Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 1): Chapter 7- Conic Sections

Class 11: Maths Chapter 7 solutions. Complete Class 11 Maths Chapter 1 Notes.

Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 1): Chapter 7- Conic Sections

Maharashtra Board 11th Maths Chapter 7, Class 11 Maths Chapter 7 solutions

Ex 7.1

ii) Given equation of the parabola is y² = -20x.
Comparing this equation with y² = -4ax, we get
⇒ 4a = 20
⇒ a = 5
Co-ordinates of focus are S(-a, 0), i.e., S(-5, 0)
Equation of the directrix is x – a = 0
⇒ x – 5 = 0
Length of latus rectum = 4a = 4(5) = 20
Co-ordinates of end points of latus rectum are (-a, 2a) and (-a, -2a),
⇒ (-5, 10) and (-5, -10).

∴ The required equation of the parabola is x² = -4(5)y, i.e., x² = -20y.

Question 4.
Find the equation of the parabola whose vertex is O(0, 0) and focus at (-7, 0).
Solution:
Focus of the parabola is S(-7, 0) and vertex is O(0, 0).
Since focus lies on X-axis, it is the axis of the parabola.
Focus S(-7, 0) lies on the left-hand side of the origin.
It is a left-handed parabola.
Required parabola is y = -4ax.
Focus is S(-a, 0).
a = 7
∴ The required equation of the parabola is y² =-4(7)x, i.e., y² = -28x.

Question 5.
Find the equation of the parabola with vertex at the origin, the axis along X-axis, and passing through the point
(i) (1, -6)
(ii) (2, 3)
Solution:
(i) Vertex of the parabola is at origin (0, 0) and its axis is along X-axis.
Equation of the parabola can be either y² = 4ax or y² = -4ax.
Since the parabola passes through (1, -6), it lies in the 4th quadrant.
Required parabola is y² = 4ax.
Substituting x = 1 and y = -6 in y² = 4ax, we get
⇒ (-6)² = 4a(1)
⇒ 36 = 4a
⇒ a = 9
∴ The required equation of the parabola is y² = 4(9)x, i.e., y² = 36x.

(ii) Vertex of the parabola is at origin (0, 0) and its axis is along X-axis.
Equation of the parabola can be either y² = 4ax or y² = -4ax.
Since the parabola passes through (2, 3), it lies in 1st quadrant.
∴ Required parabola is y² = 4ax.
Substituting x = 2 and y = 3 in y² = 4ax, we get
⇒ (3)² = 4a(2)
⇒ 9 = 8a

⇒ a = 1
Focal distance of a point = x + a
Given, focal distance = 17
⇒ x + 1 = 17
⇒ x = 16
Substituting x = 16 in y² = 4x, we get
⇒ y² = 4(16)
⇒ y² = 64
⇒ y = ±8
∴ The co-ordinates of the point on the parabola are (16, 8) or (16, -8).

Question 10.
Find the length of the latus rectum of the parabola y² = 4ax passing through the point (2, -6).
Solution:
Given equation of the parabola is y² = 4ax and it passes through point (2, -6).
Substituting x = 2 and y = -6 in y² = 4ax, we get
⇒ (-6)² = 4a(2)
⇒ 4a = 18
∴ Length of latus rectum = 4a = 18 units

Question 12.
If a parabolic reflector is 20 cm in diameter and 5 cm deep, find its focus.
Solution:

Let LOM be the parabolic reflector such that LM is the diameter and ON is its depth.
It is given that ON = 5 cm and LM = 20 cm.
LN = 10 cm
Taking O as the origin, ON along X-axis and a line through O ⊥ ON as Y-axis.
Let the equation of the reflector be y² = 4ax ……(i)
The point L has the co-ordinates (5, 10) and lies on parabola given by (i).
Substituting x = 5 and y = 10 in (i), we get
⇒ 10² = 4a(5)
⇒ 100 = 20a
⇒ a = 5
Focus is at (a, 0), i.e., (5, 0)

Question 13.
Find co-ordinates of focus, vertex, and equation of directrix and the axis of the parabola y = x² – 2x + 3.
Solution:
Given equation of the parabola is y = x² – 2x + 3
⇒ y = x² – 2x + 1 + 2
⇒ y – 2 = (x – 1)²
⇒ (x – 1)² = y – 2
Comparing this equation with X² = 4bY, we get
X = x – 1, Y = y – 2
⇒ 4b = 1

Ex 7.2

Length of minor axis = 2b = 2√3
Lengths of the principal axes are 4 and 2√3.

Since a > b,
X-axis is the major axis and Y-axis is the minor axis.

Ex 7.3

⇒ b = 5 and a = 3
Length of transverse axis = 2b = 2(5) = 10
Length of conjugate axis = 2a = 2(3) = 6
Co-ordinates of vertices are B(0, b) and B’ (0, -b),
i.e., B(0, 5) and B’ (0, -5)
We know that

Solution:
Let e be the eccentricity of a hyperbola

Question 5.
Find the equation of the hyperbola referred to its principal axes:

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Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 1): Chapter 7- Conic Sections

Download PDF: Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 1): Chapter 7- Conic Sections PDF

Chapterwise Maharashtra Board Solutions Class 11 Arts & Science Maths (Part 1) :

• Chapter 1- Angle and its Measurement
• Chapter 2- Trigonometry – I
• Chapter 3- Trigonometry – II
• Chapter 4- Determinants and Matrices
• Chapter 5- Straight Line
• Chapter 6- Circle
• Chapter 7- Conic Sections
• Chapter 8- Measures of Dispersion
• Chapter 9- Probability

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• Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 1): Chapter 6- Circle