CBSE Class 11 Mathematics Conic Sections JEE Previous Year Questions Set 01

Answer: (a) 4 ≤ x² + y² ≤ 64

Explanation:
1. Each centre is at distance 5 from the origin and each circle has radius 3.
2. A point of such a circle is at distance between 5 − 3 = 2 and 5 + 3 = 8 from the origin.
3. So 4 ≤ x² + y² ≤ 64.

Teacher's Note:
1. Points on these circles fill the ring between radii 5 − 3 and 5 + 3.
2. Square the distances because x² + y² is the square of the distance from the origin.

Answer: (c) x² + y² = 4a²

Explanation:
1. In an equilateral triangle the circumcentre is the centroid, which divides each median in the ratio 2 : 1.
2. Circumradius = \( \frac{2}{3} \times 3a = 2a \).
3. Equation: x² + y² = 4a².

Teacher's Note:
1. The centroid lies two-thirds of the way along a median from the vertex.
2. In an equilateral triangle the median, altitude and angle bisector coincide.

Answer: (b) x² + y² − 2x + 2y = 47

Explanation:
1. The centre is the intersection of the diameters: solving 2x − 3y = 5 and 3x − 4y = 7 gives (1, −1).
2. πr² = 154 with π = \( \frac{22}{7} \) gives r² = 49.
3. (x − 1)² + (y + 1)² = 49, i.e. x² + y² − 2x + 2y = 47.

Teacher's Note:
1. Two diameters always meet at the centre.
2. Take π = 22/7 when the area is given as a multiple of 22, such as 154.

Answer: (c) (x − p)² = 4qy

Explanation:
1. Let the other end be (h, k). Centre = \( \left(\frac{p + h}{2}, \frac{q + k}{2}\right) \) and radius = \( \frac{1}{2}\sqrt{(h - p)^2 + (k - q)^2} \).
2. Touching the x-axis: radius = |y-coordinate of centre| = \( \frac{|q + k|}{2} \).
3. (h − p)² + (k − q)² = (q + k)², which simplifies to (h − p)² = 4qk.
4. Locus: (x − p)² = 4qy.

Teacher's Note:
1. For a circle touching the x-axis, radius = |y-coordinate of the centre|.
2. The centre is the mid-point of the diameter and the radius is half its length.

Answer: (c) x² + y² − 2x + 2y − 23 = 0

Explanation:
1. Centre: from 3x − y − 4 = 0, y = 3x − 4; then 2x + 9x − 12 + 1 = 0 gives x = 1, y = −1.
2. 2πr = 10π, so r = 5.
3. (x − 1)² + (y + 1)² = 25, i.e. x² + y² − 2x + 2y − 23 = 0.

Teacher's Note:
1. Lines along diameters intersect at the centre.
2. Circumference 2πr gives the radius directly.

Answer: (c) x² + y² − x − y = 0

Explanation:
1. Put y = x: 2x² − 2x = 0, so x = 0 or 1. A = (0, 0), B = (1, 1).
2. Circle on AB as diameter: x(x − 1) + y(y − 1) = 0.
3. x² + y² − x − y = 0.

Teacher's Note:
1. Find the ends of the chord first, then use the diameter form.
2. The centre (1/2, 1/2) is the mid-point of AB and lies on y = x.

Answer: (d) x² + y² − 2x + 2y − 47 = 0

Explanation:
1. Solving 3x − 4y = 7 and 2x − 3y = 5 gives the centre (1, −1).
2. πr² = 49π, so r = 7.
3. (x − 1)² + (y + 1)² = 49, i.e. x² + y² − 2x + 2y − 47 = 0.

Teacher's Note:
1. The point of intersection of two diameters is the centre.
2. Constant term = h² + k² − r² = 1 + 1 − 49 = −47.

Answer: (d) \( x^2 + y^2 = \frac{9}{4} \)

Explanation:
1. The line from the centre to the mid-point M of a chord is perpendicular to it and bisects the angle 2π/3.
2. So OM = 3 cos \( \frac{\pi}{3} \) = \( \frac{3}{2} \).
3. M is always at distance 3/2 from the origin: x² + y² = \( \frac{9}{4} \).

Teacher's Note:
1. The perpendicular from the centre to a chord bisects both the chord and the angle it subtends at the centre.
2. Chords of equal length are equidistant from the centre, so their mid-points lie on a concentric circle.

Answer: (d) \( k \ge \frac{1}{2} \)

Explanation:
1. Touching the x-axis: radius = |k|. Passing through (−1, 1): (h + 1)² + (k − 1)² = k².
2. (h + 1)² + 1 − 2k = 0, so (h + 1)² = 2k − 1.
3. Since (h + 1)² ≥ 0, 2k − 1 ≥ 0, i.e. k ≥ \( \frac{1}{2} \).

Teacher's Note:
1. A real h exists only when the right-hand side 2k − 1 is not negative.
2. Since the point (−1, 1) is above the x-axis, the centre must also be above it, so k is positive.

Answer: (d) (−3, −4)

Explanation:
1. Centre = (−1, −2), which is the mid-point of P and the required point Q.
2. Q = 2(−1, −2) − (1, 0) = (−3, −4).

Teacher's Note:
1. The diametrically opposite point is the reflection of P in the centre.
2. Check that P lies on the circle: 1 + 2 − 3 = 0.

Answer: (c) x² + y² − x − y = 0

Explanation:
1. The smallest circle through two points is the one having the segment joining them as a diameter.
2. Circle: (x − 1)(x − 0) + (y − 0)(y − 1) = 0.
3. x² + y² − x − y = 0.

Teacher's Note:
1. Any circle through A and B has AB as a chord, and a chord can never be longer than the diameter.
2. So the least radius is AB/2 = 1/√2 here.

Answer: (a) \( \frac{10}{3} \)

Explanation:
1. Touching the x-axis at (1, 0): centre (1, k), radius |k|.
2. Through (2, 3): 1 + (3 − k)² = k², so 10 − 6k = 0, k = \( \frac{5}{3} \).
3. Diameter = 2k = \( \frac{10}{3} \).

Teacher's Note:
1. The centre lies directly above (or below) the point of contact with the x-axis.
2. The k² terms cancel, leaving a linear equation.

Answer: (c) (5, −2)

Explanation:
1. Centre (3, k), radius |k|. Through (1, −2): 4 + (k + 2)² = k², so 4k + 8 = 0, k = −2.
2. Circle: (x − 3)² + (y + 2)² = 4.
3. (5, −2): 4 + 0 = 4, so it lies on the circle.

Teacher's Note:
1. (1, −2) and (5, −2) are mirror images in the vertical line x = 3 through the centre.
2. Points symmetric about a diameter of a circle lie on the circle together.

Answer: (d) (25, 29)

Explanation:
1. Centre (3, 5), r² = 9 + 25 − p = 34 − p.
2. Not meeting the axes: r < 3 (the smaller distance to an axis), so 34 − p < 9, p > 25.
3. (1, 4) inside: 1 + 16 − 6 − 40 + p < 0, so p < 29.
4. Hence 25 < p < 29.

Teacher's Note:
1. A circle avoids both axes when its radius is less than the distance of the centre from the nearer axis.
2. A point is inside when S₁ < 0.

Answer: (a) x² + y² + 3x + y − 11 = 0

Explanation:
1. Put y = −3x − 5: x² + 9x² + 30x + 25 = 16, i.e. 10x² + 30x + 9 = 0, so x₁ + x₂ = −3, x₁x₂ = \( \frac{9}{10} \).
2. y₁ + y₂ = −3(x₁ + x₂) − 10 = −1 and y₁y₂ = 9x₁x₂ + 15(x₁ + x₂) + 25 = \( -\frac{119}{10} \).
3. Diameter form: x² + 3x + \( \frac{9}{10} \) + y² + y − \( \frac{119}{10} \) = 0.
4. x² + y² + 3x + y − 11 = 0.

Teacher's Note:
1. Using the sum and product of roots avoids messy surds in the end points.
2. Check: the centre (−3/2, −1/2) lies on 3x + y + 5 = 0.

Answer: (b) 3

Explanation:
1. Touching the y-axis at (0, 2): centre (h, 2), radius |h|.
2. Through (−1, 0): (h + 1)² + 4 = h², so h = \( -\frac{5}{2} \).
3. On the x-axis: \( \left(x + \frac{5}{2}\right)^2 + 4 = \frac{25}{4} \), so x = −1 or −4.
4. Chord length = 3.

Teacher's Note:
1. For a circle touching the y-axis at (0, b), the centre is (h, b) and the radius is |h|.
2. Formula check: x-intercept = 2√(g² − c) = 2√(25/4 − 4) = 3.

Answer: (a) \( 5\sqrt{3} \)

Explanation:
1. Given circle: centre C(2, −3), radius 5. The chord (a diameter of this circle) has mid-point C and half-length 5.
2. Distance from the centre of S to C = \( \sqrt{25 + 25} = 5\sqrt{2} \), and this line is perpendicular to the chord.
3. Radius of S = \( \sqrt{(5\sqrt{2})^2 + 5^2} = \sqrt{75} = 5\sqrt{3} \).

Teacher's Note:
1. The line from the centre of a circle to the mid-point of a chord is perpendicular to the chord.
2. So R² = (distance to mid-point)² + (half chord)².

Answer: (b) 2x − 3y + 10 = 0

Explanation:
1. Touching the y-axis at (0, 2): centre (h, 2), radius |h|.
2. Through (−2, 4): (h + 2)² + 4 = h², so 4h + 8 = 0, h = −2. Centre (−2, 2).
3. A diameter passes through the centre: 2(−2) − 3(2) + 10 = 0, so 2x − 3y + 10 = 0 is a diameter.

Teacher's Note:
1. Find the centre, then test which line passes through it.
2. Every line through the centre contains a diameter.

Answer: (a) 2

Explanation:
1. Let the centre be (h, 4h − 3). Equal distances: (h − 2)² + (4h − 6)² = (h − 4)² + (4h − 8)².
2. Expanding: −52h + 40 = −72h + 80, so h = 2. Centre (2, 5).
3. Radius = distance from (2, 5) to (2, 3) = 2.

Teacher's Note:
1. The centre is equidistant from every point on the circle.
2. The centre also lies on the perpendicular bisector of the chord joining (2, 3) and (4, 5), i.e. x + y = 7.

Answer: (b) 7

Explanation:
1. Hyperbola: \( \frac{x^2}{144/25} - \frac{y^2}{81/25} = 1 \), so c² = a² + b² = \( \frac{144 + 81}{25} = 9 \); foci (±3, 0).
2. Ellipse: c² = 16 − b² must also be 9.
3. Hence b² = 7.

Teacher's Note:
1. First write the hyperbola in standard form (right side 1) before reading off a² and b².
2. Ellipse uses c² = a² − b² and hyperbola uses c² = a² + b².

Answer: (c) d² + (2b + 3c)² = 0

Explanation:
1. Solving y² = 4ax and x² = 4ay gives the points (0, 0) and (4a, 4a).
2. (0, 0) on the line gives 4d = 0, so d = 0.
3. (4a, 4a) on the line gives 8ab + 12ac = 0, so 2b + 3c = 0.
4. Both conditions together: d² + (2b + 3c)² = 0.

Teacher's Note:
1. A sum of two squares is zero only when each square is zero, which is how two conditions are packed into one equation.
2. The two parabolas are mirror images in y = x, so their non-origin meeting point lies on y = x.

Answer: (b) 3x² + 4y² = 12

Explanation:
1. Directrix x = a/e = 4 with e = 1/2 gives a = 2, a² = 4.
2. b² = a²(1 − e²) = 4 × 3/4 = 3.
3. Ellipse: \( \frac{x^2}{4} + \frac{y^2}{3} = 1 \), i.e. 3x² + 4y² = 12.

Teacher's Note:
1. A directrix of the form x = constant tells you the major axis is along the x-axis.
2. For an ellipse the directrix x = a/e lies outside the curve, since a/e > a.

Answer: (d) y² − 4x + 2 = 0

Explanation:
1. Let the mid-point be (h, k); then Q = (2h − 1, 2k).
2. Q lies on y² = 8x: 4k² = 8(2h − 1), so k² = 4h − 2.
3. Locus: y² − 4x + 2 = 0.

Teacher's Note:
1. Express the moving point in terms of the mid-point, then substitute into the given curve.
2. The locus is again a parabola, with the same axis but half the size.

Answer: (b) \( \frac{1}{\sqrt{2}} \)

Explanation:
1. B = (0, b), F = (ae, 0), F′ = (−ae, 0).
2. Slopes of BF and BF′ are −b/(ae) and b/(ae); right angle gives b² = a²e².
3. a²(1 − e²) = a²e², so 2e² = 1 and \( e = \frac{1}{\sqrt{2}} \).

Teacher's Note:
1. Because OB is perpendicular to FF′ and bisects it, angle FBF′ = 90° means OB = OF, i.e. b = ae.
2. Using b = ae with b² = a² − a²e² gives the answer in one line.

Answer: (a) 3/5

Explanation:
1. 2ae = 6 gives ae = 3, and 2b = 8 gives b = 4.
2. a² = b² + a²e² = 16 + 9 = 25, so a = 5.
3. e = ae/a = 3/5.

Teacher's Note:
1. The numbers 3, 4, 5 form the familiar right triangle: the distance from a focus to an end of the minor axis is a.
2. Remember a² = b² + c² for an ellipse, where c = ae.

Answer: (b) 8/3

Explanation:
1. Distance from a focus to its corresponding directrix = \( \frac{a}{e} - ae \).
2. With e = 1/2: \( 2a - \frac{a}{2} = \frac{3a}{2} \).
3. This distance is 4, so \( \frac{3a}{2} = 4 \) and \( a = \frac{8}{3} \).

Teacher's Note:
1. The distance between a focus and its directrix does not depend on where the centre is, so the shift of origin does not matter.
2. Recall: focus at distance ae from centre, directrix at distance a/e on the same side.

Answer: (b) an ellipse of eccentricity \( \frac{\sqrt{3}}{2} \)

Explanation:
1. Let the ends be A(0, β) on the wall and B(α, 0) on the floor, with α² + β² = l².
2. P divides AB in 1 : 2 from A: \( P = \left(\frac{\alpha}{3}, \frac{2\beta}{3}\right) \), so α = 3x, β = 3y/2.
3. 9x² + \( \frac{9y^2}{4} \) = l², i.e. \( \frac{x^2}{l^2/9} + \frac{y^2}{4l^2/9} = 1 \).
4. Here the major axis is along y with \( \frac{b^2}{a^2} = \frac{1}{4} \), so \( e = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2} \).

Teacher's Note:
1. The semi-axes of the locus are the two parts of the ladder, l/3 and 2l/3.
2. For an ellipse with its major axis along the y-axis, take e² = 1 − (smaller denominator)/(larger denominator).

Answer: (c) L₁ < L₂

Explanation:
1. Substituting y² = 8x: x² + 8x − 9 = 0, so x = 1 (x = −9 is not on the parabola).
2. Then y = ±2√2, so L₁ = 4√2 ≈ 5.66.
3. Latus rectum of y² = 8x is L₂ = 8, hence L₁ < L₂.

Teacher's Note:
1. The common chord is a double ordinate, so its length is twice the |y|-value at the meeting points.
2. Discard any root that makes y² negative on the parabola.

Answer: (a) \( \frac{\sqrt{5}}{2} \)

Explanation:
1. Focus of y² = 6x is (3/2, 0); the chord is y = m(x − 3/2), i.e. 2mx − 2y − 3m = 0.
2. Distance from (0, 0): \( \frac{|3m|}{2\sqrt{1 + m^2}} = \frac{\sqrt{5}}{2} \).
3. 9m² = 5 + 5m², so m² = 5/4 and m = ±√5/2; option (a).

Teacher's Note:
1. A line through the focus with unknown slope is best written in point-slope form.
2. Squaring the distance equation removes both the modulus and the square root.

Answer: (b) x² = 2y

Explanation:
1. P(h, k) divides OQ in 1 : 3, so P = Q/4, i.e. Q = (4h, 4k).
2. Q lies on x² = 8y: 16h² = 32k, so h² = 2k.
3. Locus: x² = 2y.

Teacher's Note:
1. Points dividing segments from the vertex scale the parabola; the latus rectum shrinks by the same factor (8 to 2).
2. Always write the point on the curve in terms of the locus point before substituting.