NDA Conic Sections previous year questions with solutions

Answer: (c) e = 1

Method. every parabola has eccentricity 1

4x² + y = 0 ⇒ x² = -14y, a parabola.
For any parabola, distance from the focus = distance from the directrix, so e = 1.

Answer: (c) All three

Method. signs and equality of the two denominators decide ellipse, hyperbola or circle

k = 19: x²5 + y²3 = 2, both denominators positive and unequal: an ellipse. Statement 1 is correct.
k = 12: x²12 + y²-4 = 2, i.e. x²12 − y²4 = 2: a hyperbola. Statement 2 is correct.
k = 20: x²4 + y²4 = 2, i.e. x² + y² = 8: a circle. Statement 3 is correct.
All three are correct.

Answer: (d) 1, 2 and 3

Method. standard eccentricity ranges of the conics

Circle: e = 0 (the foci coincide with the centre). Statement 1 is correct.
Parabola: distance from focus = distance from directrix, so e = 1. Statement 2 is correct.
Ellipse: 0 < e < 1, so e < 1. Statement 3 is correct.
All three are correct.

Answer: (d) a hyperbola

Method. standard form: opposite signs of x² and y² give a hyperbola

2x² - 3y² = 6 ⇒ x²3 − y²2 = 1.
This is the standard form x²a² − y²b² = 1 with a² = 3, b² = 2: a hyperbola.

Answer: (a) A point

Method. complete the squares; a sum of squares equal to zero is a single point

x² - 2x = (x − 1)² - 1 and 4y² - 4y = 4(y − 12)² - 1.
The equation becomes (x − 1)² - 1 + 4(y − 12)² - 1 + 2 = 0, i.e. (x − 1)² + 4(y − 12)² = 0.
Both squares are ≥ 0, so both must be 0: x = 1, y = 12.
The equation represents the single point (1, 12).

Answer: (b) One

Method. compare with x² = -4ay: opens downward, axis x = 0, latus rectum y = -a

x² = -3y is of the form x² = -4ay with 4a = 3, a = 34.
I: y = -x²3 ≤ 0, so it opens downwards. Incorrect.
II: the axis is the y-axis, x = 0. Correct.
III: focus (0, -34), so the latus rectum is y = -34, i.e. 4y + 3 = 0, not 4y − 3 = 0. Incorrect.
Only one statement is correct.

Answer: (d) (1, 2)

Method. point of contact of the tangent y = mx + am is (am², 2am)

y² = 4x ⇒ a = 1; slope m = tan 45° = 1.
Point of contact (am², 2am) = (1, 2).
Check: (1, 2) is on the parabola (4 = 4), and there dydx = 2y = 1.

Answer: (b) 90°

Method. angle between AP and AQ from the dot product (or product of slopes)

Vertex A = (0, 0). Both points satisfy y² = 4kx: (± 4k)² = 16k² = 4k(4k).
Slope of AP = 4k4k = 1, slope of AQ = -4k4k = -1.
Product of slopes = -1, so AP ⊥ AQ: ∠PAQ = 90°.

Answer: (c) Both 1 and 2

Method. focal distance of (x, y) on y² = 4ax is x + a, equal to the distance from the directrix

y² = 8x ⇒ a = 2; directrix x = -2.
Focal distance x + 2 = 8 ⇒ x = 6, and y² = 48 ⇒ y = ± 4⁠√3.⁠ So P can be (6, 4⁠√3):⁠ Statement 1 is correct.
By the definition of a parabola, distance from the directrix = focal distance = 8: Statement 2 is correct.
Both statements are correct.

Answer: (a) y = -18

Method. solve the separable equation, then read the parabola x² = 4ay

dyy = 2 dxx ⇒ ln y = 2 ln x + ln C ⇒ y = Cx².
Through (1, 2): C = 2, so y = 2x², i.e. x² = 12y.
Compare with x² = 4ay: 4a = 12 ⇒ a = 18.
Directrix y = -a: y = -18.

Answer: (b) 12

Method. latus rectum of x² = 4ay is 4a

From the previous item, the parabola is x² = 12y, so 4a = 12.
Length of latus rectum = 4a = 12.

Answer: (b) x − 2 = 0

Method. choose the negative root, then the directrix of y² = 4bx is x = -b

b² + b − 2 = 0 ⇒ (b + 2)(b − 1) = 0 ⇒ b = -2 or 1; b < 0 gives b = -2.
The parabola is y² = -8x, which opens to the left with focus (-2, 0).
Directrix of y² = 4bx is x = -b = 2, i.e. x − 2 = 0.

Answer: (c) p = 2⁠√3q⁠

Method. by symmetry the other two vertices are (± h, k), each side from the vertex making 30° with the axis

The parabola x² = ⁠√3y⁠ is symmetric about the y-axis, so the other two vertices are (h, k) and (-h, k), and each side from O makes 30° with the y-axis: h = k tan 30° = k⁠√3.⁠
On the parabola: k²3 = ⁠√3k⁠ ⇒ k = 3⁠√3,⁠ h = 3.
Side p = 2h = 6 (check: OP = ⁠√(9⁠ + 27) = 6).
Latus rectum q = 4a = ⁠√3⁠ (since 4a = ⁠√3).⁠
pq = 6⁠√3⁠ = 2⁠√3,⁠ so p = 2⁠√3q.⁠

Answer: (d) y² = -12x

Method. focus–⁠directrix definition: SP = PM

For P(x, y): distance to the focus (-3, 0) equals distance to the line x = 3.
(x + 3)² + y² = (x − 3)².
x² + 6x + 9 + y² = x² - 6x + 9 ⇒ y² = -12x.

Shortcut. Focus (-a, 0) with directrix x = a gives y² = -4ax; here a = 3.

Answer: (b) Two

Method. the latus rectum fixes the focus, the axis and the value of a; only the opening direction is free

The latus rectum is the focal chord perpendicular to the axis, so the focus is its midpoint: S(-2, 0).
The latus rectum lies on x = -2, so the axis is the line y = 0 through S.
Length of latus rectum = 4 − (-4) = 8 = 4a ⇒ a = 2, so the vertex is 2 units from S on the axis: (0, 0) or (-4, 0).
The parabola can open to the left (vertex (0, 0)) or to the right (vertex (-4, 0)): exactly two parabolas.

Answer: (c) Both 1 and 2

Method. write down both parabolas and test the statements

From the previous item, the focus is (-2, 0) and a = 2 for both parabolas.
Opening left with vertex (0, 0): y² = -8x; check (-2, ± 4): 16 = 16. It passes through the origin, so Statement 1 is correct.
Opening right with vertex (-4, 0): y² = 8(x + 4); check (-2, ± 4): 16 = 16.
The focus of each is the midpoint of the latus rectum, (-2, 0), so Statement 2 is correct.
Both statements are correct.

Answer: (b) cos θ . cosec² θ

Method. polar form of the chord from the vertex: point (r cos θ, r sin θ)

A point at distance r from the vertex O(0, 0) along the chord is (r cos θ, r sin θ).
It lies on y² = x: r² sin² θ = r cos θ.
r ≠ 0 at the other end, so r = cos θsin² θ = cos θ . cosec² θ.
This r is the length of the chord.

Answer: (a) at two points on the line y = x

Method. solve the two equations simultaneously

From x² = 4ay, y = x²4a. Put into y² = 4ax: x⁴16a² = 4ax ⇒ x(x³ - 64a³) = 0.
Real roots: x = 0 or x = 4a (the other two cube roots are not real).
Points: (0, 0) and (4a, 4a), both on the line y = x.
So the parabolas meet at two points on y = x; (b) and (d) each miss one of them and (c) is false.

Answer: (b) 4/⁠5

Method. eliminate α with sin² α + cos² α = 1

x = 3 sin α, y = 5 cos α ⇒ x²9 + y²25 = 1.
The major axis is along the y-axis: semi-major axis 5, semi-minor axis 3.
e² = 1 − 925 = 1625 ⇒ e = 45.

Answer: (b) 3

Method. y² = 9 − 4x² is largest when x = 0

y² = 9 − 4x² ≤ 9, with equality at x = 0.
So the maximum value of y is 3.

Answer: (a) 94

Method. parametrise the ellipse: x = 32 cos t, y = 3 sin t

x²9/⁠4 + y²9 = 1, so take x = 32 cos t, y = 3 sin t.
xy = 92 sin t cos t = 94 sin 2t.
Maximum when sin 2t = 1: xy = 94.

Shortcut. AM ≥ GM: 9 = 4x² + y² ≥ 2(2x)(y) = 4xy, so xy ≤ 94.

Answer: (a) ⁠√32⁠

Method. substitute both points to find the axes, then e² = 1 − b²a²

Major axis on the y-axis: x²b² + y²a² = 1 with a > b. Write A = 1b², B = 1a².
(3, 2): 9A + 4B = 1; (1, 6): A + 36B = 1.
From the second, A = 1 − 36B; then 9 − 324B + 4B = 1 ⇒ B = 140, A = 110.
So a² = 40, b² = 10 and e² = 1 − 1040 = 34.
e = ⁠√32.⁠

Answer: (b) ⁠√2⁠

Method. standard form, c² = a² - b², distance between foci 2c

x² + 2y² = 1 ⇒ x²1 + y²1/⁠2 = 1, so a² = 1, b² = 12.
c² = a² - b² = 12 ⇒ c = 1⁠√2.⁠
Distance between the foci = 2c = 2⁠√2⁠ = ⁠√2.⁠

Answer: (a) x²25 + y²9 = 1

Method. vertices give a, foci give c, then b² = a² - c²

Vertices (± 5, 0) ⇒ a = 5; foci (± 4, 0) ⇒ c = 4.
b² = a² - c² = 25 − 16 = 9.
Equation: x²25 + y²9 = 1.

Answer: (a) outside the ellipse

Method. sign of S(x₁, y₁) = 2x₁² + 7y₁² - 20 decides the position of a point

Put (1, 2) in 2x² + 7y² - 20: 2(1) + 7(4) − 20 = 30 − 20 = 10 > 0.
A positive value means the point lies outside the ellipse.

Answer: (b) 49x²1024 + 7y²64 = 1

Method. latus rectum 2b²a together with b² = a²(1 − e²)

Let the ellipse be x²a² + y²b² = 1, a > b.
b² = a²(1 − 916) = 7a²16.
Latus rectum 2b²a = 4 ⇒ b² = 2a.
So 7a²16 = 2a ⇒ a = 327, a² = 102449, b² = 2a = 647.
Equation: 49x²1024 + 7y²64 = 1.

Answer: (a) x²64 + y²60 = 1

Method. ellipse with foci on the x-axis: c = ae, b² = a² - c²

Foci (± 2, 0) lie on the x-axis, so the ellipse is x²a² + y²b² = 1 with a > b and ae = 2.
e = 14 ⇒ a = 21/⁠4 = 8, so a² = 64.
b² = a² - a²e² = 64 − 4 = 60.
Equation: x²64 + y²60 = 1.

Answer: (d) a + (cx/⁠a)

Method. focal distance from the focus (-ae, 0) is a + ex

The ellipse is x²a² + y²b² = 1 with c = ae, so e = ca.
The directrix nearer to F(-c, 0) is x = -ae; by the focus–⁠directrix property PF = e(x + ae) = a + ex.
PF = a + cxa.

Answer: (b) 1

Method. sum of focal distances = length of the major axis

4x² + 9y² = 1 ⇒ x²1/⁠4 + y²1/⁠9 = 1, so the semi-major axis is 12 (along the x-axis).
PQ + PR = 2a = 2 × 12 = 1.

Answer: (d) 1⁠√2⁠

Method. right angle at the end of the minor axis: half-angle 45° gives c = b

Foci S(ae, 0), S′(-ae, 0); end of the minor axis B(0, b).
BS and BS′ are symmetric about the y-axis, so ∠SBS′ = 90° means each makes 45° with it: ae = b.
b² = a²(1 − e²) = a²e² ⇒ 1 − e² = e² ⇒ e² = 12.
e = 1⁠√2.⁠

Shortcut. Slopes of BS and BS′ are -bae and bae; product -1 ⇒ b = ae.

Answer: (b) 2

Method. sum of focal distances = length of the major axis

x² + 4y² = 1 ⇒ x²1 + y²1/⁠4 = 1, so a = 1 (major axis along the x-axis) and b = 12.
For any point P on the ellipse, PE + PF = 2a = 2.

Answer: (d) 1, 2 and 3

Method. latus rectum through the focus (c, 0) with ends (c, ± b²a)

a² = 1, b² = 14 ⇒ c² = 1 − 14 = 34, c = ⁠√32.⁠ Focus (⁠√32,⁠ 0).
The latus rectum lies on x = ⁠√32;⁠ its ends have y = ± b²a = ± 14 (check: 34 + 4 × 116 = 1).
So the latus rectum is the segment from (⁠√32,⁠ -14) to (⁠√32,⁠ 14).
Point 1 is its midpoint (the focus), points 2 and 3 are its ends: all three lie on it.

Answer: (b) 10

Method. Q and R are the foci, so PQ + PR = length of the major axis

25x² + 16y² = 400 ⇒ x²16 + y²25 = 1; the major axis is along the y-axis with semi-major axis 5 and semi-minor axis 4.
c² = 25 − 16 = 9 ⇒ foci (0, ± 3), which are exactly Q and R.
For any point of the ellipse the sum of focal distances is the major axis: PQ + PR = 2 × 5 = 10.

Answer: (b) 1⁠√2⁠

Method. right angle at the focus: the focus is equidistant along the axes, so b = ae

Ends of the minor axis B(0, b), B′(0, -b); focus S(ae, 0).
SB and SB′ are symmetric about the x-axis, so ∠BSB′ = π2 means each makes 45° with the x-axis: b = ae.
b² = a²(1 − e²) and b² = a²e² ⇒ 1 − e² = e² ⇒ e² = 12.
e = 1⁠√2.⁠

Shortcut. Slopes of SB and SB′ are -bae and bae; product -1 ⇒ b = ae directly.

Answer: (b) length of major axis

Method. definition of the ellipse: PS + PS′ = 2a

For x²a² + y²b² = 1, the focal distances of P(x, y) are a − ex and a + ex.
Their sum is (a − ex) + (a + ex) = 2a, which is the length of the major axis.

Answer: (b) 15π square metres

Method. locus with constant sum of distances from two fixed points is an ellipse; area = πab

The flag-posts are the foci and the man's path is an ellipse with 2a = 10, so a = 5.
Distance between foci 2c = 8, so c = 4.
b² = a² - c² = 25 − 16 = 9, so b = 3.
Area enclosed = πab = π × 5 × 3 = 15π square metres.

Answer: (c) 7

Method. equate the focal distance c of the hyperbola (c² = a² + b²) and the ellipse (c² = a² - b²)

Hyperbola: 81x² - 144y² = 1166425. Divide: x²144/⁠25 − y²81/⁠25 = 1 (since 11664 = 81 × 144).
c² = 14425 + 8125 = 22525 = 9 ⇒ foci (± 3, 0).
Ellipse: px² + 16y² = 16p ⇒ x²16 + y²p = 1 with p < 16, so the major axis is along the x-axis.
c² = 16 − p = 9 ⇒ p = 7.

Answer: (a) 0.5

Method. e = ca for each conic, with c = 3 from the previous item

Hyperbola: a = 125, c = 3 ⇒ e = 312/⁠5 = 54 = 1.25.
Ellipse: a = 4, c = 3 ⇒ e = 34 = 0.75.
Difference = 1.25 − 0.75 = 0.5.

Answer: (b) ⁠√5⁠

Method. standard form, c² = a² + b², distance between foci 2c

x² - 4y² = 1 ⇒ x²1 − y²1/⁠4 = 1, so a² = 1, b² = 14.
c² = 1 + 14 = 54 ⇒ c = ⁠√52.⁠
Distance between the foci = 2c = ⁠√5.⁠

Answer: (b) 4⁠√3⁠ units

Method. standard form, c² = a² + b², distance between foci 2c

25x² - 75y² = 225 ⇒ x²9 − y²3 = 1, so a² = 9, b² = 3.
c² = 9 + 3 = 12 ⇒ c = 2⁠√3.⁠
Distance between the foci = 2c = 4⁠√3⁠ units.

Answer: (c) All three

Method. c² = a² + b² for the hyperbola, with a² = cos² θ, b² = sin² θ

c² = a² + b² = cos² θ + sin² θ = 1 ⇒ foci (± 1, 0) for every θ. Statement 1 is correct.
e = ca = 1cos θ = sec θ. (The equation reads cos θ and sin θ as the semi-axes a and b, so θ is taken acute; for a general θ the eccentricity is 1|cos θ| = |sec θ|.) Statement 2 is correct.
Distance between the foci = 2c = 2. Statement 3 is correct.
All three are correct.

Answer: (d) ⁠√132⁠

Method. eliminate θ with sec² θ − tan² θ = 1

x = 3 tan θ, y = 2 sec θ ⇒ y²4 − x²9 = sec² θ − tan² θ = 1.
The transverse axis is along the y-axis with semi-transverse axis 2 and semi-conjugate axis 3.
e² = 1 + 94 = 134 ⇒ e = ⁠√132.⁠