CBSE Class 11 Mathematics Conic Sections Competitive Exam MCQs Set 02

Answer: (a) \( \sqrt{a^2 + b^2 + p^2 + q^2} \)

Explanation:
1. Circle on AB as diameter: (x² + 2ax − b²) + (y² + 2py − q²) = 0.
2. So g = a, f = p, c = −b² − q².
3. Radius = \( \sqrt{g^2 + f^2 - c} = \sqrt{a^2 + p^2 + b^2 + q^2} \).

Teacher's Note:
1. (x − x₁)(x − x₂) equals the quadratic with roots x₁, x₂, so the circle is just the sum of the two quadratics.
2. Then read g, f and c and use r = √(g² + f² − c).

Answer: (a) \( 8\sqrt{3} \) sq. units

Explanation:
1. Circles: centres (2, 0) and (−2, 0), each of radius 4. The common region lies between x = −2 and x = 2 on the x-axis.
2. Subtracting the equations: 8x = 0, so the circles meet at x = 0, y² = 12, i.e. (0, ±2√3).
3. The rhombus has vertices (−2, 0), (2, 0), (0, 2√3), (0, −2√3); its diagonals are 4 and 4√3.
4. Area = \( \frac{1}{2} \times 4 \times 4\sqrt{3} = 8\sqrt{3} \) sq. units.

Teacher's Note:
1. Area of a rhombus = half the product of its diagonals.
2. The common region on the x-axis runs from the leftmost point of the right circle to the rightmost point of the left circle.

Answer: (b) x² − y² = a² − b²

Explanation:
1. Let the centre be (h, k) and radius r. AB is a chord of length 2a on the x-axis, at distance |k| from the centre.
2. So r² = a² + k². Similarly, CD on the y-axis gives r² = b² + h².
3. a² + k² = b² + h², i.e. h² − k² = a² − b².
4. Locus: x² − y² = a² − b².

Teacher's Note:
1. The perpendicular from the centre to a chord bisects it, so r² = (half chord)² + (distance)².
2. Equate the two expressions for r² to remove r.

Answer: (b) 4/3

Explanation:
1. Put (−3, 2) in y² = 4ax: 4 = 4a(−3), so a = −1/3.
2. Length of latus rectum = |4a| = 4 × 1/3.
3. Hence the latus rectum is 4/3.

Teacher's Note:
1. A negative value of a only means the parabola opens to the left; a length is always taken as |4a|.
2. Always find a first by substituting the given point, then use LR = 4|a|.

Answer: (a) x + 3 = 0

Explanation:
1. Compare y² = 12x with y² = 4ax: 4a = 12, so a = 3.
2. The directrix of y² = 4ax is x = −a.
3. Hence the directrix is x = −3, i.e. x + 3 = 0.

Teacher's Note:
1. For y² = 4ax the focus (a, 0) and the directrix x = −a lie on opposite sides of the vertex at equal distances.
2. The directrix is always perpendicular to the axis; here the axis is the x-axis, so the directrix is a vertical line.

Answer: (c) y = −3

Explanation:
1. Compare x² = −12y with x² = −4ay: 4a = 12, so a = 3.
2. The focus of x² = −4ay is (0, −a) = (0, −3).
3. The latus rectum is the line through the focus perpendicular to the axis (y-axis), so it is y = −3.

Teacher's Note:
1. The latus rectum passes through the focus, so first locate the focus correctly with its sign.
2. For x² = −4ay the parabola opens downwards, so the focus and latus rectum lie below the x-axis.

Answer: (b) 2

Explanation:
1. The latus rectum fixes the focus (its mid-point) and its length 4a, so a is fixed.
2. The axis must be the line through the focus perpendicular to the latus rectum.
3. The vertex can lie at a distance a from the focus on either side along this axis, giving 2 parabolas.

Teacher's Note:
1. Two parabolas are equal when their latus recta are equal; here both parabolas share the same latus rectum but open in opposite directions.
2. Draw a quick sketch: the two curves are mirror images in the line of the latus rectum.

Answer: (d) x = 0

Explanation:
1. For a point P(x, y) on the parabola, distance from focus = distance from directrix x = −a.
2. So SP = x + a, where x ≥ 0 for points on y² = 4ax.
3. SP is least when x = 0, i.e. at the vertex.

Teacher's Note:
1. Using the definition (focal distance = distance from directrix) turns a distance problem into a simple linear expression.
2. The vertex is the point of the parabola closest to both the focus and the directrix.

Answer: (b) \( \frac{\pi}{2} \)

Explanation:
1. Let the double ordinate be PQ with P(h, 8p) and Q(h, −8p).
2. P lies on y² = 8px: 64p² = 8ph, so h = 8p; P = (8p, 8p), Q = (8p, −8p).
3. Slopes of OP and OQ are 1 and −1; their product is −1.
4. Hence the angle at the vertex is \( \frac{\pi}{2} \).

Teacher's Note:
1. A double ordinate is bisected by the axis, so its ends have ordinates ±(half the length).
2. Here the double ordinate is twice the latus rectum (LR = 8p); such a double ordinate always subtends a right angle at the vertex.

Answer: (b) 4a tan α sec α

Explanation:
1. The chord through the vertex is y = x tan α.
2. Put in x² = 4ay: x² = 4ax tan α, so the other end has x = 4a tan α, y = 4a tan² α.
3. Length = \( \sqrt{x^2 + y^2} = 4a\tan\alpha\sqrt{1 + \tan^2\alpha} = 4a\tan\alpha\sec\alpha \).

Teacher's Note:
1. A chord through the vertex starts at the origin, so its length is just the distance of the other end from O.
2. Use 1 + tan² α = sec² α to simplify the square root.

Answer: (a) a < 4

Explanation:
1. \( \frac{x^2}{A} + \frac{y^2}{B} = 1 \) is an ellipse when both A and B are positive.
2. 10 − a > 0 gives a < 10 and 4 − a > 0 gives a < 4.
3. Both hold when a < 4.

Teacher's Note:
1. If exactly one denominator is negative the equation becomes a hyperbola; here 4 < a < 10 gives a hyperbola.
2. Since 10 − a ≠ 4 − a for every a, the curve is never a circle.

Answer: (a) ellipse

Explanation:
1. Take the fixed lines as the axes; let the ends be A(α, 0) and B(0, β) with α² + β² = (a + b)².
2. Let P(h, k) divide AB so that PA = a and PB = b; then h = bα/(a + b), k = aβ/(a + b).
3. So α = h(a + b)/b and β = k(a + b)/a; substituting gives \( \frac{h^2}{b^2} + \frac{k^2}{a^2} = 1 \).
4. The locus \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \) is an ellipse.

Teacher's Note:
1. This is the familiar sliding-rod (ladder) problem; the point on the rod traces an ellipse with semi-axes equal to the two parts of the rod.
2. When the point is the mid-point (a = b) the locus becomes a circle, a special case of the ellipse.

Answer: (d) \( \sqrt{3}/2 \)

Explanation:
1. Latus rectum = \( \frac{2b^2}{a} \) and minor axis = 2b.
2. \( \frac{2b^2}{a} = \frac{1}{2}(2b) \) gives a = 2b, so \( \frac{b^2}{a^2} = \frac{1}{4} \).
3. \( e^2 = 1 - \frac{b^2}{a^2} = \frac{3}{4} \), so \( e = \frac{\sqrt{3}}{2} \).

Teacher's Note:
1. Convert every condition into a relation between a and b first, then use e² = 1 − b²/a².
2. A quick check: e must lie between 0 and 1 for an ellipse, and √3/2 ≈ 0.87 does.

Answer: (b) x² + 2y² = 100

Explanation:
1. Minor axis = distance between foci: 2b = 2ae, so b = ae and b² = a²e² = a² − b², i.e. a² = 2b².
2. Latus rectum \( \frac{2b^2}{a} = 10 \) gives \( \frac{a^2}{a} = 10 \), so a = 10.
3. a² = 100 and b² = 50, so the ellipse is \( \frac{x^2}{100} + \frac{y^2}{50} = 1 \), i.e. x² + 2y² = 100.

Teacher's Note:
1. Remember b² = a² − c² where c = ae is the distance of a focus from the centre.
2. Since the major axis is along the x-axis, the larger denominator must be under x².

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

Explanation:
1. \( \frac{2b^2}{a} = \frac{1}{3}(2a) \) gives \( b^2 = \frac{a^2}{3} \).
2. \( e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{1}{3} = \frac{2}{3} \).
3. Hence \( e = \sqrt{2/3} \).

Teacher's Note:
1. The major axis is 2a, not a; missing this factor of 2 is the most common slip here.
2. Write the condition, cancel a, and you get b²/a² directly.

Answer: (a) x²/9 + y²/8 = 16

Explanation:
1. Foci (±4, 0) give ae = 4; with e = 1/3, a = 12 and a² = 144.
2. b² = a² − a²e² = 144 − 16 = 128.
3. Ellipse: \( \frac{x^2}{144} + \frac{y^2}{128} = 1 \); multiplying by 16 gives \( \frac{x^2}{9} + \frac{y^2}{8} = 16 \).

Teacher's Note:
1. The options are written in a scaled form, so reduce your answer by a common factor before comparing.
2. Foci on the x-axis tell you the major axis is along the x-axis.

Answer: (c) \( \frac{1}{2} \)

Explanation:
1. SB = \( \sqrt{a^2e^2 + b^2} = \sqrt{a^2e^2 + a^2 - a^2e^2} = a \).
2. ST = distance between the foci = 2ae.
3. Equilateral triangle: 2ae = a, so e = 1/2.

Teacher's Note:
1. The distance from an end of the minor axis to either focus is always equal to a (since SB + S′B = 2a and SB = S′B).
2. Using this fact turns many focus–minor-axis questions into one-line answers.

Answer: (d) Hyperbola

Explanation:
1. From the first line: \( \sqrt{3}x - y = 4\sqrt{3}k \).
2. From the second line: \( \sqrt{3}x + y = \frac{4\sqrt{3}}{k} \).
3. Multiplying to remove k: 3x² − y² = 48, i.e. \( \frac{x^2}{16} - \frac{y^2}{48} = 1 \), a hyperbola.

Teacher's Note:
1. To find a locus, eliminate the parameter; here multiplying the two equations removes k at once.
2. An equation of the form x²/A − y²/B = 1 with A, B > 0 is a hyperbola.

Answer: (d) none of these

Explanation:
1. ae = 2 and e = 3/2 give a = 4/3, a² = 16/9.
2. b² = a²(e² − 1) = \( \frac{16}{9} \times \frac{5}{4} = \frac{20}{9} \).
3. Hyperbola: \( \frac{x^2}{16/9} - \frac{y^2}{20/9} = 1 \), i.e. \( \frac{x^2}{4} - \frac{y^2}{5} = \frac{4}{9} \).
4. This matches none of (a), (b), (c), so the answer is (d).

Teacher's Note:
1. Option (a) reduces to x²/(16/9) − y²/4 = 1, whose foci are (±2√13/3, 0), not (±2, 0).
2. Always check a candidate equation by computing its foci and eccentricity rather than by appearance.

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

Explanation:
1. \( \frac{2b^2}{a} = \frac{1}{2}(2a) \) gives 2b² = a².
2. With b² = a²(e² − 1): 2a²(e² − 1) = a², so e² = 3/2.
3. Hence \( e = \sqrt{3/2} \).

Teacher's Note:
1. For a hyperbola e > 1, so √3/2 (less than 1) can be rejected straight away.
2. Use e² = 1 + b²/a² for hyperbolas and e² = 1 − b²/a² for ellipses; mixing them is a frequent error.

Answer: (b) 3x² − y² − 12 = 0

Explanation:
1. For the ellipse, ae = \( \sqrt{25 - 9} = 4 \), so the foci are (±4, 0).
2. For the hyperbola, ae = 4 with e = 2 gives a = 2, a² = 4, and b² = a²(e² − 1) = 12.
3. Hyperbola: \( \frac{x^2}{4} - \frac{y^2}{12} = 1 \), i.e. 3x² − y² − 12 = 0.

Teacher's Note:
1. For an ellipse c² = a² − b², while for a hyperbola c² = a² + b²; here c = 4 for both curves.
2. Check: for 3x² − y² = 12, a² + b² = 4 + 12 = 16, so the foci are indeed (±4, 0).

Answer: (a) \( 8\sqrt{3} \)

Explanation:
1. By symmetry the other two vertices are (h, k) and (h, −k), and each side through O makes 30° with the axis.
2. So k = h tan 30° = h/√3, and (h, k) lies on y² = 4x: h²/3 = 4h, giving h = 12, k = 4√3.
3. Side = 2k = \( 8\sqrt{3} \).

Teacher's Note:
1. The triangle is symmetric about the axis of the parabola, so the side opposite the vertex is a double ordinate.
2. In general, for y² = 4ax the side is 8√3 a; here a = 1.

Answer: (b) |t| ≤ 1

Explanation:
1. Put x = at² in the ellipse: \( y^2 = b^2(1 - t^4) \).
2. Real points need y² ≥ 0, i.e. t⁴ ≤ 1.
3. Hence |t| ≤ 1.

Teacher's Note:
1. Every point of the ellipse has −a ≤ x ≤ a, so a vertical line x = c meets it only when |c| ≤ a.
2. At |t| = 1 the line touches the ellipse at a vertex (one repeated point), which still counts as real.

Answer: (d) 4a

Explanation:
1. By definition SP = PM; for an equilateral triangle we also need SM = PM.
2. Let P = (x, y): PM = x + a and M = (−a, y), so SM² = 4a² + y² = 4a² + 4ax.
3. (x + a)² = 4a² + 4ax gives x² − 2ax − 3a² = 0, so x = 3a.
4. SP = x + a = 4a.

Teacher's Note:
1. SP = PM is automatic for any point on a parabola, so only one more condition (SM = PM) is needed.
2. Using focal distance SP = x + a avoids square roots completely.

Answer: (a) (x − 2)² + y² = 9

Explanation:
1. For y² = 8x the vertex is (0, 0) and the focus is (2, 0).
2. The common chord is equidistant from them, so it is x = 1; it meets the parabola at (1, ±2√2).
3. Radius² = (1 − 2)² + (2√2)² = 1 + 8 = 9.
4. Circle: (x − 2)² + y² = 9.

Teacher's Note:
1. A common chord of a circle centred on the axis and the parabola is perpendicular to the axis, i.e. a vertical line.
2. Find the radius by computing the distance from the centre to any point where the curves meet.

Answer: (c) \( \frac{2A}{a} \)

Explanation:
1. QR passes through the focus F(a, 0); split triangle PQR along PF (which lies on the x-axis).
2. Area = \( \frac{1}{2} \cdot a \cdot |y_Q| + \frac{1}{2} \cdot a \cdot |y_R| = \frac{a}{2}|y_Q - y_R| \), since Q and R lie on opposite sides of the axis.
3. So \( A = \frac{a}{2}|y_Q - y_R| \), giving \( |y_Q - y_R| = \frac{2A}{a} \).

Teacher's Note:
1. Splitting the triangle along the axis gives two triangles with the common base PF = a.
2. The ends of a focal chord always lie on opposite sides of the axis, which is why the two heights add up.

Answer: (d) \( \left(\frac{2 + \sqrt{2}}{2}\right)a^2 \)

Explanation:
1. Diameter = \( \frac{3}{4}(4a) = 3a \), so the circle is \( x^2 + y^2 = \frac{9a^2}{4} \).
2. With y² = 4ax: \( x^2 + 4ax - \frac{9a^2}{4} = 0 \), so x = a/2 and y = ±√2 a; PQ = 2√2 a.
3. L₁L₂ = 4a on x = a, and the distance between x = a/2 and x = a is a/2.
4. Area = \( \frac{1}{2}(2\sqrt{2}a + 4a) \cdot \frac{a}{2} = \left(\frac{2 + \sqrt{2}}{2}\right)a^2 \).

Teacher's Note:
1. Reject the root x = −9a/2, because points of y² = 4ax (a > 0) must have x ≥ 0.
2. Both parallel sides of the trapezium are double ordinates, so their lengths come straight from the y-values.

Answer: (b) 12 sq. unit

Explanation:
1. QR = \( \sqrt{4^2 + (2\sqrt{5})^2} = \sqrt{36} = 6 \), so PQ + PR = 16 − 6 = 10.
2. So P lies on an ellipse with foci Q, R: 2a = 10, 2c = 6, giving b = \( \sqrt{25 - 9} = 4 \).
3. Area is greatest when P is at an end of the minor axis (height b = 4): area = \( \frac{1}{2} \times 6 \times 4 = 12 \).

Teacher's Note:
1. A constant sum of distances from two fixed points is exactly the definition of an ellipse.
2. With the base fixed as QR, the area is maximum when the height of P above QR is greatest, i.e. at the end of the minor axis.

Answer: (a) \( \left(\frac{p}{2}, p\right) \); (b) \( \left(\frac{p}{2}, -p\right) \)

Explanation:
1. For y² = 2px, 4a = 2p so a = p/2: focus (p/2, 0), directrix x = −p/2.
2. The circle touches the directrix, so its radius is the distance p: \( \left(x - \frac{p}{2}\right)^2 + y^2 = p^2 \).
3. With y² = 2px: x² + px − \( \frac{3p^2}{4} \) = 0, so x = p/2 (x = −3p/2 gives y² < 0).
4. Then y² = p², so the points are (p/2, p) and (p/2, −p).

Teacher's Note:
1. The circle and parabola meet at the ends of the latus rectum, because the radius p equals the semi-latus rectum 2a.
2. Points with negative x cannot lie on y² = 2px when p > 0, so (c) and (d) are ruled out at once.

Answer: (a) 16a²

Explanation:
1. Let OA have slope m; it meets y² = 4ax at A = (4a/m², 4a/m), so \( r_1 = \frac{4a\sqrt{1 + m^2}}{m^2} \).
2. OB has slope −1/m, so similarly \( r_2 = 4am\sqrt{1 + m^2} \) (taking m > 0).
3. Then \( r_1r_2 = \frac{16a^2(1 + m^2)}{m} \) and \( r_1^{2/3} + r_2^{2/3} = (4a)^{2/3}(1 + m^2)^{1/3} \cdot \frac{1 + m^2}{m^{4/3}} \).
4. Dividing, \( \frac{(r_1r_2)^{4/3}}{r_1^{2/3} + r_2^{2/3}} = \frac{(16a^2)^{4/3}}{(4a)^{2/3}} = (4a)^2 = 16a^2 \).

Teacher's Note:
1. Find where a line through the vertex meets the parabola by substituting y = mx; the non-zero root gives the other end.
2. Testing a simple case such as m = 1 (r₁ = r₂ = 4√2 a) quickly confirms the value 16a².

Answer: (a) \( \sqrt{386}/12 \); (d) \( \sqrt{386}/38 \)

Explanation:
1. Distances of the origin from the foci: \( \sqrt{25 + 144} = 13 \) and \( \sqrt{576 + 49} = 25 \).
2. Distance between the foci: \( 2c = \sqrt{19^2 + 5^2} = \sqrt{386} \).
3. If the conic is an ellipse, 2a = 13 + 25 = 38, so e = 2c/2a = √386/38.
4. If it is a hyperbola, 2a = 25 − 13 = 12, so e = √386/12; both (a) and (d) are correct.

Teacher's Note:
1. Ellipse: sum of focal distances = 2a; hyperbola: difference of focal distances = 2a.
2. In both cases e = (distance between foci) ÷ (2a).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. x = 8 meets y² = 8x at P(8, 8) and Q(8, −8); slopes of OP and OQ are 1 and −1, so ∠POQ = 90°. Assertion is true.
2. For y² = 4ax, LR = 4a; the double ordinate x = 4a has ends (4a, ±4a), whose slopes from O are ±1, so it subtends a right angle at the vertex. Reason is true.
3. Here LR = 8 and PQ = 16 = 2 × LR, so the Reason directly explains the Assertion.

Teacher's Note:
1. The double ordinate x = 4a (twice the latus rectum in length) is a handy special chord to remember.
2. In assertion–reason questions, check whether the Reason is the general rule of which the Assertion is a special case.

Answer: (c) A is true, but R is false.

Explanation:
1. BC fixed and AB + BC + CA constant means AB + AC is constant (greater than BC).
2. So A moves on an ellipse with B and C as foci; the Assertion is true.
3. The Reason is false in general: if the constant sum equals the distance between the points the locus is a line segment, and if it is smaller there is no locus.

Teacher's Note:
1. An ellipse needs the sum of distances to be greater than the distance between the two fixed points.
2. In a triangle AB + AC > BC always holds, which is why the Assertion works.