - 11
11. Conic Sections Mathematics class 11 exercise Miscellaneous Exercise on Chapter - 11 ncert book solution in english-medium
NCERT Books Subjects for class 11th Hindi Medium
Exercise 11.1
Exercise 11.1 (Conic Sections)
Q1. Find the equation of the circle with centre (0, 2) and radius 2.
Solution:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (0, 2) and radius (r) = 2.
Therefore, the equation of the circle is
(x – 0)2 + (y – 2)2 = 22
⟹ x2 + y2 + 4 – 4 y = 4
⟹ x2 + y2 – 4y = 0
Q2. Find the equation of the circle with centre (–2, 3) and radius 4.
Solution:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (–2, 3) and radius (r) = 4.
Therefore, the equation of the circle is (x + 2)2 + (y – 3)2 = (4)2
⟹ x2 + 4x + 4 + y2 – 6y + 9 = 16
⟹ x2 + y2 + 4x – 6y – 3 = 0
Q3.
Q10. Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Solution:
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the circle passes through points (4, 1) and (6, 5),
(4 – h)2 + (1 – k)2 = r2 …………………. (1)
(6 – h)2 + (5 – k)2 = r2 …………………. (2)
Since the centre (h, k) of the circle lies on line 4x + y = 16,
∴ 4h + k = 16 …………………………………… (3)
From equations (1) and (2), we obtain
(4 – h)2 + (1 – k)2 = (6 – h)2 + (5 – k)2
⇒ 16 – 8h + h2 + 1 – 2k + k2 = 36 – 12h + h2 + 25 – 10k + k2
⇒ 16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k
⇒ 4h + 8k = 44
⇒ h + 2k = 11 ………………………………… (4)
On solving equations (3) and (4), we obtain h = 3 and k = 4.
On substituting the values of h and k in equation (1), we obtain
(4 – 3)2 + (1 – 4)2 = r2
⇒ (1)2 + (– 3)2 = r2
⇒ 1 + 9 = r2
⇒ r2 = 10
⇒ 𝑟=√10
Thus, the equation of the required circle is
(x – 3)2 + (y – 4)2 = (√10)2
x2 – 6x + 9 + y2 – 8y + 16 = 10
x2 + y2 – 6x – 8y + 15 = 0
The required equation for given circle is x2 + y2 – 6x – 8y + 15 = 0
Exercise 11.2
Exercise 11.2 (Conic Sections)
Q1. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = 12x.
Solution:
The given equation is y2 = 12x.
Here, the coefficient of x is positive. Hence, the parabola opens towards the right.
On comparing this equation with y2 = 4ax, we obtain
4a = 12 ⇒ 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 direcctrix, x = –a i.e., x = – 3 i.e., x + 3 = 0
Length of latus rectum = 4a = 4 × 3 = 12
Q3. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = – 8x.
Solution:
The given equation is y2 = –8x.
Here, the coefficient of x is negative. Hence, the parabola opens towards the left.
On comparing this equation with y2 = –4ax, we obtain
–4a = –8 ⇒ a = 2
∴Coordinates of the focus = (–a, 0) = (–2, 0)
Since the given equation involves y2, the axis of the parabola is the x-axis.
Equation of directrix, x = a i.e., x = 2
Length of latus rectum = 4a = 8
Q4. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2 = – 16y.
Solution:
The given equation is x2 = –16y.
Here, the coefficient of y is negative. Hence, the parabola opens downwards.
On comparing this equation with x2 = – 4ay, we obtain
–4a = –16 ⇒ a = 4
∴Coordinates of the focus = (0, –a) = (0, –4)
Since the given equation involves x2, the axis of the parabola is the y-axis.
Equation of directrix, y = a i.e., y = 4
Length of latus rectum = 4a = 16
Q7. Find the equation of the parabola that satisfies the following conditions: Focus (6, 0); directrix x = –6.
Solution:
Focus (6, 0); directrix, x = –6
Since the focus lies on the x-axis, the x-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form y2 = 4ax or
y2 = – 4ax.
It is also seen that the directrix, x = – 6 is to the left of the y-axis, while the focus (6, 0) is to the right of the y-axis.
Hence, the parabola is of the form y2 = 4ax.
Here, a = 6
Thus, the equation of the parabola is y2 = 24x.
Exercise 11.3
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Exercise 11.4
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Miscellaneous Exercise on Chapter - 11
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Mathematics Chapter List
1. Sets
2. Relations and Functions
3. Trigonometric Functions
4. Principle Of Mathematical Induction
5. Complex Numbers and Quadratic Equations
6. Linear Inequalities
7. Permutations and Combinations
8. Binomial Theorem
9. Sequences and Series
10. Straight Lines
11. Conic Sections
12. Introduction to Three Dimensional Geometry
13. Limits and Derivatives
14. Mathematical Reasoning
15. Statistics
16. Probability
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