Question 1

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 2

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 3

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 4

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 5

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.

NDA Coordinate Geometry Questions

NDA Coordinate Geometry — Past Exam Questions

Study by Subtopic — Coordinate Geometry

Study Next — Mathematics (NDA Level) Topics