CBSE Class 11 Mathematics Conic Sections JEE Previous Year Questions Set 02

Answer: (d) 9y² = −4x

Explanation:
1. Take P(−t², 2t) and Q(−t², −2t) with t > 0.
2. R divides PQ in 2 : 1: \( R = \left(-t^2, \frac{2(-2t) + 1(2t)}{3}\right) = \left(-t^2, -\frac{2t}{3}\right) \).
3. So h = −t² and t = −3k/2, giving h = −9k²/4.
4. Locus: 9y² = −4x.

Teacher's Note:
1. A double ordinate has both ends with the same x-coordinate and opposite y-coordinates.
2. Since every point of y² = −4x has x ≤ 0, the locus must also lie in x ≤ 0, which rules out (b) and (c).

Answer: (b) \( \frac{2}{\sqrt{3}} \)

Explanation:
1. Conjugate axis = half the distance between foci: 2b = ae, so b² = a²e²/4.
2. Also b² = a²(e² − 1), so e² − 1 = e²/4.
3. 3e²/4 = 1 gives e² = 4/3 and \( e = \frac{2}{\sqrt{3}} \).

Teacher's Note:
1. The latus rectum condition is not needed for e; it would only fix the actual size of the hyperbola.
2. Eccentricity depends only on the ratio b/a, so look for the condition that gives this ratio.

Answer: (b) −7

Explanation:
1. 9e² − 18e + 5 = (3e − 1)(3e − 5) = 0; for a hyperbola e > 1, so e = 5/3.
2. Focus ae = 5 gives a = 3, a² = 9 (check: a/e = 9/5 matches the directrix x = 9/5).
3. b² = a²(e² − 1) = 9 × 16/9 = 16, so a² − b² = 9 − 16 = −7.

Teacher's Note:
1. Always reject the root e = 1/3, since a hyperbola must have e > 1.
2. The directrix x = a/e gives a useful check on the value of a.

Answer: (d) \( (5, 2\sqrt{3}) \)

Explanation:
1. The conic is \( \frac{x^2}{12} + \frac{y^2}{16} = 1 \), major axis along y, c² = 16 − 12 = 4, foci (0, ±2).
2. Hyperbola: transverse axis along y, a = 2, b² = a²(e² − 1) = 4 × 5/4 = 5, so \( \frac{y^2}{4} - \frac{x^2}{5} = 1 \).
3. Check: (0, 2) gives 1; (√5, 2√2) gives 2 − 1 = 1; (√10, 2√3) gives 3 − 2 = 1; (5, 2√3) gives 3 − 5 = −2 ≠ 1.
4. So (5, 2√3) does not lie on the hyperbola.

Teacher's Note:
1. When the transverse axis is along the y-axis, the y² term is positive in the standard form.
2. Substituting each option into the final equation is the quickest way to answer 'which point does NOT lie' questions.

Answer: (d) a hyperbola with the length of conjugate axis 3

Explanation:
1. From the first line t(x − 3) = 2y; from the second, x + 3 = 2ty.
2. Multiplying: t(x − 3)(x + 3) = 2y · 2ty, so x² − 9 = 4y².
3. \( \frac{x^2}{9} - \frac{y^2}{9/4} = 1 \): a hyperbola with a = 3, b = 3/2.
4. Conjugate axis 2b = 3 (and e = √5/2, not √5), so (d).

Teacher's Note:
1. Eliminate the parameter by combining the equations so that t cancels.
2. Compute both e and the axis lengths, because options often differ only in such details.

Answer: (d) 40

Explanation:
1. 2ae = 6 gives ae = 3; with e = 3/5, a = 5.
2. b² = a² − a²e² = 25 − 9 = 16, so b = 4.
3. The quadrilateral is a rhombus with diagonals 2a = 10 and 2b = 8; area = ½ × 10 × 8 = 40.

Teacher's Note:
1. The four vertices (±a, 0) and (0, ±b) form a rhombus whose area is half the product of its diagonals, i.e. 2ab.
2. Do not confuse this with the area of the ellipse itself, which is πab.

Answer: (c) \( \frac{\sqrt{3}}{2} \)

Explanation:
1. Let the ellipse be \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \); put u = 1/a², v = 1/b².
2. (4, −1): 16u + v = 1; (−2, 2): 4u + 4v = 1.
3. Solving: u = 1/20, v = 1/5, so a² = 20, b² = 5.
4. \( e^2 = 1 - \frac{5}{20} = \frac{3}{4} \), so \( e = \frac{\sqrt{3}}{2} \).

Teacher's Note:
1. Taking 1/a² and 1/b² as unknowns turns the problem into two linear equations.
2. After solving, decide which axis is major (larger denominator) before writing e.

Answer: (d) a hyperbola with length of its transverse axis \( 8\sqrt{2} \)

Explanation:
1. First line: \( \sqrt{2}x - y = -4\sqrt{2}k \); second line: \( \sqrt{2}x + y = \frac{4\sqrt{2}}{k} \).
2. Multiplying: 2x² − y² = −32, i.e. \( \frac{y^2}{32} - \frac{x^2}{16} = 1 \).
3. This is a hyperbola with transverse axis along y and a² = 32, so transverse axis = \( 2\sqrt{32} = 8\sqrt{2} \).
4. Its eccentricity is \( \sqrt{1 + \frac{16}{32}} = \sqrt{\frac{3}{2}} \), so (c) is wrong and (d) is correct.

Teacher's Note:
1. When the constant on the right comes out negative, rewrite the equation so that the positive term comes first; the transverse axis follows that variable.
2. Multiplying the two line equations is a standard trick to eliminate a parameter that appears as k and 1/k.

Answer: (b) \( \frac{1}{3} \)

Explanation:
1. \( \frac{2b^2}{a} = 4 \) gives b² = 2a, i.e. a²(1 − e²) = 2a, so a(1 − e)(1 + e) = 2.
2. Focus to nearest vertex: a − ae = a(1 − e) = 3/2.
3. Dividing: \( 1 + e = \frac{2}{3/2} = \frac{4}{3} \), so \( e = \frac{1}{3} \).

Teacher's Note:
1. The nearest vertex is on the same side as the focus, at distance a − ae; the farther one is at a + ae.
2. Dividing the two equations removes a and leaves an equation in e alone.

Answer: (a) the area of △ABC is maximum when it is isosceles

Explanation:
1. ∠ACB = 90° (angle in a semicircle). Let AB = 2r and ∠CAB = θ; then AC = 2r cos θ, BC = 2r sin θ.
2. Area = \( \frac{1}{2} \)·AC·BC = r² sin 2θ, which is greatest when θ = 45°, i.e. AC = BC.
3. So the area is maximum when the triangle is isosceles.

Teacher's Note:
1. The angle in a semicircle is a right angle, so AB is the hypotenuse.
2. Equivalently, the height from C to AB is greatest (equal to r) when C is at the top of the circle.

Answer: (c) x² + y² = 2

Explanation:
1. The chord subtends 90° at the centre, so the triangle formed with the centre is right isosceles with equal sides 2.
2. Distance of the mid-point from the centre = 2 cos 45° = √2.
3. Locus: x² + y² = 2.

Teacher's Note:
1. The perpendicular from the centre bisects the chord and the angle at the centre.
2. All such chords are equal, so their mid-points lie on a concentric circle.

Answer: (c) \( \frac{\pi}{4} \)

Explanation:
1. Slope of OQ = \( \frac{4}{3} \), slope of OR = \( -\frac{3}{4} \); their product is −1, so ∠QOR = \( \frac{\pi}{2} \).
2. The angle subtended at the circumference is half the angle at the centre.
3. ∠QPR = \( \frac{\pi}{4} \).

Teacher's Note:
1. The angle at the centre is twice the angle at the circumference on the same arc.
2. P on the major arc gives half of ∠QOR; on the minor arc it would be π − π/4.

Answer: (b) a circle

Explanation:
1. Let A, B be fixed and P = (r cos θ, r sin θ). Centroid G = \( \frac{A + B + P}{3} \).
2. Writing A + B = (p, q) (fixed), G = \( \left(\frac{p + r\cos\theta}{3}, \frac{q + r\sin\theta}{3}\right) \).
3. Eliminating θ: \( \left(x - \frac{p}{3}\right)^2 + \left(y - \frac{q}{3}\right)^2 = \frac{r^2}{9} \), a circle.

Teacher's Note:
1. When one vertex moves on a circle and the others are fixed, the centroid moves on a circle one-third the size.
2. Eliminate θ using cos²θ + sin²θ = 1.

Answer: (a) (4, 7)

Explanation:
1. x² − 8x + 12 = 0 gives x = 2 and x = 6; y² − 14y + 45 = 0 gives y = 5 and y = 9.
2. The square has vertices (2, 5), (6, 5), (6, 9), (2, 9).
3. The centre of the inscribed circle is the centre of the square: (4, 7).

Teacher's Note:
1. Factorise each quadratic to get the pairs of parallel lines.
2. The inscribed circle of a square has the same centre as the square and radius equal to half the side.

Answer: (c) 3

Explanation:
1. Given circle: centre (1, 3), radius \( \sqrt{1 + 9 - 6} = 2 \). The chord has mid-point (1, 3) and half-length 2.
2. Distance from (2, 1) to (1, 3) = √5, perpendicular to the chord.
3. Radius = \( \sqrt{5 + 4} = 3 \).

Teacher's Note:
1. The mid-point of the chord is the centre of the first circle.
2. Use R² = d² + (half chord)², where d is the distance from the centre to the chord.

Answer: (a) 0.75

Explanation:
1. Take A(0, 0), B(2, 0), C(2, 2), D(0, 2), centre O(1, 1). For any point X, ΣXA² = 4·OX² + ΣOA² = 4·OX² + 8.
2. P on C₁ (radius 1): ΣPA² = 4 + 8 = 12.
3. Q on C₂ (radius √2): ΣQA² = 8 + 8 = 16.
4. Ratio = \( \frac{12}{16} = 0.75 \).

Teacher's Note:
1. The incircle of a square has radius half the side; the circumcircle has radius half the diagonal.
2. Writing the point as (1 + r cos θ, 1 + r sin θ) and expanding shows the sum does not depend on θ.

Answer: (a) x² + y² − 6x + 8y + 9 = 0; (c) x² + y² − 6x − 8y + 9 = 0

Explanation:
1. Touching the x-axis at (±3, 0): centre (±3, k), radius |k|; the equation is x² + y² ∓ 6x − 2ky + 9 = 0.
2. Intercept on the y-axis = \( 2\sqrt{k^2 - 9} = 2\sqrt{7} \), so k² = 16, k = ±4.
3. With centre (3, ±4): x² + y² − 6x ∓ 8y + 9 = 0.
4. So options (a) and (c) are correct.

Teacher's Note:
1. For a circle touching the x-axis at (a, 0), the constant term is a² and the radius is |k|.
2. Both signs of k are allowed because the circle may lie above or below the x-axis.

Answer: 2

Explanation:
1. Centre (−1, −2), r² = 5 + p, so p > −5. On the x-axis: x² + 2x − p = 0; on the y-axis: y² + 4y − p = 0.
2. Case 1, circle through the origin (p = 0): x-axis points 0, −2; y-axis points 0, −4. The common points are (0, 0), (−2, 0), (0, −4), three in all.
3. Case 2, touches the x-axis (4 + 4p = 0, p = −1): y² + 4y + 1 = 0 has two roots, giving 1 + 2 = 3 points.
4. Case 3, touches the y-axis (16 + 4p = 0, p = −4): x² + 2x + 4 = 0 has no real roots, only 1 point.
5. Hence p = 0 or p = −1: 2 values.

Teacher's Note:
1. Exactly three common points happen either when the circle passes through the origin or when it touches one axis and cuts the other.
2. Always check the other axis in each case; touching the y-axis here gives only one common point.

Answer: (b) a hyperbola

Explanation:
1. |r| < 1 means −1 < r < 1, so 1 − r > 0 and 1 + r > 0.
2. The equation has the form \( \frac{x^2}{A} - \frac{y^2}{B} = 1 \) with A, B > 0.
3. Hence it represents a hyperbola.

Teacher's Note:
1. Check the signs of the denominators first; the minus sign between the terms then decides the curve.
2. Its foci are fixed: a² + b² = (1 − r) + (1 + r) = 2 for every r.

Answer: (c) 10

Explanation:
1. 16x² + 25y² = 400 is \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \): a = 5, b = 4.
2. c = \( \sqrt{25 - 16} = 3 \), so F₁ and F₂ are the foci.
3. For any point on an ellipse, PF₁ + PF₂ = 2a = 10.

Teacher's Note:
1. The sum of the focal distances is the defining property of an ellipse and equals the major axis.
2. Always check that the given fixed points really are the foci before using this property.

Answer: (b) abscissae of foci

Explanation:
1. Here a² = cos² α and b² = sin² α.
2. a²e² = a² + b² = cos² α + sin² α = 1, so the foci are (±1, 0) for every α.
3. Vertices (±cos α, 0), e = sec α and directrices x = ±cos² α all change with α.

Teacher's Note:
1. For a hyperbola, the distance of a focus from the centre is \( \sqrt{a^2 + b^2} \).
2. When a² + b² simplifies to a constant, the foci stay fixed even though the shape changes.

Answer: (b) \( \frac{x^2}{16} - \frac{y^2}{9} = -1 \)

Explanation:
1. Ellipse: major axis along y, e₁² = 1 − 16/25 = 9/25, so e₁ = 3/5; foci (0, ±3).
2. Hence e₂ = 5/3; the hyperbola passes through (0, ±3), so take it as \( \frac{y^2}{9} - \frac{x^2}{b^2} = 1 \).
3. e₂² = 1 + b²/9 = 25/9 gives b² = 16: \( \frac{y^2}{9} - \frac{x^2}{16} = 1 \), i.e. \( \frac{x^2}{16} - \frac{y^2}{9} = -1 \).

Teacher's Note:
1. Since the hyperbola passes through points on the y-axis, its transverse axis must be along the y-axis.
2. Options (a) and (c) have vertices on the x-axis, so they cannot pass through (0, ±3).

Answer: (a) x² cosec² θ − y² sec² θ = 1

Explanation:
1. Ellipse \( \frac{x^2}{4} + \frac{y^2}{3} = 1 \) has c² = 4 − 3 = 1, foci (±1, 0).
2. Hyperbola: a = sin θ and a² + b² = 1, so b² = cos² θ.
3. \( \frac{x^2}{\sin^2\theta} - \frac{y^2}{\cos^2\theta} = 1 \), i.e. x² cosec² θ − y² sec² θ = 1.

Teacher's Note:
1. Confocal means the curves have the same foci, so equate their values of c.
2. Dividing by sin² θ is the same as multiplying by cosec² θ, which explains the form of the answer.

Answer: (a) x² + y² − 12x + 24 = 0

Explanation:
1. From the circle y² = 8x − x²; put in 4x² − 9y² = 36: 13x² − 72x − 36 = 0.
2. (13x + 6)(x − 6) = 0; x = −6/13 gives y² < 0, so x = 6 and y² = 12.
3. A = (6, 2√3), B = (6, −2√3); circle on AB as diameter: (x − 6)² + y² = 12.
4. That is x² + y² − 12x + 24 = 0.

Teacher's Note:
1. Always check each x-root by finding y²; a negative value means no real intersection.
2. The circle on AB as diameter has the mid-point of AB as centre and AB/2 as radius.

Answer: (c) y² = x

Explanation:
1. P(h, k) divides the segment from O in 1 : 3, so P = (x/4, y/4), i.e. x = 4h, y = 4k.
2. (4h, 4k) lies on y² = 4x: 16k² = 16h.
3. Locus: y² = x.

Teacher's Note:
1. Write the point of the curve in terms of the locus point, then substitute.
2. The new parabola has a latus rectum one-fourth as long (4 to 1), as expected from scaling by 1/4.

Answer: (a) \( \left(-\frac{9}{10}, 0\right) \)

Explanation:
1. c² = 9 − 8 = 1, so F₁ = (−1, 0), F₂ = (1, 0); the parabola is y² = 4x.
2. With the ellipse: \( \frac{x^2}{9} + \frac{4x}{8} = 1 \Rightarrow 2x^2 + 9x - 18 = 0 \), so x = 3/2; M = (3/2, √6), N = (3/2, −√6).
3. By symmetry the orthocentre is on the x-axis; the altitude from M is perpendicular to F₁N (slope −2√6/5), so its slope is 5/(2√6).
4. y − √6 = \( \frac{5}{2\sqrt{6}} \)(x − 3/2); at y = 0, x = 3/2 − 12/5 = −9/10. Orthocentre (−9/10, 0).

Teacher's Note:
1. Triangle F₁MN is symmetric about the x-axis, so the orthocentre must lie on the x-axis; this rules out (d).
2. One altitude plus the axis of symmetry is enough to locate the orthocentre.

Answer: (b) a focus of the hyperbola is (2, 0); (d) the equation of the hyperbola is x² − 3y² = 3

Explanation:
1. Ellipse \( \frac{x^2}{4} + y^2 = 1 \): e = √3/2, foci (±√3, 0).
2. Hyperbola: e = 2/√3; it passes through (√3, 0), so a² = 3.
3. b² = a²(e² − 1) = 3(4/3 − 1) = 1, so \( \frac{x^2}{3} - y^2 = 1 \), i.e. x² − 3y² = 3.
4. Focus (ae, 0) = (√3 × 2/√3, 0) = (2, 0). So (b) and (d) are correct.

Teacher's Note:
1. A hyperbola of the form x²/a² − y²/b² = 1 meets the x-axis only at its vertices, so the focus of the ellipse becomes the vertex of the hyperbola.
2. Reciprocal eccentricities: an ellipse e < 1 pairs with a hyperbola e > 1.

Answer: (a) \( (4, 2\sqrt{2}) \); (d) \( (1, \sqrt{2}) \)

Explanation:
1. Let P = (p²/2, p), Q = (q²/2, q). The circle on PQ passes through O, so ∠POQ = 90°: \( \frac{p^2q^2}{4} + pq = 0 \), giving pq = −4.
2. Area = \( \frac{1}{2}\left|\frac{p^2}{2}q - \frac{q^2}{2}p\right| = \frac{1}{4}|pq||p - q| = |p - q| = 3\sqrt{2} \).
3. With q = −4/p and p > 0: p + 4/p = 3√2, so p² − 3√2p + 4 = 0, giving p = 2√2 or √2.
4. P = (4, 2√2) or (1, √2); options (a) and (d).

Teacher's Note:
1. An angle in a semicircle is a right angle, so 'circle on PQ as diameter passes through O' means OP ⊥ OQ.
2. Writing a point of y² = 2x as (y²/2, y) keeps only one unknown per point.

Answer: (c) A–S, B–R, C–P, D–Q

Explanation:
1. L(0, b), M(0, −b), N(a, 0): LN = MN, so with 60° at N the triangle is equilateral and tan 30° = b/a, i.e. a = √3 b.
2. Area = ½ × 2b × a = ab = √3 b² = 4√3, so b = 2 and a = 2√3.
3. (A) conjugate axis 2b = 4 → S; (B) e² = 1 + b²/a² = 4/3, e = 2/√3 → R.
4. (C) distance between foci 2ae = 2 × 2√3 × 2/√3 = 8 → P; (D) latus rectum 2b²/a = 8/(2√3) = 4/√3 → Q.

Teacher's Note:
1. The vertex N lies on the perpendicular bisector of the conjugate axis, so triangle LMN is isosceles; a 60° apex angle makes it equilateral.
2. Find a and b first; every entry in Column I then follows from a standard formula.