Question 1

The equation of the parabola with vertex at the origin, axis along the positive x-axis, and passing through the point (2, 4) is:

Accepted Answer

Setup: Standard form of parabola with vertex at origin and axis along x-axis is y² = 4ax, where a > 0 is the focal parameter.

Solution:
Step 1: The standard form is y² = 4ax.
Step 2: The parabola passes through (2, 4), so substitute x = 2 and y = 4:
  (4)² = 4a(2)
  16 = 8a
  a = 2
Step 3: Therefore, the equation is y² = 4(2)x = 8x.

Verification: At (2, 4): (4)² = 16 and 8(2) = 16 ✓

Trap Explanation: Students often confuse the standard form y² = 4ax with y = ax² (which is a parabola opening upward). They may also forget that a is the focal parameter, not the coefficient directly in the equation, leading to y² = 2x or y² = 4x.

Question 2

The foci of the ellipse 9x² + 16y² = 144 are located at:

Accepted Answer

Setup: Standard form of ellipse is x²/a² + y²/b² = 1. For an ellipse, if a > b, foci lie on the x-axis at (±c, 0) where c² = a² - b².

Solution:
Step 1: Rewrite the equation in standard form by dividing by 144:
  9x²/144 + 16y²/144 = 1
  x²/16 + y²/9 = 1
Step 2: Identify a² = 16 and b² = 9, so a = 4 and b = 3.
Step 3: Since a > b, the major axis is along the x-axis.
Step 4: Calculate c using c² = a² - b²:
  c² = 16 - 9 = 7
  c = √7
Step 5: The foci are at (±√7, 0).

Verification: c² = 7 < a² = 16 ✓ and c² = 7 > 0 ✓

Trap Explanation: Students often confuse which axis is major (forgetting to compare a and b), or incorrectly compute c² = a² + b² (which is for hyperbola), or forget to reduce the equation to standard form first.

Question 3

The equation of the tangent to the ellipse x²/25 + y²/9 = 1 at the point (5, 0) is:

Accepted Answer

Setup: For an ellipse x²/a² + y²/b² = 1, the equation of the tangent at a point (x₀, y₀) on the ellipse is (x·x₀)/a² + (y·y₀)/b² = 1.

Solution:
Step 1: Identify a² = 25 and b² = 9.
Step 2: The point of tangency is (x₀, y₀) = (5, 0).
Step 3: Verify that (5, 0) lies on the ellipse:
  (5)²/25 + (0)²/9 = 25/25 + 0 = 1 ✓
Step 4: Apply the tangent formula:
  (x·5)/25 + (y·0)/9 = 1
  5x/25 = 1
  x/5 = 1
  x = 5
Step 5: The equation of the tangent is x = 5.

Alternative (using calculus): Differentiate implicitly:
  2x/25 + 2y·(dy/dx)/9 = 0
  dy/dx = -(9x)/(25y)
  At (5, 0): dy/dx is undefined (vertical tangent), so the tangent is vertical: x = 5.

Trap Explanation: Students often forget the tangent formula for ellipse and try to use the circle formula or differentiate incorrectly. They may also confuse the point (5, 0) with the semi-major axis length and write y = 0 instead of x = 5.

Question 4

The equation of the hyperbola with foci at (±5, 0) and the difference of distances from any point on the hyperbola to the two foci equal to 6 is:

Accepted Answer

Setup: Standard form of hyperbola with foci on x-axis is x²/a² - y²/b² = 1, where 2a is the difference of distances and c² = a² + b² with c being the focal distance.

Solution:
Step 1: Given foci at (±5, 0), so c = 5.
Step 2: The difference of distances is 2a = 6, so a = 3.
Step 3: Use the relationship c² = a² + b²:
  25 = 9 + b²
  b² = 16
  b = 4
Step 4: The equation is x²/9 - y²/16 = 1.

Verification: For any point P on the hyperbola, |PF₁ - PF₂| = 2a = 6 ✓, and c² = 9 + 16 = 25 ✓

Trap Explanation: Students often confuse the definition of hyperbola (difference of distances = 2a) with ellipse (sum of distances = 2a). They may also use c² = a² - b² (ellipse formula) instead of c² = a² + b² (hyperbola formula), or forget that the given difference is 2a, not a.

Question 5

A point P moves such that its distance from the line x = 4 equals its distance from the point F(1, 0). The locus of P is:

Accepted Answer

Setup: By definition of a parabola, the locus of a point equidistant from a fixed point (focus) and a fixed line (directrix) is a parabola.

Solution:
Step 1: Let P(x, y) be any point on the locus.
Step 2: Distance from P to the line x = 4 is |x - 4|.
Step 3: Distance from P to the point F(1, 0) is √[(x - 1)² + y²].
Step 4: By the given condition:
  |x - 4| = √[(x - 1)² + y²]
Step 5: Square both sides:
  (x - 4)² = (x - 1)² + y²
Step 6: Expand:
  x² - 8x + 16 = x² - 2x + 1 + y²
Step 7: Simplify:
  -8x + 16 = -2x + 1 + y²
  -6x + 15 = y²
  y² = -6x + 15
  y² = -6(x - 2.5)
Step 8: This is a parabola with vertex at (2.5, 0), opening leftward.

Alternatively, rewrite as: y² = -6(x - 5/2) or 6x + y² = 15.

Trap Explanation: Students often forget to square the distance equation correctly, or they confuse the focus and directrix, or they fail to recognize the parabola definition.

Question 6

The equation of the circle with center (−2, 3) and radius 5 is:

Accepted Answer

Setup: Standard form of a circle with center (h, k) and radius r is (x − h)² + (y − k)² = r².

Solution:
Step 1: Identify center (h, k) = (−2, 3) and radius r = 5.
Step 2: Substitute into the standard form:
  (x − (−2))² + (y − 3)² = 5²
  (x + 2)² + (y − 3)² = 25
Step 3: Expand to general form (optional):
  x² + 4x + 4 + y² − 6y + 9 = 25
  x² + y² + 4x − 6y + 13 = 25
  x² + y² + 4x − 6y − 12 = 0

Verification: Center from general form: (−4/2, −(−6)/2) = (−2, 3). ✓ Radius: √(4 + 9 + 12) = √25 = 5. ✓

Question 7

The eccentricity of the hyperbola 16x² - 9y² = 144 is:

Accepted Answer

Setup: Standard form of hyperbola is x²/a² - y²/b² = 1. The eccentricity is e = √(1 + b²/a²) = c/a, where c² = a² + b².

Solution:
1. Rewrite the equation in standard form:
   16x² - 9y² = 144
   x²/9 - y²/16 = 1
2. Here a² = 9 and b² = 16, so a = 3 and b = 4.
3. Calculate c: c² = a² + b² = 9 + 16 = 25, so c = 5.
4. Eccentricity: e = c/a = 5/3.

Verification: e > 1 ✓ (required for hyperbola)
Alternative: e = √(1 + b²/a²) = √(1 + 16/9) = √(25/9) = 5/3 ✓

Question 8

The eccentricity of the ellipse 9x² + 16y² = 144 is:

Accepted Answer

Setup: Standard form of ellipse is x²/a² + y²/b² = 1. Eccentricity e = √(1 − b²/a²) when a > b. Solution: Step 1: Rewrite 9x² + 16y² = 144 in standard form by dividing by 144: 9x²/144 + 16y²/144 = 1 x²/16 + y²/9 = 1 Step 2: Identify a² = 16 and b² = 9, so a = 4 and b = 3. Step 3: Since a > b, the major axis is along the x-axis. Step 4: Calculate eccentricity: e = √(1 − b²/a²) = √(1 − 9/16) = √(7/16) = √7/4 Verification: e² = 7/16, so e = √7/4 ≈ 0.66 (valid for ellipse: 0 < e < 1) ✓

Question 9

The length of the latus rectum of the hyperbola x²/25 − y²/9 = 1 is:

Accepted Answer

Setup: Standard form of hyperbola is x²/a² − y²/b² = 1. The latus rectum has length 2b²/a.

Solution:
Step 1: Identify a² = 25 and b² = 9, so a = 5 and b = 3.
Step 2: The latus rectum of a hyperbola x²/a² − y²/b² = 1 has length:
  L = 2b²/a = 2(9)/5 = 18/5

Verification: The latus rectum is a chord through the focus perpendicular to the transverse axis. For this hyperbola, it has length 18/5 = 3.6 units. ✓

Question 10

The equation of the circle with centre (−2, 3) and radius 5 is:

Accepted Answer

Setup: Standard form of circle with centre (h, k) and radius r is (x − h)² + (y − k)² = r².

Solution:
Step 1: Identify centre (h, k) = (−2, 3) and radius r = 5.
Step 2: Substitute into the standard form:
  (x − (−2))² + (y − 3)² = 5²
  (x + 2)² + (y − 3)² = 25
Step 3: Expand to general form (optional):
  x² + 4x + 4 + y² − 6y + 9 = 25
  x² + y² + 4x − 6y + 13 = 25
  x² + y² + 4x − 6y − 12 = 0

Verification: Centre from x² + y² + 4x − 6y − 12 = 0 is (−2, 3) and radius = √(4 + 9 + 12) = √25 = 5 ✓

Question 11

The foci of the ellipse 4x² + 9y² = 36 are located at:

Accepted Answer

Setup: Standard form of ellipse is x²/a² + y²/b² = 1. Foci are at (±c, 0) when a > b, where c² = a² − b². Solution: Step 1: Rewrite 4x² + 9y² = 36 in standard form by dividing by 36: 4x²/36 + 9y²/36 = 1 x²/9 + y²/4 = 1 Step 2: Identify a² = 9 and b² = 4, so a = 3 and b = 2. Step 3: Since a > b, the major axis is along the x-axis, and foci are at (±c, 0). Step 4: Calculate c: c² = a² − b² = 9 − 4 = 5 c = √5 Step 5: Foci are at (±√5, 0), or (√5, 0) and (−√5, 0). Verification: For an ellipse, 0 < c < a, so 0 < √5 ≈ 2.24 < 3 ✓

Question 12

The length of the latus rectum of the parabola y² = 12x is:

Accepted Answer

Setup: For a parabola y² = 4ax, the latus rectum is the chord through the focus perpendicular to the axis. Its length is 4a.

Solution:
1. Compare y² = 12x with the standard form y² = 4ax:
   4a = 12
   a = 3
2. The length of the latus rectum is 4a = 4(3) = 12.

Verification: The focus is at (a, 0) = (3, 0). The latus rectum passes through (3, 0) perpendicular to the x-axis, so it is the vertical line x = 3. Points on the parabola with x = 3: y² = 12(3) = 36 → y = ±6. The chord endpoints are (3, 6) and (3, -6), with length |6 - (-6)| = 12 ✓

Question 13

The eccentricity of the hyperbola x²/25 − y²/9 = 1 is:

Accepted Answer

Setup: Standard form of hyperbola with transverse axis along x-axis is x²/a² − y²/b² = 1. The eccentricity is e = √(1 + b²/a²) = c/a, where c² = a² + b².

Solution:
Step 1: From x²/25 − y²/9 = 1, identify a² = 25 and b² = 9.
Step 2: So a = 5 and b = 3.
Step 3: Calculate c² = a² + b² = 25 + 9 = 34, so c = √34.
Step 4: Eccentricity e = c/a = √34/5.

Alternative method:
Step 1: e² = 1 + b²/a² = 1 + 9/25 = 25/25 + 9/25 = 34/25.
Step 2: e = √(34/25) = √34/5.

Verification: For a hyperbola, e > 1. Since √34 ≈ 5.83, we have e ≈ 1.166 > 1. ✓

Question 14

The equation of the tangent to the circle x² + y² − 4x − 6y + 9 = 0 at the point (2, 1) is:

Accepted Answer

Setup: For a circle in general form x² + y² + 2gx + 2fy + c = 0, the tangent at point (x₀, y₀) on the circle is given by: xx₀ + yy₀ + g(x + x₀) + f(y + y₀) + c = 0.

Solution:
Step 1: Rewrite the circle equation in standard form by completing the square:
  x² − 4x + y² − 6y + 9 = 0
  (x² − 4x + 4) + (y² − 6y + 9) + 9 − 4 − 9 = 0
  (x − 2)² + (y − 3)² = 4
  Centre: (2, 3), Radius: 2
Step 2: Verify that (2, 1) lies on the circle:
  (2 − 2)² + (1 − 3)² = 0 + 4 = 4 ✓
Step 3: Use the tangent formula at (x₀, y₀) = (2, 1):
  (x − 2)(2 − 2) + (y − 3)(1 − 3) = 4
  0 + (y − 3)(−2) = 4
  −2(y − 3) = 4
  −2y + 6 = 4
  −2y = −2
  y = 1

Alternatively, using the general tangent formula with g = −2, f = −3, c = 9:
  2x + 1y − 2(x + 2) − 3(y + 1) + 9 = 0
  2x + y − 2x − 4 − 3y − 3 + 9 = 0
  −2y + 2 = 0
  y = 1

Verification: The tangent at (2, 1) is perpendicular to the radius from (2, 3) to (2, 1). The radius is vertical (undefined slope), so the tangent must be horizontal (y = constant). ✓

Question 15

The equation of the tangent to the parabola y² = 12x at the point (3, 6) is:

Accepted Answer

Setup: For a parabola y² = 4ax, the equation of the tangent at point (x₀, y₀) on the curve is yy₀ = 2a(x + x₀).

Solution:
Step 1: Verify that (3, 6) lies on y² = 12x:
  6² = 36 and 12(3) = 36 ✓
Step 2: From y² = 12x, we have 4a = 12, so a = 3.
Step 3: Apply the tangent formula at (x₀, y₀) = (3, 6):
  yy₀ = 2a(x + x₀)
  y(6) = 2(3)(x + 3)
  6y = 6(x + 3)
  6y = 6x + 18
  y = x + 3
Step 4: Rearrange to standard form: x − y + 3 = 0.

Verification: At (3, 6): 3 − 6 + 3 = 0 ✓

Question 16

The eccentricity of the hyperbola 16x² − 9y² = 144 is:

Accepted Answer

Setup: Standard form of hyperbola is x²/a² − y²/b² = 1. For a hyperbola, eccentricity e = c/a where c² = a² + b².

Solution:
Step 1: Rewrite the equation in standard form:
  16x² − 9y² = 144
  16x²/144 − 9y²/144 = 1
  x²/9 − y²/16 = 1
Step 2: Identify a² = 9 and b² = 16, so a = 3 and b = 4.
Step 3: Calculate c:
  c² = a² + b² = 9 + 16 = 25
  c = 5
Step 4: Calculate eccentricity:
  e = c/a = 5/3

Verification: For a hyperbola, e > 1. Here 5/3 ≈ 1.67 > 1 ✓

Question 17

A circle passes through the points (1, 0), (0, 1), and (−1, 0). The equation of the circle is:

Accepted Answer

Setup: General equation of circle is x² + y² + 2gx + 2fy + c = 0. A circle passing through three non-collinear points is uniquely determined by substituting each point into the general equation.

Solution:
Step 1: Substitute (1, 0) into x² + y² + 2gx + 2fy + c = 0:
  1 + 0 + 2g(1) + 2f(0) + c = 0
  1 + 2g + c = 0 ... (i)
Step 2: Substitute (0, 1):
  0 + 1 + 2g(0) + 2f(1) + c = 0
  1 + 2f + c = 0 ... (ii)
Step 3: Substitute (−1, 0):
  1 + 0 + 2g(−1) + 2f(0) + c = 0
  1 − 2g + c = 0 ... (iii)
Step 4: From (i) and (iii):
  (1 + 2g + c) − (1 − 2g + c) = 0
  4g = 0 → g = 0
Step 5: From (i) with g = 0:
  1 + c = 0 → c = −1
Step 6: From (ii) with c = −1:
  1 + 2f − 1 = 0 → f = 0
Step 7: The equation is x² + y² − 1 = 0, or x² + y² = 1.

Verification: (1, 0): 1 + 0 = 1 ✓; (0, 1): 0 + 1 = 1 ✓; (−1, 0): 1 + 0 = 1 ✓

Question 18

For the hyperbola x²/25 − y²/16 = 1, the length of the latus rectum is:

Accepted Answer

Setup: Standard form of hyperbola with transverse axis along x-axis is x²/a² − y²/b² = 1. The latus rectum is a chord through a focus perpendicular to the transverse axis, with length L = 2b²/a.

Solution:
Step 1: From the equation x²/25 − y²/16 = 1, identify a² = 25 and b² = 16.
Step 2: Therefore, a = 5 and b = 4.
Step 3: The latus rectum length is given by:
  L = 2b²/a = 2(16)/5 = 32/5
Step 4: Convert to decimal if needed: 32/5 = 6.4.

Verification: The formula L = 2b²/a applies to hyperbolas with transverse axis along x-axis. ✓

Question 19

The eccentricity of the ellipse 4x² + 9y² = 36 is:

Accepted Answer

Setup: Standard form of ellipse is x²/a² + y²/b² = 1 (with a > b). Eccentricity is e = √(1 − b²/a²) = c/a, where c² = a² − b². Solution: Step 1: Convert 4x² + 9y² = 36 to standard form by dividing by 36: 4x²/36 + 9y²/36 = 1 x²/9 + y²/4 = 1 Step 2: Identify a² = 9 and b² = 4, so a = 3 and b = 2. Step 3: Since a > b, the major axis is along the x-axis. Step 4: Calculate c using c² = a² − b²: c² = 9 − 4 = 5 c = √5 Step 5: Eccentricity e = c/a = √5/3. Alternative calculation: e = √(1 − b²/a²) = √(1 − 4/9) = √(5/9) = √5/3 Verification: 0 < e < 1 for an ellipse. √5 ≈ 2.236, so e ≈ 0.745. ✓

Question 20

The eccentricity of the hyperbola 16x² - 9y² = 144 is:

Accepted Answer

Setup: Standard form of hyperbola is x²/a² - y²/b² = 1. For a hyperbola, eccentricity e = √(1 + b²/a²) = c/a, where c² = a² + b².

Solution:
Step 1: Rewrite the equation in standard form by dividing by 144:
  16x²/144 - 9y²/144 = 1
  x²/9 - y²/16 = 1
Step 2: Identify a² and b²:
  a² = 9 (so a = 3)
  b² = 16 (so b = 4)
Step 3: Calculate c using the hyperbola relation c² = a² + b²:
  c² = 9 + 16 = 25
  c = 5
Step 4: Calculate eccentricity:
  e = c/a = 5/3

Verification: For a hyperbola, e > 1. Here e = 5/3 ≈ 1.67 > 1 ✓

Question 21

The length of the latus rectum of the ellipse x²/25 + y²/9 = 1 is:

Accepted Answer

Setup: For an ellipse x²/a² + y²/b² = 1 with a > b, the latus rectum is a chord through a focus perpendicular to the major axis. Its length is L = 2b²/a.

Solution:
Step 1: Identify a² and b² from the standard form x²/25 + y²/9 = 1:
  a² = 25 (so a = 5)
  b² = 9 (so b = 3)
Step 2: Verify a > b: 5 > 3 ✓, so the major axis is along the x-axis.
Step 3: Apply the latus rectum formula:
  L = 2b²/a = 2(9)/5 = 18/5

Verification: The latus rectum is always less than the minor axis 2b = 6. Here 18/5 = 3.6 < 6 ✓

Question 22

An ellipse has semi-major axis a = 5 and semi-minor axis b = 3. A point P on the ellipse is such that the sum of its distances from the two foci is S. If another point Q on the ellipse has distances d₁ and d₂ from the two foci respectively, where d₁ = 8, find d₂.

Accepted Answer

By the definition of an ellipse, the sum of distances from any point on the ellipse to the two foci is constant and equals 2a. Here, a = 5, so 2a = 10. For point P, the sum S = 10. For point Q with distances d₁ = 8 and d₂ from the two foci, we apply the same property: d₁ + d₂ = 2a = 10. Therefore, 8 + d₂ = 10, which gives d₂ = 2. This is the fundamental definition of an ellipse and does not depend on the specific location of the point on the ellipse.

Question 23

A parabola has its vertex at the origin and focus at (0, 3). A line passing through the focus intersects the parabola at points P and Q. If the length of the chord PQ is 12, find the sum of the ordinates of P and Q.

Accepted Answer

The parabola with vertex at origin and focus at (0, 3) has the form x² = 4ay where a = 3, so x² = 12y.

For a focal chord of a parabola x² = 4ay, if the chord makes an angle θ with the axis, the length of the focal chord is 4a/sin²θ.

Alternatively, using the focal chord property: if P(x₁, y₁) and Q(x₂, y₂) lie on x² = 12y and pass through focus F(0, 3), then y₁ + y₂ = 2a + 2a·sec²θ for a focal chord, but more directly:

For any focal chord of x² = 4ay, if the endpoints are P and Q, then y₁·y₂ = a² and the focal chord length is y₁ + y₂ + 2a.

Given focal chord length PQ = 12:
y₁ + y₂ + 2a = 12
y₁ + y₂ + 2(3) = 12
y₁ + y₂ = 6

Verification: For focal chord, y₁·y₂ = a² = 9. If y₁ + y₂ = 6 and y₁·y₂ = 9, then y₁ and y₂ are roots of t² - 6t + 9 = 0, giving (t - 3)² = 0, so y₁ = y₂ = 3 (the latus rectum case). Focal chord length = 4a = 12. ✓

Question 24

An ellipse has the equation 9x² + 16y² = 144. A point P on the ellipse is such that the sum of its distances from the two foci is k times the length of the major axis. Find k.

Accepted Answer

Rewrite the ellipse equation in standard form:
9x² + 16y² = 144
x²/16 + y²/9 = 1

Here a² = 16, b² = 9, so a = 4, b = 3.
Major axis is along the x-axis with length 2a = 8.

By the definition of an ellipse, for any point P on the ellipse, the sum of distances from P to the two foci is constant and equals 2a.

Sum of distances from P to foci = 2a = 2(4) = 8.
Length of major axis = 2a = 8.

Therefore: k = (sum of distances)/(length of major axis) = 8/8 = 1.

Question 25

A parabola has its vertex at the origin and its focus at (3, 0). A line passing through the focus intersects the parabola at points P and Q. If the length of the focal chord PQ is 12, find the sum of the ordinates of P and Q.

Accepted Answer

Setup: For a parabola y² = 4ax with vertex at origin and focus at (a, 0), we use the focal chord property.

Given: Focus at (3, 0), so a = 3. The parabola equation is y² = 12x.

For a focal chord of a parabola y² = 4ax, if the chord makes an angle θ with the positive x-axis, the length of the focal chord is 4a/sin²θ.

Alternatively, using the focal chord property: If P(x₁, y₁) and Q(x₂, y₂) lie on y² = 4ax and PQ passes through focus (a, 0), then:
- x₁ · x₂ = a² (focal chord property)
- y₁ · y₂ = -4a² (focal chord property)

Given: Length PQ = 12
Using focal chord length formula: PQ = x₁ + x₂ + 2a = 12
So x₁ + x₂ = 12 - 6 = 6

From y₁² = 12x₁ and y₂² = 12x₂:
y₁² + y₂² = 12(x₁ + x₂) = 12(6) = 72

Now, (y₁ + y₂)² = y₁² + y₂² + 2y₁y₂ = 72 + 2(-36) = 72 - 72 = 0

Therefore, y₁ + y₂ = 0

NDA Conic Sections

NDA Conic Sections — Past Exam Questions

Conic Sections— Rules & Concept

Key Points to Remember

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Conic Sections — Practice Questions

60-Second Revision — Conic Sections