Conic Sections — Class 11 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Conic Sections" — 8 important questions with detailed answers for CBSE board exam preparation.
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TL;DR: Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Conic Sections" — 8 important questions with detailed answers for CBSE board exam…
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Key Questions Covered:
- Find the equation of the circle with center (2, -3) and radius 5.
- Find the center and radius of the circle x² + y² - 6x + 8y - 11 = 0.
- Find the equation of the parabola with vertex at origin and focus at (2, 0).
- Find the vertices and foci of the ellipse x²/25 + y²/16 = 1.
- Find the equation of the hyperbola with vertices at (±3, 0) and foci at (±5, 0).
- Find the equation of the parabola with focus at (-2, 0) and directrix x = 2.
- + 2 more questions in the full chapter
Solutions Summary:
| Question | Status |
|---|---|
| Find the equation of the circle with center (2, -3) and r… | ✓ Solved |
| Find the center and radius of the circle x² + y² - 6x + 8… | ✓ Solved |
| Find the equation of the parabola with vertex at origin a… | ✓ Solved |
| Find the vertices and foci of the ellipse x²/25 + y²/16 = 1. | ✓ Solved |
| Find the equation of the hyperbola with vertices at (±3, … | ✓ Solved |
| Find the equation of the parabola with focus at (-2, 0) a… | ✓ Solved |
Showing 6 of 8 questions
Q1: Find the equation of the circle with center (2, -3) and radius 5.
Given: Center C(2, -3) and radius r = 5
Step 1: Use the standard form of circle equation:
(x - h)² + (y - k)² = r²
Step 2: Substitute h = 2, k = -3, r = 5:
(x - 2)² + (y - (-3))² = 5²
(x - 2)² + (y + 3)² = 25
Step 3: Expand the equation:
x² - 4x + 4 + y² + 6y + 9 = 25
x² + y² - 4x + 6y + 13 = 25
x² + y² - 4x + 6y - 12 = 0
Final Answer: Circle equation is (x - 2)² + (y + 3)² = 25 or x² + y² - 4x + 6y - 12 = 0
Q2: Find the center and radius of the circle x² + y² - 6x + 8y - 11 = 0.
Given circle equation: x² + y² - 6x + 8y - 11 = 0
Step 1: Rearrange:
x² - 6x + y² + 8y = 11
Step 2: Complete the square for x terms:
x² - 6x + 9 - 9 + y² + 8y = 11
(x - 3)² - 9 + y² + 8y = 11
Step 3: Complete the square for y terms:
(x - 3)² + y² + 8y + 16 - 16 = 11 + 9
(x - 3)² + (y + 4)² - 16 = 20
(x - 3)² + (y + 4)² = 36
Step 4: Identify center and radius:
Comparing with (x - h)² + (y - k)² = r²
Center = (3, -4)
Radius = √36 = 6
Final Answer: Center (3, -4), Radius = 6
Q3: Find the equation of the parabola with vertex at origin and focus at (2, 0).
Given: Vertex at origin (0, 0) and focus at (2, 0)
Step 1: Determine the orientation:
Focus is on the positive x-axis, so the parabola opens rightward.
Standard form: y² = 4ax
Step 2: Find the value of a:
Focus is at (a, 0), so a = 2
Step 3: Write the equation:
y² = 4(2)x
y² = 8x
Step 4: Verify:
For parabola y² = 8x:
Vertex (0, 0) ✓
Focus (a, 0) = (2, 0) ✓
Directrix x = -a = -2
Final Answer: The equation of the parabola is y² = 8x
Q4: Find the vertices and foci of the ellipse x²/25 + y²/16 = 1.
Given ellipse: x²/25 + y²/16 = 1
Compare with standard form: x²/a² + y²/b² = 1
Step 1: Identify a and b:
a² = 25, so a = 5
b² = 16, so b = 4
Since a > b, the major axis is along the x-axis.
Step 2: Find vertices:
Vertices are (±a, 0) = (±5, 0)
Vertices: (5, 0) and (-5, 0)
Step 3: Find the relationship c² = a² - b²:
c² = 25 - 16 = 9
c = 3
Step 4: Find foci:
Foci are (±c, 0) = (±3, 0)
Foci: (3, 0) and (-3, 0)
Step 5: Find eccentricity:
e = c/a = 3/5
Final Answer: Vertices (±5, 0), Foci (±...
Q5: Find the equation of the hyperbola with vertices at (±3, 0) and foci at (±5, 0).
Given: Vertices at (±3, 0) and foci at (±5, 0)
Step 1: Identify a and c:
Since vertices and foci are on the x-axis:
a = 3, c = 5
Step 2: Find b using c² = a² + b²:
5² = 3² + b²
25 = 9 + b²
b² = 16
b = 4
Step 3: Write the standard form:
x²/a² - y²/b² = 1
x²/9 - y²/16 = 1
Step 4: Verify:
Vertices: (±a, 0) = (±3, 0) ✓
Foci: (±c, 0) = (±5, 0) where c = √(9 + 16) = √25 = 5 ✓
Final Answer: The equation is x²/9 - y²/16 = 1
Q6: Find the equation of the parabola with focus at (-2, 0) and directrix x = 2.
Given: Focus F(-2, 0) and directrix x = 2
Step 1: Use the definition of parabola:
A parabola is the locus of points equidistant from the focus and directrix.
Let P(x, y) be any point on the parabola.
Step 2: Distance from P to focus F(-2, 0):
√[(x - (-2))² + (y - 0)²] = √[(x + 2)² + y²]
Step 3: Distance from P to directrix x = 2:
|x - 2|
Step 4: Equate the distances:
√[(x + 2)² + y²] = |x - 2|
Step 5: Square both sides:
(x + 2)² + y² = (x - 2)²
x² + 4x + 4 + y² = x² - 4x + 4
4x + y² = -4x
y...
Showing 6 of 8 questions. Visit the full page for complete solutions.
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