68 free MCQs on Conic Sections with worked answers and explanations. Study circles, parabolas, ellipses, and hyperbolas as curves formed by intersecting a plane with a double cone, with their standard equations and key properties.
Take the timed Conic Sections chapterwise test →Below are 68 practice questions on Conic Sections, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Conic Sections notes.
All four conics form a single family distinguished only by eccentricity: a circle is the most "closed" (e=0), an ellipse is an elongated closed curve, a parabola is the borderline open curve (e=1), and a hyperbola has two separate open branches (e>1).
Q1.
A conic section is formed by the intersection of a plane with:
Answer: B. A double-napped right circular cone
Why: Conic sections (circle, parabola, ellipse, hyperbola) are obtained by intersecting a plane with a double-napped right circular cone at different angles.
Q2.
The eccentricity of a circle is:
Answer: A. 0
Why: A circle is a special conic with eccentricity e = 0, since it has no distinct directrix and constant radius.
Q3.
The eccentricity of a parabola is always:
Answer: B. 1
Why: A parabola is defined as the locus of points equidistant from focus and directrix, giving eccentricity exactly equal to 1.
Q4.
For an ellipse, the eccentricity e satisfies:
Answer: C. 0 < e < 1
Why: An ellipse has eccentricity strictly between 0 and 1.
Q5.
For a hyperbola, the eccentricity e satisfies:
Answer: D. e > 1
Why: A hyperbola always has eccentricity greater than 1.
Q6.
The standard equation of a parabola opening to the right with vertex at the origin is:
Answer: B. y<sup>2</sup> = 4ax
Why: y<sup>2</sup> = 4ax is the standard form of a parabola opening rightward with vertex at the origin and axis along the x-axis.
Q7.
For the parabola y<sup>2</sup> = 4ax, the coordinates of the focus are:
Answer: B. (a, 0)
Why: For y<sup>2</sup> = 4ax, the focus lies on the axis at (a, 0).
Q8.
The standard equation of an ellipse with major axis along the x-axis is:
Answer: A. x<sup>2</sup>/a<sup>2</sup> + y<sup>2</sup>/b<sup>2</sup> = 1, a > b
Why: x<sup>2</sup>/a<sup>2</sup> + y<sup>2</sup>/b<sup>2</sup> = 1 with a > b > 0 is the standard ellipse equation with the major axis along the x-axis.
Q9.
The standard equation of a hyperbola with transverse axis along the x-axis is:
Answer: B. x<sup>2</sup>/a<sup>2</sup> - y<sup>2</sup>/b<sup>2</sup> = 1
Why: x<sup>2</sup>/a<sup>2</sup> - y<sup>2</sup>/b<sup>2</sup> = 1 is the standard hyperbola equation with the transverse axis along the x-axis.
Q10.
For an ellipse x<sup>2</sup>/a<sup>2</sup> + y<sup>2</sup>/b<sup>2</sup> = 1 with a > b, the relationship between a, b, and c (distance to focus) is:
Answer: B. c<sup>2</sup> = a<sup>2</sup> - b<sup>2</sup>
Why: For an ellipse, c<sup>2</sup> = a<sup>2</sup> - b<sup>2</sup>, since the foci lie inside the ellipse.
Q11.
For a hyperbola x<sup>2</sup>/a<sup>2</sup> - y<sup>2</sup>/b<sup>2</sup> = 1, the relationship between a, b, and c (distance to focus) is:
Answer: B. c<sup>2</sup> = a<sup>2</sup> + b<sup>2</sup>
Why: For a hyperbola, c<sup>2</sup> = a<sup>2</sup> + b<sup>2</sup>, since the foci lie outside the curve's vertices, making c always greater than a.
Q12.
The length of the latus rectum of the parabola y<sup>2</sup> = 4ax is:
Answer: C. 4a
Why: The latus rectum of y<sup>2</sup> = 4ax has length 4a, the focal chord perpendicular to the axis.
Q13.
The general equation of a circle with center (h, k) and radius r is:
Answer: A. (x-h)<sup>2</sup> + (y-k)<sup>2</sup> = r<sup>2</sup>
Why: The standard circle equation centered at (h,k) with radius r is (x-h)<sup>2</sup> + (y-k)<sup>2</sup> = r<sup>2.</sup>
Q14.
A circle, ellipse, parabola and hyperbola are together known as:
Answer: A. conic sections
Why: These curves arise from slicing a cone at different angles, hence "conic sections".
Q15.
The standard equation of a circle with centre at the origin and radius r is:
Answer: A. x² + y² = r²
Why: Every point at distance r from the origin satisfies x² + y² = r².
Q16.
The fixed point used to define a parabola is called its:
Answer: A. focus
Why: A parabola is the locus of points equidistant from the focus and the directrix.
Q17.
The standard equation of a parabola opening to the right is:
Answer: A. y² = 4ax
Why: y² = 4ax opens rightward with vertex at the origin.
Q18.
An ellipse has how many foci?
Answer: A. 2
Why: An ellipse has two foci; the sum of distances from any point to them is constant.
Q19.
The longest diameter of an ellipse is called the:
Answer: A. major axis
Why: The major axis is the longest chord, passing through both foci.
Q20.
A hyperbola consists of two branches and has how many foci?
Answer: A. 2
Why: A hyperbola has two foci, one associated with each branch.
Q21.
Find the focus of the parabola y<sup>2</sup> = 12x.
Answer: A. (3, 0)
Why: Comparing with y<sup>2</sup> = 4ax gives 4a = 12, so a = 3. The focus is (a, 0) = (3, 0).
Q22.
Find the length of the latus rectum of the parabola y<sup>2</sup> = 12x.
Answer: C. 12
Why: 4a = 12 directly gives the latus rectum length as 12.
Q23.
Find the equation of the directrix of the parabola y<sup>2</sup> = 12x.
Answer: A. x = -3
Why: Since 4a = 12, a = 3. The directrix of y<sup>2</sup> = 4ax is x = -a, so x = -3.
Q24.
For the ellipse x<sup>2</sup>/25 + y<sup>2</sup>/9 = 1, find the eccentricity.
Answer: B. 4/5
Why: a<sup>2</sup>=25, b<sup>2</sup>=9, so c<sup>2</sup> = 25-9 = 16, c=4. Eccentricity e = c/a = 4/5.
Q25.
For the ellipse x<sup>2</sup>/25 + y<sup>2</sup>/9 = 1, find the coordinates of the foci.
Answer: B. (±4, 0)
Why: From a<sup>2</sup>=25, b<sup>2</sup>=9, c<sup>2</sup>=a<sup>2</sup>-b<sup>2</sup>=16, c=4. Foci are at (±c, 0) = (±4, 0).
Q26.
For the ellipse x<sup>2</sup>/25 + y<sup>2</sup>/9 = 1, find the length of the latus rectum.
Answer: B. 18/5
Why: Latus rectum = 2b<sup>2</sup>/a = 2(9)/5 = 18/5.
Q27.
For the hyperbola x<sup>2</sup>/16 - y<sup>2</sup>/9 = 1, find the eccentricity.
Answer: C. 5/4
Why: a<sup>2</sup>=16, b<sup>2</sup>=9, c<sup>2</sup>=a<sup>2</sup>+b<sup>2</sup>=25, c=5. Eccentricity e = c/a = 5/4.
Q28.
For the hyperbola x<sup>2</sup>/16 - y<sup>2</sup>/9 = 1, find the equations of the asymptotes.
Answer: B. y = ±(3/4)x
Why: Asymptotes of x<sup>2</sup>/a<sup>2</sup> - y<sup>2</sup>/b<sup>2</sup> = 1 are y = ±(b/a)x = ±(3/4)x since a=4, b=3.
Q29.
Find the equation of a parabola with vertex at the origin and focus at (5, 0).
Answer: C. y<sup>2</sup> = 20x
Why: Focus (a,0) = (5,0) gives a=5. Equation: y<sup>2</sup> = 4ax = 20x.
Q30.
An ellipse has vertices (±5, 0) and foci (±3, 0). Find its equation.
Answer: B. x<sup>2</sup>/25 + y<sup>2</sup>/16 = 1
Why: a=5, c=3, so b<sup>2</sup> = a<sup>2</sup>-c<sup>2</sup> = 25-9 = 16. Equation: x<sup>2</sup>/25 + y<sup>2</sup>/16 = 1.
Q31.
A hyperbola has vertices (±4, 0) and foci (±6, 0). Find the value of b<sup>2.</sup>
Answer: B. 20
Why: a=4, c=6, so b<sup>2</sup> = c<sup>2</sup> - a<sup>2</sup> = 36 - 16 = 20.
Q32.
Find the directrices of the ellipse x<sup>2</sup>/25 + y<sup>2</sup>/9 = 1.
Answer: A. x = ±25/4
Why: Directrices are x = ±a/e. With a=5 and e=4/5, x = ±5/(4/5) = ±25/4.
Q33.
Identify the conic represented by 9x<sup>2</sup> + 25y<sup>2</sup> = 225 and find its eccentricity.
Answer: A. Ellipse, e = 4/5
Why: Dividing by 225: x<sup>2</sup>/25 + y<sup>2</sup>/9 = 1, an ellipse with a<sup>2</sup>=25, b<sup>2</sup>=9. c<sup>2</sup>=16, c=4, e=c/a=4/5.
Q34.
For the ellipse x²/a² + y²/b² = 1 with a > b, the eccentricity e is:
Answer: A. less than 1
Why: An ellipse always has eccentricity between 0 and 1.
Q35.
For a parabola, the eccentricity e equals:
Answer: A. 1
Why: A parabola has eccentricity exactly 1.
Q36.
For a hyperbola, the eccentricity e is:
Answer: A. greater than 1
Why: A hyperbola always has eccentricity greater than 1.
Q37.
The equation x²/16 + y²/9 = 1 represents an:
Answer: A. ellipse
Why: A sum of two squared terms equal to 1 with unequal denominators is an ellipse.
Q38.
The equation x²/16 − y²/9 = 1 represents a:
Answer: A. hyperbola
Why: A difference of two squared terms equal to 1 is a hyperbola.
Q39.
For the parabola y² = 4ax, the focus is located at:
Answer: A. (a, 0)
Why: The focus of y² = 4ax is the point (a, 0).
Q40.
The length of the latus rectum of the parabola y² = 4ax is:
Answer: A. 4a
Why: The latus rectum of y² = 4ax has length 4a.
Q41.
Find the equation of the parabola with vertex at the origin, axis along the x-axis, and passing through the point (2, 4).
Answer: A. y<sup>2</sup> = 8x
Why: Standard form y²=4ax. Substituting point (2,4): 4²=4a(2) → 16=8a → a=2. So 4a=8. Equation: y²=8x. Verify: (2,4)→16=8(2)=16. ✓ Ans: y²=8x.
Q42.
An ellipse has eccentricity 3/5 and its foci at (±3, 0). Find the equation of the ellipse.
Answer: A. x<sup>2</sup>/25 + y<sup>2</sup>/16 = 1
Why: Foci at (±3,0) means c=3. Eccentricity e=c/a=3/5, so a=c/e=3÷(3/5)=5; a²=25. Then b²=a²−c²=25−9=16. Equation: x²/25+y²/16=1. Ans: x²/25+y²/16=1.
Q43.
Find the length of the latus rectum and the eccentricity of the hyperbola 9x<sup>2</sup> - 16y<sup>2</sup> = 144.
Answer: A. LR = 4.5, e = 5/4
Why: Dividing by 144: x<sup>2</sup>/16 - y<sup>2</sup>/9 = 1, so a<sup>2</sup>=16, b<sup>2</sup>=9, a=4, b=3. LR = 2b<sup>2</sup>/a = 2(9)/4 = 4.5. Also c<sup>2</sup> = a<sup>2</sup>+b<sup>2</sup> = 25, c=5, so e = c/a = 5/4.
Q44.
A point on a parabola y<sup>2</sup> = 8x is at a distance of 6 units from the focus. Find its distance from the directrix.
Answer: B. 6 units
Why: By the defining property of a parabola, the distance from any point on it to the focus equals its distance to the directrix. So the distance to the directrix is also 6 units.
Q45.
Find the equation of an ellipse whose major axis is along the y-axis, with semi-major axis 5 and semi-minor axis 3.
Answer: A. x<sup>2</sup>/9 + y<sup>2</sup>/25 = 1
Why: Major axis along y-axis: standard form x²/b²+y²/a²=1 with a=5 (semi-major) and b=3 (semi-minor). So denominators: b²=9 under x², a²=25 under y². Equation: x²/9+y²/25=1. Ans: x²/9+y²/25=1.
Q46.
Find the distance between the directrices of the hyperbola x<sup>2</sup>/9 - y<sup>2</sup>/16 = 1.
Answer: A. 18/5
Why: a=3, b=4, so c<sup>2</sup>=a<sup>2</sup>+b<sup>2</sup>=25, c=5, e=c/a=5/3. Directrices are at x=±a/e=±3/(5/3)=±9/5. Distance between the two directrices = 2(9/5) = 18/5.
Q47.
The latus rectum of an ellipse is half of its minor axis. Find the eccentricity of the ellipse.
Answer: B. sqrt(3)/2
Why: Latus rectum = 2b<sup>2</sup>/a. Minor axis = 2b. Given 2b<sup>2</sup>/a = (1/2)(2b) = b, so 2b<sup>2</sup>/a = b gives 2b = a, b = a/2. Then e<sup>2</sup> = 1 - b<sup>2</sup>/a<sup>2</sup> = 1 - 1/4 = 3/4, so e = sqrt(3)/2.
Q48.
Find the eccentricity of the hyperbola whose latus rectum is equal to half of its transverse axis.
Answer: B. e = sqrt(6)/2
Why: Latus rectum = 2b<sup>2</sup>/a, transverse axis = 2a. Given 2b<sup>2</sup>/a = (1/2)(2a) = a, so 2b<sup>2</sup> = a<sup>2</sup>, meaning b<sup>2</sup> = a<sup>2</sup>/2. Then e<sup>2</sup> = 1 + b<sup>2</sup>/a<sup>2</sup> = 1 + 1/2 = 3/2, so e = sqrt(3/2) = sqrt(6)/2.
Q49.
A line passes through the focus of the parabola y<sup>2</sup> = 4ax and is perpendicular to its axis. The chord it cuts on the parabola is called the latus rectum. If a = 4, find the endpoints of the latus rectum.
Answer: A. (4, 8) and (4, -8)
Why: With a=4, parabola is y²=16x; focus at (a,0)=(4,0). At x=4: y²=16(4)=64, y=±8. Latus rectum endpoints are (4,8) and (4,−8), and its length=2(2a)=16. Ans: (4,8) and (4,−8).
Q50.
Find the equation of the hyperbola with foci (±5, 0) and transverse axis of length 8.
Answer: A. x<sup>2</sup>/16 - y<sup>2</sup>/9 = 1
Why: Foci (±5,0) means c=5. Transverse axis length=2a=8, so a=4, a²=16. b²=c²−a²=25−16=9. Standard form (foci on x-axis): x²/16−y²/9=1. Ans: x²/16−y²/9=1.
Q51.
An ellipse and a hyperbola have the same foci. If the ellipse has eccentricity 3/5 and the hyperbola has eccentricity 5/3, and the ellipse's semi-major axis is 10, find c (distance from center to focus).
Answer: B. 6
Why: For the ellipse: e=c/a=3/5, a=10. So c=(3/5)×10=6. Both conics share the same foci, so the common focal distance is c=6. Verify: hyperbola e=c/a<sub>h</sub>=5/3 → a<sub>h</sub>=c×3/5=18/5. Ans: c=6.
Q52.
The eccentricity of a conic is found to be exactly 1. Which type of conic must it be, and what defines its shape uniquely?
Answer: C. Parabola, defined by equal distance to focus and directrix
Why: Eccentricity exactly equal to 1 uniquely identifies a parabola, where every point is equidistant from the focus and the directrix.
Q53.
For the ellipse x²/25 + y²/16 = 1, the value of a (semi-major axis) is:
Answer: A. 5
Why: a² = 25, so a = 5.
Q54.
For the ellipse x²/25 + y²/16 = 1, the eccentricity e = √(1 − b²/a²) equals:
Answer: A. 3/5
Why: e = √(1 − 16/25) = √(9/25) = 3/5.
Q55.
The directrix of the parabola y² = 4ax is the line:
Answer: A. x = −a
Why: For y² = 4ax the directrix is the vertical line x = −a.
Q56.
A general second-degree equation represents a circle when the coefficients of x² and y² are equal and the xy term is:
Answer: A. absent (zero)
Why: A circle needs equal x² and y² coefficients and no xy (cross) term.
Q57.
The vertices of the hyperbola x²/a² − y²/b² = 1 are located at:
Answer: A. (±a, 0)
Why: The transverse axis lies along the x-axis, giving vertices (±a, 0).
Q58.
The centre of the circle x² + y² − 6x + 4y − 12 = 0 is:
Answer: A. (3, −2)
Why: With 2g = −6 and 2f = 4, the centre (−g, −f) is (3, −2).
Q59.
The radius of the circle x² + y² = 49 is:
Answer: A. 7
Why: r² = 49, so the radius is 7.
Q60.
For an ellipse, the sum of the distances from any point on it to the two foci equals:
Answer: A. 2a, a constant
Why: This defining property gives a constant sum equal to 2a, the length of the major axis.
Q61.
The eccentricity of the ellipse x²/25 + y²/16 = 1 is:
Answer: A. 3/5
Why: e = √(1 − 16/25) = √(9/25) = 3/5.
Q62.
The length of the latus rectum of the parabola y² = 12x is:
Answer: C. 12
Why: For y² = 4ax the latus rectum is 4a; here 4a = 12.
Q63.
The foci of the hyperbola x²/9 − y²/16 = 1 are:
Answer: A. (±5, 0)
Why: c = √(9 + 16) = 5, so the foci are (±5, 0).
Q64.
The directrix of the parabola x² = 8y is:
Answer: B. y = −2
Why: For x² = 4ay with 4a = 8, a = 2, so the directrix is y = −2.
Q65.
The eccentricity of a rectangular hyperbola (asymptotes y = ±x) is:
Answer: B. √2
Why: For a rectangular hyperbola a = b, so e = √(1 + b²/a²) = √2.
Q66.
The line y = x + c is tangent to the circle x² + y² = 2 when c equals:
Answer: B. ±2
Why: Tangency requires c² = r²(1 + m²) = 2·2 = 4, so c = ±2.
Q67.
The centre of the hyperbola (x − 1)²/4 − (y − 2)²/9 = 1 is:
Answer: A. (1, 2)
Why: The centre is read directly as (1, 2).
Q68.
The sum of the focal distances of any point on the ellipse x²/16 + y²/9 = 1 is:
Answer: B. 8
Why: The constant sum equals 2a = 2·4 = 8.