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.
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…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 11 Mathematics — Conic Sections. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Conic Sections — All 8 Questions Solved
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 (±3, 0), Eccentricity e = 3/5
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
y² = -8x
Step 6: Verify:
Vertex (0, 0) ✓
Focus at (-2, 0) for y² = -8x where 4a = 8, a = 2, so focus = (-2, 0) ✓
Directrix x = 2 ✓
Final Answer: The equation is y² = -8x
Q7. An ellipse has eccentricity 1/2 and the distance between its foci is 4. Find the equation of the ellipse.
Given: Eccentricity e = 1/2, distance between foci = 4
Step 1: Find c (semi-focal distance):
Distance between foci = 2c = 4
c = 2
Step 2: Find a using eccentricity:
e = c/a
1/2 = 2/a
a = 4
Step 3: Find b using b² = a² - c²:
b² = 4² - 2²
b² = 16 - 4
b² = 12
Step 4: Assume major axis along x-axis:
Ellipse equation: x²/a² + y²/b² = 1
x²/16 + y²/12 = 1
Step 5: Verify:
Eccentricity e = c/a = 2/4 = 1/2 ✓
Foci distance 2c = 4 ✓
Final Answer: The equation is x²/16 + y²/12 = 1
Q8. Find the length of the latus rectum of the parabola y² = 12x.
Given parabola: y² = 12x
Step 1: Compare with standard form y² = 4ax:
y² = 4ax
12x = 4ax
4a = 12
a = 3
Step 2: The latus rectum is a chord through the focus perpendicular to the axis of the parabola.
Step 3: Length of latus rectum:
Length = 4a = 4(3) = 12
Step 4: Verification:
Focus of y² = 12x is at (a, 0) = (3, 0)
At x = 3: y² = 12(3) = 36, so y = ±6
Latus rectum passes through (3, 6) and (3, -6)
Length = 6 - (-6) = 12 ✓
Final Answer: Length of latus rectum = 12
Conic Sections — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. The equation of an ellipse whose center is (1, 2), major axis is 10, and minor axis is 8, with major axis parallel to the x-axis, is:
- A. $\frac{(x-1)^2}{25} + \frac{(y-2)^2}{16} = 1$
- B. $\frac{(x-1)^2}{16} + \frac{(y-2)^2}{25} = 1$
- C. $\frac{(x-1)^2}{25} + \frac{(y-2)^2}{9} = 1$
- D. $\frac{(x-1)^2}{100} + \frac{(y-2)^2}{64} = 1$
Answer: A. $\frac{(x-1)^2}{25} + \frac{(y-2)^2}{16} = 1$ — The center is $(h,k) = (1,2)$. Major axis length is $2a = 10 \Rightarrow a=5$. Minor axis length is $2b = 8 \Rightarrow b=4$. Since the major axis is parallel to the x-axis, the equation is $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$. Substituting the values, we get $\frac{(x-1)^2}{5^2} + \frac{(y-2)^2}{4^2} = 1 \Rightarrow \frac{(x-1)^2}{25} + \frac{(y-2)^2}{16} = 1$.
MCQ 2. The focus of the parabola $y^2 = -8x$ is:
- A. (2, 0)
- B. (-2, 0)
- C. (0, 2)
- D. (0, -2)
Answer: B. (-2, 0) — The given equation $y^2 = -8x$ is of the form $y^2 = -4ax$. Comparing, $-4a = -8 \Rightarrow a = 2$. For this form, the focus is $(-a, 0)$. Thus, the focus is $(-2, 0)$.
MCQ 3. The eccentricity of an ellipse whose latus rectum is half of its major axis is:
- A. $\frac{1}{\sqrt{2}}$
- B. $\frac{1}{2}$
- C. $\frac{\sqrt{3}}{2}$
- D. $\frac{1}{\sqrt{3}}$
Answer: A. $\frac{1}{\sqrt{2}}$ — For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the length of the latus rectum is $\frac{2b^2}{a}$ and the length of the major axis is $2a$. Given $\frac{2b^2}{a} = \frac{1}{2}(2a) \Rightarrow 2b^2 = a^2$. Using $b^2 = a^2(1 - e^2)$, we get $2a^2(1 - e^2) = a^2 \Rightarrow 2(1 - e^2) = 1 \Rightarrow e^2 = \frac{1}{2} \Rightarrow e = \frac{1}{\sqrt{2}}$.
MCQ 4. For the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$, the length of the transverse axis is:
- A. 3
- B. 4
- C. 6
- D. 8
Answer: D. 8 — The equation is of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Here, $a^2 = 16 \Rightarrow a = 4$. The length of the transverse axis is $2a$. So, $2a = 2(4) = 8$.
MCQ 5. The equation of the circle passing through the origin (0,0) and having intercepts 3 and 4 on the x and y axes respectively is:
- A. $x^2 + y^2 - 3x - 4y = 0$
- B. $x^2 + y^2 + 3x + 4y = 0$
- C. $x^2 + y^2 - 6x - 8y = 0$
- D. $x^2 + y^2 + 6x + 8y = 0$
Answer: A. $x^2 + y^2 - 3x - 4y = 0$ — A circle passing through the origin and having intercepts 'a' and 'b' on the x and y axes respectively has the equation $x^2 + y^2 - ax - by = 0$. Here, $a=3$ and $b=4$. So, the equation is $x^2 + y^2 - 3x - 4y = 0$.
MCQ 6. A conic section is formed when a plane intersects a double-napped cone. What type of conic section is formed when the plane is parallel to a generator of the cone?
- A. Circle
- B. Ellipse
- C. Parabola
- D. Hyperbola
Answer: C. Parabola — When the cutting plane is parallel to a generator of the cone, it intersects only one nappe of the cone and forms a parabola.
MCQ 7. The directrix of the parabola $x^2 = 12y$ is:
- A. $y = -3$
- B. $y = 3$
- C. $x = -3$
- D. $x = 3$
Answer: A. $y = -3$ — The given equation $x^2 = 12y$ is of the form $x^2 = 4ay$. Comparing, $4a = 12 \Rightarrow a = 3$. For this form, the directrix is $y = -a$. Thus, the directrix is $y = -3$.
MCQ 8. If the distance between the foci of an ellipse is 8 and the distance between the directrices is 18, then the equation of the ellipse (centered at origin) is:
- A. $\frac{x^2}{36} + \frac{y^2}{20} = 1$
- B. $\frac{x^2}{20} + \frac{y^2}{36} = 1$
- C. $\frac{x^2}{36} + \frac{y^2}{28} = 1$
- D. $\frac{x^2}{28} + \frac{y^2}{36} = 1$
Answer: A. $\frac{x^2}{36} + \frac{y^2}{20} = 1$ — The distance between foci is $2ae = 8 \Rightarrow ae = 4$. The distance between directrices is $\frac{2a}{e} = 18 \Rightarrow \frac{a}{e} = 9$. Multiplying these gives $a^2 = 36 \Rightarrow a = 6$. Then $e = \frac{4}{a} = \frac{4}{6} = \frac{2}{3}$. For an ellipse, $b^2 = a^2(1 - e^2) = 36(1 - (\frac{2}{3})^2) = 36(\frac{5}{9}) = 20$. The equation is $\frac{x^2}{36} + \frac{y^2}{20} = 1$.
MCQ 9. The foci of the hyperbola $9x^2 - 16y^2 = 144$ are:
- A. $(\pm 5, 0)$
- B. $(0, \pm 5)$
- C. $(\pm 4, 0)$
- D. $(0, \pm 4)$
Answer: A. $(\pm 5, 0)$ — Divide the equation by 144 to get $\frac{x^2}{16} - \frac{y^2}{9} = 1$. Here $a^2 = 16, b^2 = 9$. For a hyperbola, $c^2 = a^2 + b^2 = 16 + 9 = 25 \Rightarrow c = 5$. The foci are $(\pm c, 0)$, so they are $(\pm 5, 0)$.
MCQ 10. The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. This equation represents a point if:
- A. $g^2 + f^2 - c > 0$
- B. $g^2 + f^2 - c < 0$
- C. $g^2 + f^2 - c = 0$
- D. $c = 0$
Answer: C. $g^2 + f^2 - c = 0$ — The radius of a circle is given by $r = \sqrt{g^2 + f^2 - c}$. For the circle to be a point, its radius must be zero, which means $g^2 + f^2 - c = 0$.
MCQ 11. The length of the latus rectum of the parabola $x^2 = -16y$ is:
- A. 4
- B. 8
- C. 12
- D. 16
Answer: D. 16 — The equation $x^2 = -16y$ is of the form $x^2 = -4ay$. Comparing, $-4a = -16 \Rightarrow a = 4$. The length of the latus rectum for any standard parabola is $4a$. So, the length is $4(4) = 16$.
MCQ 12. The equation of the circle whose center is (3, -1) and which passes through the point (2, 4) is:
- A. $x^2 + y^2 - 6x + 2y - 15 = 0$
- B. $x^2 + y^2 - 6x + 2y - 16 = 0$
- C. $x^2 + y^2 + 6x - 2y - 15 = 0$
- D. $x^2 + y^2 + 6x - 2y - 16 = 0$
Answer: B. $x^2 + y^2 - 6x + 2y - 16 = 0$ — The center is $(h, k) = (3, -1)$. The radius squared $r^2$ is the distance squared between $(3, -1)$ and $(2, 4)$, so $r^2 = (2-3)^2 + (4-(-1))^2 = (-1)^2 + 5^2 = 1 + 25 = 26$. The equation is $(x-3)^2 + (y+1)^2 = 26$, which simplifies to $x^2 - 6x + 9 + y^2 + 2y + 1 = 26$, or $x^2 + y^2 - 6x + 2y - 16 = 0$.
MCQ 13. The length of the latus rectum of the parabola $y^2 = 4ax$ is:
- A. $a$
- B. $2a$
- C. $4a$
- D. $8a$
Answer: C. $4a$ — The latus rectum is a chord passing through the focus perpendicular to the axis of the parabola. For $y^2=4ax$, the focus is $(a,0)$. Substituting $x=a$ into the equation gives $y^2 = 4a(a) = 4a^2 \Rightarrow y = \pm 2a$. The endpoints of the latus rectum are $(a, 2a)$ and $(a, -2a)$. The distance between these points is $2a - (-2a) = 4a$.
MCQ 14. The length of the minor axis of the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$ is:
- A. 4
- B. 5
- C. 8
- D. 10
Answer: C. 8 — The equation is of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Here $a^2 = 25 \Rightarrow a = 5$ and $b^2 = 16 \Rightarrow b = 4$. Since $a > b$, the major axis is along the x-axis and the minor axis is along the y-axis. The length of the minor axis is $2b = 2(4) = 8$.
MCQ 15. Find the equation of the hyperbola whose foci are $(\pm 4, 0)$ and eccentricity is 2.
- A. $\frac{x^2}{4} - \frac{y^2}{12} = 1$
- B. $\frac{x^2}{12} - \frac{y^2}{4} = 1$
- C. $\frac{x^2}{4} - \frac{y^2}{8} = 1$
- D. $\frac{x^2}{8} - \frac{y^2}{4} = 1$
Answer: A. $\frac{x^2}{4} - \frac{y^2}{12} = 1$ — Foci are $(\pm 4, 0)$, so $c = 4$. Eccentricity $e = 2$. We know $e = \frac{c}{a}$, so $2 = \frac{4}{a} \Rightarrow a = 2$. For a hyperbola, $b^2 = a^2(e^2 - 1) = 2^2(2^2 - 1) = 4(3) = 12$. The equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, which is $\frac{x^2}{4} - \frac{y^2}{12} = 1$.
MCQ 16. The vertex of the parabola $y^2 = 16x$ is:
- A. (4, 0)
- B. (-4, 0)
- C. (0, 0)
- D. (0, 4)
Answer: C. (0, 0) — All standard parabolas of the form $y^2 = 4ax$, $y^2 = -4ax$, $x^2 = 4ay$, $x^2 = -4ay$ have their vertex at the origin (0, 0).
MCQ 17. The number of common tangents to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 6x - 8y + 16 = 0$ is:
- A. 1
- B. 2
- C. 3
- D. 4
Answer: C. 3 — For $C_1: x^2 + y^2 = 4$, center $C_1 = (0, 0)$, radius $r_1 = 2$. For $C_2: x^2 + y^2 - 6x - 8y + 16 = 0$, center $C_2 = (3, 4)$, radius $r_2 = \sqrt{3^2 + 4^2 - 16} = 3$. The distance between centers $d = \sqrt{(3-0)^2 + (4-0)^2} = 5$. Since $d = r_1 + r_2 = 2 + 3 = 5$, the circles touch externally, resulting in 3 common tangents.
MCQ 18. The equation of the parabola with vertex at origin, axis along the x-axis and passing through the point (2, 3) is:
- A. $y^2 = \frac{9}{2}x$
- B. $y^2 = \frac{2}{9}x$
- C. $x^2 = \frac{9}{2}y$
- D. $x^2 = \frac{2}{9}y$
Answer: A. $y^2 = \frac{9}{2}x$ — Since the vertex is at the origin and the axis is along the x-axis, the equation is $y^2 = 4ax$. Substituting the point (2, 3) into the equation: $3^2 = 4a(2) \Rightarrow 9 = 8a \Rightarrow a = \frac{9}{8}$. So the equation of the parabola is $y^2 = 4(\frac{9}{8})x \Rightarrow y^2 = \frac{9}{2}x$.
MCQ 19. The value of 'a' for the hyperbola $\frac{y^2}{25} - \frac{x^2}{16} = 1$ is:
- A. 4
- B. 5
- C. 16
- D. 25
Answer: B. 5 — The equation is of the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$. Here, $a^2 = 25$, so $a = 5$.
MCQ 20. What conic section is represented by the equation $x^2 - y^2 = 0$?
- A. Circle
- B. Parabola
- C. Pair of straight lines
- D. Point
Answer: C. Pair of straight lines — The equation $x^2 - y^2 = 0$ can be factored as $(x-y)(x+y) = 0$. This implies either $x-y = 0$ or $x+y = 0$, which are two distinct straight lines passing through the origin. Hence, it represents a pair of straight lines.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.