Master coordinate geometry concepts tested in NDA Paper 1
Apply formulas for straight lines, circles, and conic sections
Solve NDA-level problems with speed and accuracy using shortcut techniques
1. Straight Lines
Slope & Equation Forms
The slope (or gradient) of a line measures its steepness. If a line passes through A(x₁, y₁) and B(x₂, y₂), the slope m = tan θ = (y₂ − y₁)/(x₂ − x₁), where θ is the inclination (0 ≤ θ < π). A vertical line has undefined slope; a horizontal line has slope 0.
NDA Shortcut — Slope from Two Points
In NDA, always write slope as difference in y over difference in x. Subtract in the same order for numerator and denominator. If the line is vertical, slope is undefined. For horizontal lines, slope = 0. NDA often tests this distinction.
Different Forms of Straight Lines
Form
Equation
Parameters
Slope‑Intercept
y = mx + c
m = slope, c = y‑intercept
Point‑Slope
y − y₁ = m(x − x₁)
(x₁, y₁) = known point
Two‑Point
(y−y₁)/(y₂−y₁) = (x−x₁)/(x₂−x₁)
(x₁,y₁), (x₂,y₂) = two points
Intercept
x/a + y/b = 1
a = x‑intercept, b = y‑intercept
Normal / Perpendicular
x cos α + y sin α = p
p = perpendicular from origin, α = normal angle
Parametric
x = x₁ + r cos θ, y = y₁ + r sin θ
r = distance, θ = direction angle
Example 1 — Equation of a Line (NDA 2020 Pattern)
Find the equation of the line passing through (2, −3) and (−1, 5). Express in slope‑intercept form and find its intercepts on the axes.
Solution: Slope m = (5 + 3)/(−1 − 2) = 8/(−3) = −8/3. Using point‑slope with (2,−3): y + 3 = (−8/3)(x − 2) → y = (−8/3)x + 16/3 − 3 → y = (−8/3)x + 7/3. Slope‑intercept form: y = (−8/3)x + 7/3. Intercepts: set x = 0 → y = 7/3; set y = 0 → x = 7/8. So x‑intercept = 7/8, y‑intercept = 7/3.
NDA Exam Tip — Intercept Form
When a line cuts both axes, use x/a + y/b = 1. For a line with x‑intercept a and y‑intercept b, the equation is x/a + y/b = 1. NDA questions often give intercepts directly. Example: line with intercepts 3 and 4 → x/3 + y/4 = 1 → 4x + 3y = 12.
Angle Between Lines
If two lines have slopes m₁ and m₂, the acute angle θ between them is:
tan θ = |(m₁ − m₂) / (1 + m₁·m₂)|
Two lines are parallel if m₁ = m₂, and perpendicular if m₁·m₂ = −1. For lines in general form a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, parallelism requires a₁/a₂ = b₁/b₂, perpendicularity requires a₁a₂ + b₁b₂ = 0.
Condition
Slope Form
General Form
Parallel
m₁ = m₂
a₁/a₂ = b₁/b₂ (provided a₂,b₂ ≠ 0)
Perpendicular
m₁·m₂ = −1
a₁a₂ + b₁b₂ = 0
Acute Angle
tan θ = |(m₁−m₂)/(1+m₁m₂)|
cos θ = |a₁a₂+b₁b₂|/√((a₁²+b₁²)(a₂²+b₂²))
Example 2 — Angle Between Lines (NDA 2021)
Find the acute angle between the lines 3x + 4y = 10 and 4x − 3y = 5. Determine if they are perpendicular.
Solution: Slope m₁ = −3/4, slope m₂ = 4/3. Check m₁·m₂ = (−3/4)(4/3) = −1. Since product is −1, the lines are perpendicular. The acute angle is 90°. Alternatively, using a₁a₂ + b₁b₂ = 0: 3·4 + 4·(−3) = 12 − 12 = 0, confirming perpendicularity.
NDA Shortcut — Perpendicular Lines
For lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, they are perpendicular iff a₁a₂ + b₁b₂ = 0. In NDA, quickly check this sum — if zero, the lines are perpendicular. No need to compute slopes. This saves 30−40 seconds per question.
Distance of a Point from a Line
The perpendicular distance from point (x₁, y₁) to line ax + by + c = 0 is:
d = |ax₁ + by₁ + c| / √(a² + b²)
The distance between two parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0 is:
d = |c₁ − c₂| / √(a² + b²)
Example 3 — Distance Between Parallel Lines (NDA 2022)
Find the distance between the parallel lines 2x − 3y + 6 = 0 and 4x − 6y + 15 = 0.
Solution: Make coefficients proportional. Divide second by 2: 2x − 3y + 7.5 = 0. Now c₁ = 6, c₂ = 7.5. Distance = |6 − 7.5| / √(2² + (−3)²) = 1.5 / √13 = (1.5√13)/13 = (3√13)/26 units.
NDA Exam Tip — Distance Shortcut
For parallel lines, always make coefficients of x and y identical before applying the distance formula. NDA questions intentionally give lines with coefficients that need scaling. Failure to do so yields a wrong answer. Also note that distance is always non‑negative, so use absolute value.
Family of Lines
Any line passing through the intersection of L₁ ≡ a₁x + b₁y + c₁ = 0 and L₂ ≡ a₂x + b₂y + c₂ = 0 can be written as L₁ + λL₂ = 0, where λ is a real parameter. This family represents all lines through the intersection point (except L₂ itself).
Example 4 — Family of Lines (NDA 2021)
Find the equation of the line passing through the intersection of 2x + 3y − 5 = 0 and x − 2y + 1 = 0, and also through the point (1, 2).
Use the family method L₁ + λL₂ = 0 instead of solving for the intersection point. This saves 2−3 minutes. In NDA, this technique is especially useful when the line also satisfies an additional condition (like passing through a point or being parallel to a given line).
2. Circles
Standard Equation
A circle with centre C(h, k) and radius r has the standard equation (x − h)² + (y − k)² = r². The general equation is x² + y² + 2gx + 2fy + c = 0, where centre = (−g, −f) and radius = √(g² + f² − c). The circle is real if g² + f² − c ≥ 0, a point circle if g² + f² − c = 0, and imaginary if negative.
Parameter
Expression
Centre (general form)
(−g, −f)
Radius (general form)
r = √(g² + f² − c)
Condition for real circle
g² + f² − c ≥ 0
Point lies on circle
x₁² + y₁² + 2gx₁ + 2fy₁ + c = 0
Diameter form
(x−x₁)(x−x₂) + (y−y₁)(y−y₂) = 0
Example 5 — Circle Centre and Radius (NDA 2022)
Find the centre and radius of the circle 2x² + 2y² − 8x + 12y − 6 = 0.
Solution: Divide by 2: x² + y² − 4x + 6y − 3 = 0. Compare with x² + y² + 2gx + 2fy + c = 0: 2g = −4 → g = −2, 2f = 6 → f = 3, c = −3. Centre = (−g, −f) = (2, −3). Radius = √(g² + f² − c) = √(4 + 9 + 3) = √16 = 4 units.
NDA Shortcut — Centre from General Form
For x² + y² + 2gx + 2fy + c = 0, centre = (−g, −f). But if coefficients of x² and y² are not 1, first divide by the common factor. In NDA, this is a common trap — they give 2x² + 2y² + ... and many students forget to divide.
Tangents & Normals
The tangent to the circle x² + y² + 2gx + 2fy + c = 0 at point (x₁, y₁) on the circle is given by T = 0:
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
For the standard circle x² + y² = a², the tangent at (x₁, y₁) is xx₁ + yy₁ = a². The slope of the tangent is −x₁/y₁, and the slope of the normal is y₁/x₁. The normal always passes through the centre of the circle.
Example 6 — Tangent to a Circle (NDA 2023)
Find the equation of the tangent to the circle x² + y² = 25 at the point (3, 4). Also find the equation of the normal.
Solution: Using xx₁ + yy₁ = a²: 3x + 4y = 25. Slope of tangent = −3/4. Normal slope = 4/3, and it passes through centre (0,0): equation y = (4/3)x or 4x − 3y = 0. The normal is the line joining the centre to the point of contact.
NDA Exam Tip — Tangent Shortcut (T = 0)
To write the tangent equation at (x₁, y₁) on any circle, use the T = 0 method: replace x² → xx₁, y² → yy₁, x → (x + x₁)/2, y → (y + y₁)/2, and xy → (xy₁ + x₁y)/2. Constant term remains unchanged. This works for all conics, not just circles.
Condition for Tangency
A line y = mx + c is tangent to the circle x² + y² = a² if the distance from the centre (0,0) to the line equals the radius a:
c = ± a √(1 + m²)
The length of the tangent from an external point (x₁, y₁) to the circle is √(S₁), where S₁ = x₁² + y₁² + 2gx₁ + 2fy₁ + c. The chord of contact from (x₁, y₁) is T = 0, which has the same form as the tangent equation.
Concept
Formula / Condition
Tangency: y = mx + c to x² + y² = a²
c = ± a√(1 + m²)
Tangency: y = mx + c to (x−h)²+(y−k)² = r²
Distance from centre to line = r
Chord of contact (from external point)
T = 0 (same form as tangent)
Length of tangent from (x₁, y₁)
√(S₁) where S₁ = x₁² + y₁² + 2gx₁ + 2fy₁ + c
Director circle (locus of perpendicular tangents)
x² + y² = 2a² (for x² + y² = a²)
Example 7 — Tangency Condition (NDA 2020)
Find the value of k if the line y = 2x + k is tangent to the circle x² + y² = 20.
Solution: For circle x² + y² = a², a = √20 = 2√5. Tangency: k = ± a√(1 + m²) = ± 2√5 · √(1 + 4) = ± 2√5 · √5 = ± 2 · 5 = ± 10. So k = 10 or k = −10. Both lines y = 2x + 10 and y = 2x − 10 are tangent to the circle.
NDA Shortcut — Two Tangents
For a given slope m, there are always two parallel tangents to a circle: c = ± a√(1 + m²). NDA questions often ask for the positive value of c. Remember the ± sign — both are valid tangents.
3. Conic Sections
Parabola
Standard Forms & Properties
A parabola is the locus of a point equidistant from a fixed point (focus) and a fixed line (directrix). The standard parabola y² = 4ax has focus at (a, 0), directrix x = −a, vertex at (0, 0), and axis along the x‑axis. The length of the latus rectum is 4a. The parametric form is (at², 2at).
Equation
Axis
Focus
Directrix
Latus Rectum
y² = 4ax
x‑axis (right)
(a, 0)
x = −a
x = a, length 4a
y² = −4ax
x‑axis (left)
(−a, 0)
x = a
x = −a, length 4a
x² = 4ay
y‑axis (up)
(0, a)
y = −a
y = a, length 4a
x² = −4ay
y‑axis (down)
(0, −a)
y = a
y = −a, length 4a
Example 8 — Parabola Focus (NDA 2021)
Find the focus, directrix, and length of latus rectum of the parabola 2y² = 16x.
Solution: Rewrite as y² = 8x. Compare with y² = 4ax: 4a = 8 → a = 2. Focus = (2, 0). Directrix: x = −2. Length of latus rectum = 4a = 8. Endpoints of latus rectum: (2, ±4). Parametric coordinates: (2t², 4t).
NDA Shortcut — Parabola Identification
Identify which variable is squared: if y², axis is horizontal (opens left/right); if x², axis is vertical (opens up/down). The sign determines direction: positive opens towards positive axis, negative towards negative. In NDA, the most common form is y² = 4ax (opens right).
Ellipse
Foci, Eccentricity, Latus Rectum
The standard ellipse is x²/a² + y²/b² = 1 where a > b. The foci are at (±ae, 0), eccentricity e = √(1 − b²/a²) where 0 < e < 1, and the directrices are x = ±a/e. The length of the latus rectum is 2b²/a. The sum of distances from any point on the ellipse to the two foci is constant (= 2a). If b > a, the major axis is along the y‑axis.
Property
x²/a² + y²/b² = 1 (a > b)
x²/b² + y²/a² = 1 (a > b)
Centre
(0, 0)
(0, 0)
Major Axis
x‑axis (length 2a)
y‑axis (length 2a)
Foci
(±ae, 0)
(0, ±ae)
Eccentricity e
√(1 − b²/a²)
√(1 − b²/a²)
Directrices
x = ±a/e
y = ±a/e
Latus Rectum length
2b²/a
2b²/a
Example 9 — Ellipse Eccentricity (NDA 2022)
For the ellipse 9x² + 16y² = 144, find the eccentricity, foci, and length of latus rectum.
Solution: Divide by 144: x²/16 + y²/9 = 1. So a² = 16, b² = 9, a = 4, b = 3. e = √(1 − 9/16) = √(7/16) = √7/4. Foci = (±4·√7/4, 0) = (±√7, 0). Length of latus rectum = 2b²/a = 2·9/4 = 9/2 = 4.5 units.
NDA Exam Tip — Eccentricity Range
Remember the eccentricity ranges: ellipse → 0 < e < 1, parabola → e = 1, hyperbola → e > 1. For NDA MCQs, if the answer choices include values both less than and greater than 1, use this to eliminate options. For ellipse, e < 1 always.
Hyperbola
Rectangular Hyperbola
The standard hyperbola is x²/a² − y²/b² = 1. The foci are at (±ae, 0), eccentricity e = √(1 + b²/a²) where e > 1, and the directrices are x = ±a/e. The asymptotes are y = ±(b/a)x. A rectangular hyperbola has a = b, giving asymptotes at right angles (y = ±x) and eccentricity e = √2. Its equation is x² − y² = a² or, in rotated form, xy = c²/2.
Property
x²/a² − y²/b² = 1
y²/a² − x²/b² = 1
Centre
(0, 0)
(0, 0)
Transverse Axis
x‑axis (length 2a)
y‑axis (length 2a)
Foci
(±ae, 0)
(0, ±ae)
Eccentricity e
√(1 + b²/a²)
√(1 + b²/a²)
Asymptotes
y = ±(b/a)x
y = ±(a/b)x
Rectangular Hyperbola
x² − y² = a², e = √2
xy = c² (rotated form)
Example 10 — Hyperbola Asymptotes (NDA 2023)
Find the asymptotes and eccentricity of the hyperbola 4x² − 9y² = 36.
Solution: Divide by 36: x²/9 − y²/4 = 1. Here a² = 9 → a = 3, b² = 4 → b = 2. Asymptotes: y = ±(b/a)x = ±(2/3)x. Eccentricity: e = √(1 + b²/a²) = √(1 + 4/9) = √(13/9) = √13/3. Since e > 1, it confirms a hyperbola.
Conic
Eccentricity
Standard Equation
Focus / Foci
Directrix / Directrices
Parabola
e = 1
y² = 4ax
(a, 0)
x = −a
Ellipse
0 < e < 1
x²/a² + y²/b² = 1
(±ae, 0)
x = ±a/e
Hyperbola
e > 1
x²/a² − y²/b² = 1
(±ae, 0)
x = ±a/e
NDA Shortcut — Conic Identification
Quickly identify conics by signs: both x² and y² positive → ellipse (circle if coefficients equal); one positive, one negative → hyperbola; only one squared term → parabola. For eccentricity: e = 1 → parabola, e < 1 → ellipse, e > 1 → hyperbola. NDA often tests this classification.
4. Locus & Transformation of Axes
Locus
The locus of a point is the path traced when it moves under a given condition. To find the locus, assume the moving point as P(h, k) or P(x, y), apply the geometric condition, and eliminate any parameters to obtain an equation in x and y. The resulting equation represents the locus.
Example 11 — Locus of a Point (NDA 2022)
A point P moves such that its distance from A(2, 3) is always equal to its distance from B(6, 5). Find the locus of P.
Solution: Let P = (x, y). Condition: PA = PB → (x − 2)² + (y − 3)² = (x − 6)² + (y − 5)². Expand: x² − 4x + 4 + y² − 6y + 9 = x² − 12x + 36 + y² − 10y + 25. Cancel common terms: −4x − 6y + 13 = −12x − 10y + 61 → 8x + 4y − 48 = 0 → 2x + y − 12 = 0. The locus is a straight line — the perpendicular bisector of AB.
NDA Exam Tip — Standard Loci to Memorise
Know these standard loci: PA = PB → perpendicular bisector (line). PA² + PB² = constant → circle. PA/PB = constant (≠1) → circle (Apollonius). |PA − PB| = constant → hyperbola. PA + PB = constant → ellipse. Distance from fixed point = distance from fixed line → parabola. These are directly tested in NDA.
Transformation of Axes
Transformation of axes involves shifting the origin or rotating the coordinate axes. Shifting the origin to (h, k) replaces x by X + h and y by Y + k. Rotation of axes by angle θ replaces x by X cos θ − Y sin θ and y by X sin θ + Y cos θ. This simplifies equations — for example, shifting the origin to the centre of a circle eliminates the linear terms.
Example 12 — Shift of Origin (NDA 2021)
Find the new equation of the circle (x − 3)² + (y + 2)² = 16 when the origin is shifted to (3, −2).
Solution: Let X = x − 3, Y = y + 2. Substituting: X² + Y² = 16. The circle's centre (3, −2) becomes the origin in the new coordinate system, and the equation simplifies to X² + Y² = 16 — a circle centred at the new origin with radius 4.
NDA Shortcut — Shift of Origin
When shifting origin to (h, k), substitute x = X + h and y = Y + k. The new equation has no linear terms if (h, k) is the centre of a circle or vertex of a parabola. This simplifies the equation for faster solving. In NDA, this technique is tested with circles and parabolas.