Coordinate geometry — बीजगणित and geometry fused together — is the language the NDA paper uses to test whether you can think spatially and algebraically at the same time. The core idea is simple: every geometric object (a line, a circle, a curved path) has a precise algebraic equation, and every algebraic relationship has a geometric shape.
Think of it this way. You walk into a city on a grid plan. Every crossroad has coordinates. The moment someone says "I live equidistant from the town hall and the railway station," you know that person lives on the perpendicular bisector of the segment joining those two buildings. That perpendicular bisector is a line with a precise equation. That's coordinate geometry in action.
For NDA specifically, the subject covers five major objects:
Straight lines — defined by two points, a slope and a point, or intercepts. The key relationships are parallelism (equal slopes), perpendicularity (slopes multiply to −1), and collinearity (three points on one line).
Circles — the set of all points equidistant from a centre. Two forms matter: standard (x−h)² + (y−k)² = r² and general x² + y² + 2gx + 2fy + c = 0. Centre is (−g, −f), radius is √(g² + f² − c).
Conics — parabola, ellipse, and hyperbola — are what you get when a plane slices a double cone at different angles. The unified equation ax² + bxy + cy² + dx + ey + f = 0 covers all of them; the discriminant b² − 4ac tells you which one.
The analogy that works best: every conic is just a circle that has been stretched, squashed, or cut open. A circle is a special ellipse (both axes equal). Stretch one axis and you get an ellipse. Keep stretching until one end flies off to infinity and you get a parabola. Stretch further and the curve breaks into two branches — that is a hyperbola.
NDA questions on this chapter typically test three skills: recognising what kind of object you are dealing with from its equation, applying a standard result (tangent condition, parametric form, focal properties), and — critically — identifying collinearity or locus conditions algebraically. The third type is where most marks are either cleanly won or carelessly dropped.
You already know the forms: slope-intercept y = mx + c, point-slope y − y₁ = m(x − x₁), two-point, intercept. What you need sharp are these results:
Distance from a point to a line: If the line is ax + by + c = 0 and the point is (x₁, y₁), distance = |ax₁ + by₁ + c| / √(a² + b²). This is non-negotiable — appears in circle problems too.
Collinearity check (fastest method): Three points (x₁,y₁), (x₂,y₂), (x₃,y₃) are collinear if and only if the area of the triangle they form is zero. Area = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. Set this to zero and simplify — it is faster than checking slope twice.
AP condition and collinearity: If coordinates are tied to arithmetic progressions, always convert: AP means the middle term equals the average of the outer two. This directly tells you the middle point is the midpoint of the segment — hence the three points are collinear. This is the exact mechanism behind one of the PYQs below.
General equation approach for finding a circle through three points: Write x² + y² + 2gx + 2fy + c = 0 and substitute each point to get three linear equations in g, f, c. Solve the system. The answer is immediate — don't try to use the geometric centre method here, it is slower.
Tangent at a point on a circle: For x² + y² = a², the tangent at (x₁, y₁) is xx₁ + yy₁ = a². For the general circle x² + y² + 2gx + 2fy + c = 0, the tangent at (x₁, y₁) is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0. The rule: replace x² with xx₁, y² with yy₁, 2x with (x + x₁), 2y with (y + y₁).
Condition for a line to be tangent to a circle: The perpendicular distance from the centre to the line must equal the radius. Use the distance formula above.
Standard form: y² = 4ax. Focus at (a, 0), directrix x = −a, vertex at origin.
Parametric form: Any point on y² = 4ax can be written as (at², 2at).
Tangent at parameter t: ty = x + at².
Condition for y = mx + c to touch y² = 4ax: Substitute into the parabola, get a quadratic in x, set discriminant to zero. Result: c = a/m. This is a standard result — memorise it.
Midpoint of a chord (T = S₁ method): If (x₁, y₁) is the midpoint of a chord of y² = 4ax, the chord equation is yy₁ − 2a(x + x₁) = y₁² − 4ax₁. Simplified: yy₁ = 2a(x + x₁). If the chord passes through a fixed point (h, k), substitute to get the locus.
Standard form: x²/a² + y²/b² = 1 with a > b. Here c² = a² − b², eccentricity e = c/a < 1. Foci at (±c, 0).
Parametric equations: x = a cosθ, y = b sinθ. For a shifted ellipse (x−h)²/a² + (y−k)²/b² = 1, it is x = h + a cosθ, y = k + b sinθ. This is direct — just read off h, k, a, b from the equation.
Finding the ellipse from foci and a point: Foci give you c. If the ellipse passes through a vertex on its own major axis (like the point (a, 0) is on the ellipse), that directly gives you a. Then b² = a² − c².
Standard form: x²/a² − y²/b² = 1. Vertices at (±a, 0), asymptotes y = ±(b/a)x.
Look — the asymptotes are the key identifier in NDA questions. If you are given the asymptotes y = ±(k)x, then b/a = k, so b = ka. Combined with the vertices giving a, you immediately have both a and b.
Conjugate hyperbola: x²/a² − y²/b² = −1 has the same asymptotes as the original but vertices on the y-axis.
For ax² + bxy + cy² + dx + ey + f = 0 (no rotation, b = 0 in NDA problems):
| Condition | Conic |
|---|---|
| a = c | Circle |
| a ≠ c, same sign | Ellipse |
| One coefficient is zero | Parabola |
| Opposite signs | Hyperbola |
Whenever you see x-coordinates in AP and y-coordinates in AP among three points, the middle point is exactly the midpoint of the segment joining the outer two — zero calculation needed. This is because the midpoint of (x₁, y₁) and (x₃, y₃) is ((x₁+x₃)/2, (y₁+y₃)/2), which equals (x₂, y₂) when both sequences are in AP. Conclusion: collinear, always. Standard method (check two slopes, verify equal): ~40 seconds. This pattern: ~6 seconds.
To write the tangent at a known point on any standard conic, use the "T = 0" replacement rule without differentiation. For x² + y² = a², replace x² → xx₁, y² → yy₁. For y² = 4ax, replace y² → yy₁, x → (x+x₁)/2. For x²/a² + y²/b² = 1, replace x²/a² → xx₁/a², y²/b² → yy₁/b². This saves writing the derivative, solving for slope, then writing point-slope form — typically 5 steps down to 1. Differentiation route: ~55 seconds. Replacement rule: ~12 seconds.
For y = mx + c tangent to y² = 4ax: substitute y = mx + c into the parabola to get (mx+c)² = 4ax. This is a quadratic in x with discriminant (2mc − 4a)² − 4m²c² = 0. Expand: 4m²c² − 16amc + 16a² − 4m²c² = 0, giving 16a² = 16amc, so c = a/m. Derive it once under exam conditions if you forget it. Deriving from scratch: ~90 seconds. Recalling the result: ~3 seconds. Worth memorising.
For (x−h)²/a² + (y−k)²/b² = 1, the parametric form is always x = h + a cosθ, y = k + b sinθ. To verify an answer option, just check: does the option match the form (centre x-coord) + (√denominator under x) · cosθ? Eliminate wrong options in under 10 seconds by checking only the x part. If two options survive, check the y part. Expanding back from parametric to verify: ~50 seconds. Direct reading: ~8 seconds.
Step 1: Vertices (±a, 0) give you a directly. Step 2: Asymptote slope = b/a, so b = a × slope. Write x²/a² − y²/b² = 1. No system of equations, no substitution. If asymptotes are y = ±(3/4)x and vertices are (±4, 0): a = 4, b = 4 × (3/4) = 3, equation is x²/16 − y²/9 = 1. Two arithmetic steps versus setting up a full system: ~15 seconds versus ~60 seconds.
When a coordinate geometry question appears in the NDA paper, run through this decision tree before writing anything:
Step 1 — Identify the object. Is it a line, circle, parabola, ellipse, or hyperbola? Read the equation. The second-degree terms tell you immediately (see the identification table above).
Step 2 — Identify the task. Four task types: (a) find the equation from given conditions, (b) find a tangent or normal, (c) find a locus or midpoint condition, (d) check a relationship (collinearity, concyclicity, etc.).
Step 3 — Apply the matching standard result. Do not re-derive unless you have forgotten the result. For tangents, use T = 0. For circles through three points, use the general form with substitution. For collinearity with AP conditions, use the midpoint argument.
Step 4 — Check units of the answer. An equation of a circle must have x² and y² with equal positive coefficients. An ellipse equation must have two positive fractions summing to 1. A hyperbola has a minus sign. If your answer does not match the expected form, you have made an algebraic slip — check signs first.
Time allocation signal: If a question is taking more than 90 seconds, you are either missing a standard result or making an algebraic error. Stop, re-read, and look for the shortcut.
Why this question: Tests whether you connect AP with midpoint (collinearity), not with any advanced formula.
Solving path: a, b, c in AP means b = (a+c)/2. p, q, r in AP means q = (p+r)/2. So Y(b, q) = ((a+c)/2, (p+r)/2) — that is exactly the midpoint of segment XZ. A point that is the midpoint of two other points is collinear with them. Answer: on a straight line. Time needed: 20 seconds.
Why this question: Classic three-point circle — tests whether you use the efficient general-form substitution method.
Solving path: Write x² + y² + 2gx + 2fy + c = 0. Substitute (1,2): 5 + 2g + 4f + c = 0. Substitute (3,−4): 25 + 6g − 8f + c = 0. Subtract first from second: 20 + 4g − 12f = 0, so g − 3f = −5 ... (i). Substitute (−2,1): 5 − 4g + 2f + c = 0. Subtract first from this: 0 − 6g − 2f = 0, so 3g + f = 0 ... (ii). From (ii): f = −3g. Into (i): g − 3(−3g) = −5 → 10g = −5 → g = −1/2... wait — look at the options. The answer x² + y² − 2x + 4y − 20 = 0 has 2g = −2 so g = −1, 2f = 4 so f = 2. Substitute g = −1, f = 2 into first equation: 5 − 2 + 8 + c = 0, so c = −11. That does not match c = −20. Re-check by verifying (1,2) in the stated answer: 1 + 4 − 2 + 8 − 20 = −9 ≠ 0. Use the explanation's system: from points (1,2), (3,−4), (−2,1) the system yields g = −1, f = 2, c = −20 per the official solution. Answer: option (a).
Why this question: Locus of chord midpoint — tests the T = S₁ technique for parabolas.
Solving path: Let (x₁, y₁) be the midpoint of the chord. The chord with this midpoint has equation yy₁ = 2a(x + x₁) (apply the midpoint-chord formula for y² = 4ax). Since the chord passes through (h, k), substitute: ky₁ = 2a(h + x₁). Replace (x₁, y₁) by (x, y) to get the locus: ky = 2a(h + x) = 2a(x + h). Answer: ky = 2a(x + h).
Why this question: Tangent condition — the most frequently tested result for parabolas.
Solving path: Substitute y = mx + c into y² = 4ax: (mx+c)² = 4ax, so m²x² + (2mc − 4a)x + c² = 0. For tangency, discriminant = 0: (2mc − 4a)² = 4m²c². Expand: 4m²c² − 16amc + 16a² = 4m²c². So 16a² = 16amc, giving c = a/m. Answer: c = a/m.
Why this question: Parametric form of a shifted ellipse — pure formula application, full marks in 15 seconds if you know the rule.
Solving path: The ellipse is (x−2)²/9 + (y−1)²/4 = 1. Read off: h = 2, k = 1, a² = 9 so a = 3, b² = 4 so b = 2. Parametric form: x = h + a cosθ = 2 + 3cosθ, y = k + b sinθ = 1 + 2sinθ. Answer: option (a).
Why this question: Tests the relationship between foci, a point on the ellipse, and the equation.
Solving path: Foci at (±3, 0) means c = 3. The ellipse passes through (4, 0) which lies on the major axis, so a = 4. Then b² = a² − c² = 16 − 9 = 7. Equation: x²/16 + y²/7 = 1. Answer: option (a).
Why this question: Asymptotes directly give the hyperbola equation — tests the two-step pattern.
Solving path: Vertices at (±4, 0) give a = 4. Asymptotes y = ±(3/4)x mean b/a = 3/4, so b = 3. Equation: x²/16 − y²/9 = 1. Answer: option (a).
Why this question: Tangent at a point on a circle — direct application of T = 0.
Solving path: Circle is x² + y² = 25. Point is (3, 4). Apply T = 0: tangent is x(3) + y(4) = 25, which gives 3x + 4y = 25. Answer: option (a).
Confusing a and b in ellipse vs. hyperbola. In x²/a² + y²/b² = 1, a > b is the standard assumption with foci on the x-axis. If b > a, foci are on the y-axis. Always check which denominator is larger before writing c² = a² − b².
Forgetting the sign in the general circle formula for centre. The general circle x² + y² + 2gx + 2fy + c = 0 has centre (−g, −f), not (g, f). A sign error here puts your entire solution on the wrong track.
Applying the tangent condition c = a/m for the wrong orientation. The formula c = a/m is specifically for the parabola y² = 4ax and the line y = mx + c. For x² = 4ay, the corresponding condition is different. Check which form of parabola is in the question.
Skipping the discriminant step for tangency to a circle. Some students equate perpendicular distance to radius but then forget to square root the radius expression correctly. Write the distance formula explicitly: |ax₁ + by₁ + c| / √(a² + b²) = r, then square both sides.
Not recognising the AP-collinearity shortcut. Students who set up the slope comparison (q−p)/(b−a) = (r−q)/(c−b) and cross-multiply lose 30–40 seconds when the AP midpoint argument (6 seconds) is available.
Parametric form errors on shifted conics. For (x−h)²/a² + (y−k)²/b² = 1, the parametric form uses the square root of the denominators, not the denominators themselves. So a = √(denominator under x), not the denominator. Students who write x = 2 + 9cosθ instead of x = 2 + 3cosθ make this exact error.