JEE Mathematics Parabola MCQs Set 16

Answer: (a) \( y^2=2a(x-a) \)

Explanation:
1. The centre is the mid-point of the focal chord with ends t and −1/t, namely (at², 2at) and (a/t², −2a/t): x = (a/2)(t² + 1/t²) and y = a(t − 1/t).
2. y² = a²(t² − 2 + 1/t²) = a²(2x/a − 2) = 2a(x − a).

Teacher's Note:
1. Take the focal chord with parameters t and −1/t.

Answer: (c) 6

Explanation:
1. A circle and a parabola can meet in at most 4 points.
2. Any two of the four points determine a chord, so the number of common chords is 4C2 = 6.

Teacher's Note:
1. Count the pairs of intersection points.

Answer: (c) \( a\in(-1,\,-5+2\sqrt6) \)

Explanation:
1. The smaller region lies inside the circle and inside the parabola (x > 0), so −2a > 0, i.e. a < 0.
2. Inside the circle: 4a² + (a + 1)² < 4, i.e. (5a − 3)(a + 1) < 0, so −1 < a < 3/5.
3. Inside the parabola: (a + 1)² < −8a, i.e. a² + 10a + 1 < 0, so −5 − 2√6 < a < −5 + 2√6 ≈ −0.1.
4. Intersection: −1 < a < −5 + 2√6.

Teacher's Note:
1. Combine the three conditions: x > 0, inside the circle, inside the parabola.

Answer: (b) \( \dfrac23 \)

Explanation:
1. Let P = (at², 2at), N = (at², 0). The line y = at meets the curve at Q = (at²/4, at).
2. NQ has slope at/(at²/4 − at²) = −4/(3t). It meets the tangent at the vertex (x = 0) at T with y = (−4/(3t))(0 − at²) = 4at/3.
3. AT = 4at/3 and NP = 2at, so k = 2/3.

Teacher's Note:
1. Use the parameter t for P and find T from the line NQ.

Answer: (b) GP

Explanation:
1. Take the parabola y² = 4ax with B and C having parameters t₁, t₂ and the other tangent x − ty + at² = 0. A = (at₁t₂, a(t₁ + t₂)).
2. The distances from the tangent: p₁ = a|t − t₁||t − t₂|/√(1 + t²), p₂ = a(t − t₁)²/√(1 + t²), p₃ = a(t − t₂)²/√(1 + t²).
3. So p₁² = p₂p₃: p₂, p₁, p₃ are in GP.

Teacher's Note:
1. Compute the three distances with the parameter form of the tangent.

Answer: (c) 24

Explanation:
1. For y² = 4Ax with A = −2, the normal of slope m is y = mx − 2Am − Am³ = mx + 4m + 2m³.
2. The line is y = −2x − λ, so m = −2 and −λ = 4(−2) + 2(−8) = −24.
3. λ = 24.

Teacher's Note:
1. Use the slope form of the normal.

Answer: (b) 2

Explanation:
1. For the normal chord, q = −p − 2/p.
2. OP ⟂ OQ gives pq = −4, so p(−p − 2/p) = −4, i.e. −p² − 2 = −4.
3. p² = 2.

Teacher's Note:
1. Use q = −p − 2/p together with the condition pq = −4.

Answer: (a) \( \tan^{-1}\sqrt2 \)

Explanation:
1. For y² = 4ax, the normal at t₁ has slope −t₁ and meets the curve again at t₂ = −t₁ − 2/t₁.
2. OP ⟂ OQ gives t₁t₂ = −4, so −t₁² − 2 = −4 and t₁² = 2.
3. The slope is −t₁ = ∓√2, so the acute angle is θ = tan⁻¹√2.

Teacher's Note:
1. The slope of the normal chord is −t₁.

Answer: (c) \( \left(\dfrac12,\dfrac12\right) \)

Explanation:
1. The equation is (x − y)² = 4(x + y − 1). The axis is x − y = 0 and the tangent at the vertex is x + y − 1 = 0.
2. The vertex is their intersection: x = y and x + y = 1, so x = y = 1/2.

Teacher's Note:
1. The vertex is where the axis meets the tangent at the vertex.

Answer: (b) 4

Explanation:
1. The circle has its centre (h, 0) on the axis and passes through the focus (1, 0): radius r = h − 1.
2. Touching the parabola: (x − h)² + 4x = r² must have a double root, i.e. (4 − 2h)² − 4(h² − r²) = 0, so r² = 4(h − 1).
3. (h − 1)² = 4(h − 1) gives h − 1 = 4, so r = 4.

Teacher's Note:
1. The largest such circle touches the parabola.

Answer: (d) hyperbola

Explanation:
1. At a point of contact the slopes are equal: from the first, dy/dx = 2a/y; from the second, dy/dx = x/(2a).
2. So 2a/y = x/(2a), i.e. xy = 4a².
3. The locus of the points of contact is the hyperbola xy = 4a².

Teacher's Note:
1. Equate the slopes at the point of contact.

Answer: (c) \( a^2\ge8b^2 \)

Explanation:
1. A tangent y = mx + a/m is a normal to x² = 4by at (2bp, bp²) if its slope is m = −1/p and its intercept is a/m = 2b + bp².
2. So −ap = 2b + bp², i.e. bp² + ap + 2b = 0.
3. Real p needs a² − 8b² ≥ 0, i.e. a² ≥ 8b².

Teacher's Note:
1. Write the normal to the second parabola in terms of its parameter.

Answer: (b) \( 4a(2-\sqrt3) \)

Explanation:
1. SA and SB make angles of 30° with the negative x-axis (the axis direction towards the vertex). In polar form from the focus, r = 2a/(1 − cos φ), with φ measured from the positive x-axis.
2. For φ = 150°: r = 2a/(1 + √3/2) = 4a/(2 + √3) = 4a(2 − √3).
3. The side equals SA = 4a(2 − √3).

Teacher's Note:
1. Use the focal polar equation of the parabola.

Answer: (a) \( x^2+y^2-20x+8y-12=0 \)

Explanation:
1. With a = 2, P = (2t², 4t) gives t = 1. The normal meets the parabola again at t₂ = −t − 2/t = −3, so Q = (18, −12).
2. The circle on PQ as diameter: (x − 2)(x − 18) + (y − 4)(y + 12) = 0.
3. Expanding: x² + y² − 20x + 8y − 12 = 0.

Teacher's Note:
1. The circle on the diameter PQ is (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0.

Answer: (d) \( (-2,-8) \)

Explanation:
1. The closest point has a tangent parallel to the line: 2x + 7 = 3, so x = −2.
2. y = 4 − 14 + 2 = −8, so the point is (−2, −8).

Teacher's Note:
1. The nearest point is where the tangent is parallel to the line.

Answer: (a) x = 0

Explanation:
1. (y − 2)² = 4(x + 1): a parabola with vertex (−1, 2), axis y = 2 (horizontal) and a = 1, so the focus is (0, 2).
2. The incident ray y = 4 is parallel to the axis, so the reflected ray passes through the focus. It hits the curve at (0, 4) (put y = 4: 16 = 4(x + 4), x = 0).
3. The reflected ray goes through (0, 4) and (0, 2): the line x = 0.

Teacher's Note:
1. Rays parallel to the axis reflect through the focus.

Answer: (c) \( 6\sqrt3\,a \)

Explanation:
1. As before, t₁² = 2 and t₂ = −t₁ − 2/t₁ = −2√2 (taking t₁ = √2).
2. P = (2a, 2√2a) and Q = (8a, −4√2a).
3. PQ = √(36a² + 72a²) = √108 a = 6√3 a.

Teacher's Note:
1. Find the parameters of the end points first.

Answer: (c) \( 5\left(2^{1/3}\right) \)

Explanation:
1. The vertex is (a², 0). On the y-axis, y² = 8a², so y = ±2√2 a and the base is 4√2 a.
2. Area = ½ · 4√2 a · a² = 2√2 a³ = 500√2, so a³ = 250.
3. a = 250^1/3 = 5 · 2^1/3.

Teacher's Note:
1. Base times height: the base is the chord on the y-axis, the height is a².

Answer: (a) 2√2 y = x + 8

Explanation:
1. The circle has centre (4, 0) and radius 4. The tangent y = mx + 1/m is at distance 4 from the centre: |4m + 1/m|/√(1 + m²) = 4.
2. 16m² + 8 + 1/m² = 16m² + 16, so m² = 1/8 and m = 1/(2√2).
3. y = x/(2√2) + 2√2, i.e. 2√2 y = x + 8.

Teacher's Note:
1. Distance from the centre = radius.

Answer: (a) \( \dfrac7{4\sqrt2} \)

Explanation:
1. A point of the curve is (t² + 2, t). Its distance from x − y = 0 is (t² − t + 2)/√2.
2. The minimum of t² − t + 2 is 7/4 at t = 1/2.
3. Shortest distance = 7/(4√2).

Teacher's Note:
1. Parametrise the curve and minimise the distance.

Answer: (c) \( \left(\dfrac34,\dfrac74\right) \)

Explanation:
1. x² + 4x = 5 gives x = 1 (x > 0), y = 2: the point (1, 2).
2. The tangent at (1, 2) to y² = 4x is 2y = 2(x + 1), i.e. y = x + 1.
3. Of the options only (3/4, 7/4) satisfies y = x + 1.

Teacher's Note:
1. Find the point of intersection and write the tangent at it.

Answer: (a) 25

Explanation:
1. Here a = 4. The end (1, 4) = (4t², 8t) has t = 1/2, and the other end of the focal chord has parameter −1/t = −2.
2. The length of a focal chord is a(t + 1/t)² = 4(1/2 + 2)² = 4 · 25/4 = 25.

Teacher's Note:
1. For a focal chord with parameters t and −1/t, the length is a(t + 1/t)².

Answer: (c) \( \left(\dfrac52,-1\right) \)

Explanation:
1. (x − 2)² − 1 = x − 3 gives x² − 5x + 6 = 0, so x = 2 or 3: points (2, −1) and (3, 0).
2. The tangent at (2, −1) has slope 2(x − 2) = 0: y = −1. At (3, 0) the slope is 2: y = 2(x − 3).
3. Solving: −1 = 2x − 6, so x = 5/2: the point (5/2, −1).

Teacher's Note:
1. Find the slope 2(x − 2) at each point.

Answer: (c) \( 64\sqrt2 \)

Explanation:
1. The vertices are (±x, 24 − x²), so the area is 2x(24 − x²) = 48x − 2x³.
2. Setting the derivative 48 − 6x² = 0 gives x = 2√2.
3. Area = 4√2(24 − 8) = 64√2.

Teacher's Note:
1. Maximise the area with respect to the half-width.