• Skip to main content
  • Skip to secondary menu
  • Skip to primary sidebar
  • Skip to footer
  • RBSE Model Papers
    • RBSE Class 12th Board Model Papers 2022
    • RBSE Class 10th Board Model Papers 2022
    • RBSE Class 8th Board Model Papers 2022
    • RBSE Class 5th Board Model Papers 2022
  • RBSE Books
  • RBSE Solutions for Class 10
    • RBSE Solutions for Class 10 Maths
    • RBSE Solutions for Class 10 Science
    • RBSE Solutions for Class 10 Social Science
    • RBSE Solutions for Class 10 English First Flight & Footprints without Feet
    • RBSE Solutions for Class 10 Hindi
    • RBSE Solutions for Class 10 Sanskrit
    • RBSE Solutions for Class 10 Rajasthan Adhyayan
    • RBSE Solutions for Class 10 Physical Education
  • RBSE Solutions for Class 9
    • RBSE Solutions for Class 9 Maths
    • RBSE Solutions for Class 9 Science
    • RBSE Solutions for Class 9 Social Science
    • RBSE Solutions for Class 9 English
    • RBSE Solutions for Class 9 Hindi
    • RBSE Solutions for Class 9 Sanskrit
    • RBSE Solutions for Class 9 Rajasthan Adhyayan
    • RBSE Solutions for Class 9 Physical Education
    • RBSE Solutions for Class 9 Information Technology
  • RBSE Solutions for Class 8
    • RBSE Solutions for Class 8 Maths
    • RBSE Solutions for Class 8 Science
    • RBSE Solutions for Class 8 Social Science
    • RBSE Solutions for Class 8 English
    • RBSE Solutions for Class 8 Hindi
    • RBSE Solutions for Class 8 Sanskrit
    • RBSE Solutions

RBSE Solutions

Rajasthan Board Textbook Solutions for Class 5, 6, 7, 8, 9, 10, 11 and 12

  • RBSE Solutions for Class 7
    • RBSE Solutions for Class 7 Maths
    • RBSE Solutions for Class 7 Science
    • RBSE Solutions for Class 7 Social Science
    • RBSE Solutions for Class 7 English
    • RBSE Solutions for Class 7 Hindi
    • RBSE Solutions for Class 7 Sanskrit
  • RBSE Solutions for Class 6
    • RBSE Solutions for Class 6 Maths
    • RBSE Solutions for Class 6 Science
    • RBSE Solutions for Class 6 Social Science
    • RBSE Solutions for Class 6 English
    • RBSE Solutions for Class 6 Hindi
    • RBSE Solutions for Class 6 Sanskrit
  • RBSE Solutions for Class 5
    • RBSE Solutions for Class 5 Maths
    • RBSE Solutions for Class 5 Environmental Studies
    • RBSE Solutions for Class 5 English
    • RBSE Solutions for Class 5 Hindi
  • RBSE Solutions Class 12
    • RBSE Solutions for Class 12 Maths
    • RBSE Solutions for Class 12 Physics
    • RBSE Solutions for Class 12 Chemistry
    • RBSE Solutions for Class 12 Biology
    • RBSE Solutions for Class 12 English
    • RBSE Solutions for Class 12 Hindi
    • RBSE Solutions for Class 12 Sanskrit
  • RBSE Class 11

RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise

June 17, 2019 by Fazal Leave a Comment

Rajasthan Board RBSE Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise

Question 1.
Radius of a circle
9x2 +y2 + 8x = 4(x2 – y2) is :
(A) 1
(B) 2
(C) 4 / 5
(D) 5 / 4
Solution:
Equation of circle,
9x2 + y2 + 8x = 4(x2 – y2)
⇒ 9x2 + y2 + 8x = 4x2 – 4y2
⇒ 9x2 – 4x2 + 8x + y2 + 4y2 = 0
⇒ 5x2 + 5y2 + 8x = 0
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise

Question 2.
Equation of a circle whose centre is in I quadrant as (α, β) and touches x-axis will be :
(A) x2 + y2 – 2αx – 2βy + α2 = 0
(B) x2 + y2 + 2αx – 2βy + α2 = 0
(C) x2 + y2 – 2αx + 2βy + α2 = 0
(D) x2 + y2 + 2αx + 2βy + α2 = 0
Solution:
Centre of circle,
(- g, -f) = α, β
g = – α
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise

Question 3.
If liney = mx + c touches the circle x2 + y2 = 4y, then value of c is :
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
Solution:
Equation of line
y = mx + c …..(i)
and equation of circle
x2 + y2 = 4y
⇒ x2 + y2 – 4y = 0
⇒ x2 + y2 – 2.2.y + 4 = 4
⇒ x2 + (y – 2)2 = 4
⇒ x2 + y2 = 22
where X = x, Y – y – 2, a = 2
According to condition of tangency
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
Thus option (C) is correct.

Question 4.
Line 3x + 4y = 25 touches the circle x2 + y2 = 25 at the point :
(A) (4, 3)
(B) (3, 4)
(C) (-3, -4)
(D) (3, -4)
Solution:
Equation of line,
3x + 4y = 25
and equation of circle
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise

Question 5.
A conic section will be parabola if:
(A) e = 0
(B) e < 0
(C) e > 0
(D) e = 1
Solution:
Option (C) is correct.

Question 6.
Equation of directrix of parabola x2 =- 8y is :
(A) y = -2
(B) y = 2
(C) x = 2
(D) x = – 2
Solution:
Equation of parabola
x2 = – 8y
⇒ x2 = -4 × 2 × y
Here a = 2
By equation of directrix y = a
y = 2
Thus, option (B) is correct.

Question 7.
Vertex of parabola x2 + 4x + 2y = 0 is:
(A) (0, 0)
(B) (2, – 2)
(C) (- 2, -2)
(D) (-2, 2)
Solution:
Equation of parabola is :
x2 + 4x + 2y = 0
⇒ x2 + 4x + 4 = – 2y + 4
⇒ (x + 2)2 = – 2(y – 2)
⇒ x2 = – 2y
Here X = x + 2
and Y = y – 2
Vertex of parabola is (0, 0).
then x = x + 2 = 0
⇒ y = -2
y = y -2 = 0
⇒ y = 2
So vertex is (- 2, 2).
Thus, option (D) is correct.

Question 8.
If focus of any parabola is (-3,0) and directrix is x + 5 = 0, then its equation will be :
(A) y2 = 4(x + 4)
(B) y2 + 4x + 16 = 0
(C) y2 + 4x = 16
(D) x2 = 4(y + 4)
Solution:
Let P(x, y) be any point on parabola. By definition of parabola,
Distance between (x, y) and focus (- 3, 0)
= Length of ⊥ drawn from P(x, y) to directrix x + 5 = 0.
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise

Question 9.
If vertex and focus of any parabola are (2,0) and (5, 0) respectively, then its equation will be :
(A) y2 = 12x + 24
(B) y2 = 12x – 24
(C) y2 = – 12x – 24
(D) y2 = – 12x + 24
Solution:
Vertex = (2, 0),
Focus = (5, 0)
Distance between vertex and focus
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
∴ X = x – 2
and Y = y
Thus equation of parabola
Y2 = 4aX
⇒ y2 = 4 × 3 × (x – 2)
⇒ y2 = 12x – 24
Thus, optional (B) is correct.

Question 10.
Focus of parabola x2 = – 8y is :
(A) (2, 0)
(B) (0, 2)
(C) (-2, 0)
(D) (0, – 2)
Solution:
Equation of parabola,
x2 = – 8y
⇒ x2 = -4 × 2 × y
Coordinates of focus = (0, – a) – (0, – 2)
Thus, optional (D) is correct.

Question 11.
Equation of any tangent at parabola y2 = x is :
(A) y = mx + 1/m
(B) y = mx + 1 /4m
(C) y = mx + 4/m
(D) y = mx + 4m
Solution:
Equation of parabola,
y2 = x
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
Thus, optional (B) is correct.

Question 12.
If line 2y – x = 2 touches the parabola y2 = 2x then tangent point is :
(A) (4, 3)
(B) (-4, 1)
(C) (2, 2)
(D) (1, 4)
Solution:
Equation of line,
2y – x = 2
x = 2y – 2 ….(i)
Equation of parabola,
y2 = 2x
From equation (i) and (ii),
y2 = 2(2y – 2)
⇒ y2 = 4y – 4
⇒ y2 – 4y + 4 = 0
⇒ (y – 2)2 = 0
⇒ y – 2 = 0
⇒ y = 2
Put value ofy in equation (i),
x = 2 × 2 – 2 = 2
Thus tangent point, (x, y) = (2, 2)
Thus, optional (C) is correct.

Question 13.
Tangent equation of parabola x2 = 8y parallel to line x + 2y + 1 = 0 is :
(A) x + 2y + 1 = 0
(B) x – 2y + 1 = 0
(C) x + 2y – 1 = 0
(D) x – 2y – 1 = 0
Solution:
Equation of parabola,
x2 = 8y
⇒ x2 = 4 × 2 × y
here a = 2
Equation of line,
x + 2y + 1 = 0
Equation of its parallel line.
x + 2y + λ = 0
According to condition of tandency
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
Thus, equation of tangent line x + 2y – 4 = 0
Thus, optional (A) is correct.

Question 14.
A normal of parabola y2 = 4x is :
(A) y = x + 4
(B) y + x = 3
(C) y + x = 2
(D) y + x = 1
Solution:
Equation of parabola
y2 = 4x
y2 = 4 × 1 × x
Here a = 1
To find normal, a point should be given which is not given here. So Question is incomplete.

Question 15.
Length of semi latus rectum of ellipse x2 + 4y2 = 12 will be :
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
Solution:
Education of ellipse
3x2 + 4y2 =12
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
Thus option (A) is correct.

Question 16.
Eccentricity of ellipse 3x2 + 4y2 = 12 will be :
(A) -2
(B) \(\frac { 1 }{ 2 } \)
(C) 1
(D) 2
Solution:
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise

Question 17.
If line y = mx + c touches the ellipse
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
then value of c will be:
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
Solution:
Option (C) is correct.

Question 18.
Coordinates of focus of ellipse
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
(A) (± ae, 0)
(B) (± be, 0)
(C) (0, ± ae)
(D) (0, ± be)
Solution:
Option (D) is correct.

Question 19.
Eccentricity of rectangular hyperbola will be:
(A) 0
(B) 1
(C) \(\sqrt { 2 }\)
(D) 2
Solution:
Option (C) is correct.

Question 20.
Eccentricity of hyperbola 9×2 – 16y2 = 144 will be :
(A) 1
(B) 0
(C) 5/16
(D) 5 / 4
Solution:
Equation of hyperbola,
9x2 – 16y2 = 144
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
Thus Option (D) is correct.

Question 21.
Write the equation of circle whose centre is (a cos α, a sin α) and radius is a.
Solution:
Centre of circle
(- g, -f) – (a cos α, a sin α)
or (g,f) = (-a cos α, – a sin α)
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
Then equation of circle,
x2 + y2 + 2gx + 2fy + c = 0
⇒ x2 + y2 – 2a cos α . x – 2a sin α. y = 0
This is required equation.

Question 22.
If tangents at points (x1, y1) and (x2, y2) of circle x2 + y2 + 2gx + 2fy + c = 0 are perpendicular to each other than prove that
x1x2 + y1y2 + g(x1 + x2) + f(y1 + y2) + g2 + f2 = 0
Solution:
Given equation of circle
x2 + y2 + 2gx + 2fy + c = 0 …..(i)
Equation of tangent at point (x1, y1)
xx1 + yy1 + g(x + x1) +f(y + y1) + c = o
⇒ x6x + y6y + g(x1 + x) +f(y1 + y) + c = 0 …..(ii)

RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise

Question 23.
Find the equation of circle with radius r, whose centre lies in 1st quadrant and touches y-axis at a distance of h from the origin. Find the equation of other tangent which passes through origin.
Solution:
Centre of circle = (r, h)
Radius of circle = r
Thus, equation of circle
(x – r)2 + (y – h)2 = r2
Let tangent OB touches the circle at B. Tangent passes through origin. Let tangent touches the circle at point (x1,y1).
∴ Equation of tangent at point of circle,
xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0
xx1 + yy1 – r(x + x1) + h(y + y1) + h2 = 0
Since the line passes through origin.
∴ x × 0 + y × 0 – r(x + 0) – h(y + 0) + h2 = 0
⇒ – rx – hy + h2 = 0
⇒ rx + hy – h2 = 0
This is required equation.

Question 24.
Tangents drawn at point (α, β) of circle x2 + y2 = a2 meets the axis at points A and B respectively. Prove that area of ∆OAB will be,
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
where O is origin.
Solution:
Tangent equation of circle
x2 + y2 = a2
Equation of tangent at point (α, β)
xx1 + yy1 = a2
At x-axis x × a + y × β = a2
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise

Question 25.
Find the equation of tangent at circle x2 + y2 = a2 which makes a triangle of area a2 with axis.
Solution:
Equation of circle,
x2 + y2 = a2
Area of ∆ABO = a2
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise

Question 26.
Write the coordinates of the focus of parabola x2 – 4x – 8y = 4.
Solution:
Equation of parabola,
x2 – 4x – 8y = 4
⇒ x2 – 2.2x + (2)2 = 8y + 4 + 22
⇒ (x – 2)2 = 8y + 8
= 8(y + 1)
(x-2)2 = 4.2.(y + 1)
X2 = 4.2.Y
Where x = x – 2 and Y = y + 1
a = 2
Coordinates of focus = (0, a)
x = x- 2 = 0
⇒ x = 2
Y = y + 1 = 2
⇒ y = 1
Thus coordinates of focus = (2, 1)

Question 27.
Write the eccentricity of parabola
x2 – 4x – 4y + 4 = 0.
Solution:
Eccentricity of parabola is 1.

Question 28.
Write the condition for which line lx + my + n = 0 touches the parabola y2 = 4ax
Solution:
Required condition: In = am2

Question 29.
Write the equation of parabola whose vertex is (0, 0) and focus is (0, – a).
Solution:
Vertex (0, 0) and focus is at (0, – a) which lies at y-axis.
Thus axis of parabola is -ve y-axis.
Equation of parabola is of the form x2 = – 4ay
Thus equation of parabola,
x2 = 4(-a)y
⇒ x2 = – 4ay
This is required equation.

Question 30.
Write the equation of axis of parabola
9y2 – 16x – 12y – 57 = 0.
Solution:
Equation of parabola,
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise

Question 31.
Write the coordinates of centre of ellipse
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
Solution:
Equation of ellipse
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise

Question 32.
Write the condition for which line x cos α + y sin α = p touches the ellipse
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
Solution:
Putting value ofy form line x cos a + y sin α =p in the ellipse.
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
or x2(α2 cos2 α + b2 sin2 α) – 2a2 px cos α
+ (α2p2 – a2b2 sin2 α) = 0 …(i)
Given line will touch ellipse if eqn. (i) has equal roots.
(- 2a2 p cos α)2 – 4(α2 cos2 α + b2 sin2 α)
(α2p2 – α2b2 sin2 α) = 0
or 4α2b2 sin2 α[α2 cos2 α – p2 + b2 sinz α) = 0
Thus p2 = a2 cos2 α + b2 sin2 α

Question 33.
Write the equation of hyperbola whose trans¬verse axis and conjugate axis are 4 and 5 respectively.
Solution:
Equation of hyperbola,
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise

Question 34.
Write the coordinates of centre of hyperbola
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
Solution:
Equation of hyperbola
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise
Comparing eqn. (1) by standard equation of hyperbola, coordinates of centre.
X = x – 1 = 0
⇒ x = 1
Y = y + 2 = 0
⇒ y = – 2
Thus, coordinates of centre = (1, – 2).

RBSE Solutions for Class 11 Maths

Share this:

  • Click to share on WhatsApp (Opens in new window)
  • Click to share on Twitter (Opens in new window)
  • Click to share on Facebook (Opens in new window)

Related

Filed Under: Class 11

Reader Interactions

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Primary Sidebar

Rajasthan Board Questions and Answers

Recent Posts

  • RBSE Solutions for Class 11 Drawing in Hindi Medium & English Medium
  • RBSE Solutions for Class 11 Hindi Sahitya कक्षा 11 हिंदी साहित्य अन्तरा, अन्तराल
  • RBSE Solutions for Class 11 English Literature Woven Words, Julius Caesar & The Guide
  • RBSE Solutions for Class 11 English Compulsory (English Course) Hornbill & Snapshots
  • RBSE Solutions for Class 11 Geography in Hindi Medium & English Medium
  • RBSE Solutions for Class 12 Accountancy in Hindi Medium & English Medium
  • RBSE Solutions for Class 11 Home Science in Hindi Medium & English Medium
  • RBSE Solutions for Class 8 Our Rajasthan in Hindi Medium & English Medium
  • RBSE Solutions for Class 7 Our Rajasthan in Hindi Medium & English Medium
  • RBSE Solutions for Class 6 Our Rajasthan in Hindi Medium & English Medium
  • RBSE Solutions for Class 11 History in Hindi Medium & English Medium

Footer

RBSE Solutions for Class 12
RBSE Solutions for Class 11
RBSE Solutions for Class 10
RBSE Solutions for Class 9
RBSE Solutions for Class 8
RBSE Solutions for Class 7
RBSE Solutions for Class 6
RBSE Solutions for Class 5
RBSE Solutions for Class 12 Maths
RBSE Solutions for Class 11 Maths
RBSE Solutions for Class 10 Maths
RBSE Solutions for Class 9 Maths
RBSE Solutions for Class 8 Maths
RBSE Solutions for Class 7 Maths
RBSE Solutions for Class 6 Maths
RBSE Solutions for Class 5 Maths
Target Batch
RBSE Class 11 Political Science Notes
RBSE Class 11 Geography Notes
RBSE Class 11 History Notes

Copyright © 2022 RBSE Solutions

 

Loading Comments...