JEE Mathematics Parabola MCQs Set 15

Answer: (a) 32

Explanation:
1. Let the vertices on the parabola be (±x, 12 − x²). Area A = 2x(12 − x²) = 24x − 2x³.
2. A' = 24 − 6x² = 0 gives x = 2 (a maximum, since A'' < 0).
3. A = 2 · 2 · 8 = 32.

Teacher's Note:
1. Express the area through the half-width x and set dA/dx = 0.

Answer: (d) \( \dfrac{125}{4} \)

Explanation:
1. P = (t² , 2t) with t = −2 and Q with t = 3. The area is largest where the tangent at X is parallel to PQ.
2. Slope of PQ = (6 + 4)/(9 − 4) = 2, and the slope of the tangent at (t², 2t) is 1/t, so t = 1/2 and X = (1/4, 1).
3. Area = ½|(9 − 4)(1 + 4) − (1/4 − 4)(6 + 4)| = ½(25 + 37.5) = 125/4.

Teacher's Note:
1. The farthest point of an arc from a chord is where the tangent is parallel to the chord.

Answer: (b) \( (1,1,3) \)

Explanation:
1. With slope m, the normals are y = m(x − c) − 2bm − bm³ and y = mx − 4am − 2am³. Equating: (b − 2a)m² + (c + 2b − 4a) = 0.
2. So m² = (4a − 2b − c)/(b − 2a), and a normal other than the axis exists only if this is positive.
3. For (1, 1, 3): m² = (4 − 2 − 3)/(1 − 2) = 1 > 0. For (1/2, 2, 0): −2; for (1/2, 2, 3): −5; for (1, 1, 0): −2: all negative.

Teacher's Note:
1. A common normal with non-zero slope needs m² > 0.

Answer: (b) 5

Explanation:
1. The vertex is (a², 0). On the y-axis (x = 0), y² = 4a², so y = ±2a.
2. The triangle has base 4a and height a², so its area is ½ · 4a · a² = 2a³.
3. 2a³ = 250 gives a³ = 125 and a = 5.

Teacher's Note:
1. Find the intercepts on the y-axis; the base is 4a and the height is a².

Answer: (a) \( (8,6) \)

Explanation:
1. The vertex is (2, 0) and the focal length is a = 4 − 2 = 2, so the parabola is y² = 8(x − 2).
2. (4, −4): 16 = 8 · 2 ✓. (5, 2√6): 24 = 8 · 3 ✓. (6, 4√2): 32 = 8 · 4 ✓.
3. (8, 6): 36 ≠ 8 · 6 = 48, so (8, 6) is not on the parabola.

Teacher's Note:
1. Write y² = 4a(x − h) from the vertex and focus, then test each point.

Answer: (d) √3 y = x + 3

Explanation:
1. A tangent to the parabola is y = mx + 1/m, i.e. mx − y + 1/m = 0. The circle has centre (3, 0) and radius 3.
2. Distance: |3m + 1/m|/√(m² + 1) = 3, so 9m² + 6 + 1/m² = 9m² + 9, giving m² = 1/3.
3. For m = 1/√3: y = x/√3 + √3, i.e. √3 y = x + 3.

Teacher's Note:
1. Use the condition that the distance from the circle's centre equals its radius.

Answer: (c) 6√3

Explanation:
1. Substituting y = (x + 4√2)/√2 in x² = 4y: x² = 2√2 x + 16, i.e. x² − 2√2 x − 16 = 0.
2. The roots differ by √(8 + 64) = 6√2, and the line has slope 1/√2.
3. Length = √(1 + 1/2) · 6√2 = √(3/2) · 6√2 = 6√3.

Teacher's Note:
1. Chord length = √(1 + m²) |x₁ − x₂|.

Answer: (b) \( x=y\cot\theta+2\tan\theta \)

Explanation:
1. For x² = 4ay with a = 2, the tangent of slope m is y = mx − am² = mx − 2m².
2. With m = tan θ: y = x tan θ − 2tan²θ.
3. Rewriting for x: x tan θ = y + 2tan²θ, so x = y cot θ + 2 tan θ.

Teacher's Note:
1. Use the slope form of the tangent to x² = 4ay.

Answer: (d) a parabola

Explanation:
1. The given circle has centre (4, 4) and radius 6. Let the required circle have centre (h, k) and radius |k| (it touches the x-axis).
2. External contact: √((h − 4)² + (k − 4)²) = 6 + |k|. For k > 0 this gives (h − 4)² = 20k + 20, and for k < 0 it gives (h − 4)² = 20 − 4k.
3. In both cases the centre lies on a parabola.

Teacher's Note:
1. Equate the distance between centres to the sum of the radii and simplify.

Answer: (d) x² = 2y

Explanation:
1. Let Q = (x₁, y₁) with x₁² = 8y₁. P = (x₁/4, y₁/4) = (h, k), so x₁ = 4h and y₁ = 4k.
2. 16h² = 32k, so h² = 2k.
3. The locus is x² = 2y.

Teacher's Note:
1. Use the section formula and eliminate the coordinates of Q.

Answer: (a) \( \dfrac{3\sqrt2}{8} \)

Explanation:
1. A point of the curve is (y², y), whose distance from x − y + 1 = 0 is |y² − y + 1|/√2.
2. y² − y + 1 has minimum 3/4 at y = 1/2.
3. Shortest distance = (3/4)/√2 = 3√2/8.

Teacher's Note:
1. Minimise the distance expression over the parameter y.

Answer: (c) x = −1

Explanation:
1. Tangents y = mx + 1/m through P(h, k): m²h − mk + 1 = 0. The product of the slopes is 1/h.
2. Perpendicular tangents need m₁m₂ = −1, so h = −1.
3. The locus is x = −1, the directrix.

Teacher's Note:
1. The product of slopes of the two tangents from (h, k) is a/h.

Answer: (c) \( (1,0) \)

Explanation:
1. The axis is the perpendicular from the focus to the directrix, i.e. the x-axis.
2. The vertex is midway between the focus (0, 0) and the foot (2, 0) of the directrix, so it is (1, 0).

Teacher's Note:
1. The vertex is the midpoint of the focus and the foot of the directrix.

Answer: (d) \( y^2-4x+2=0 \)

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

Teacher's Note:
1. Express Q in terms of the mid-point and substitute in the parabola.

Answer: (a) a parabola whose latus rectum is 2l

Explanation:
1. Let the fixed parabola be y² = 4ax and the moving one (y − k)² = −4a(x − h). At the point of contact, the slopes must be equal: if the fixed point has parameter t, the moving point has parameter −t.
2. Equal points: at² = h − at² gives h = 2at², and 2at = k − 2at gives k = 4at.
3. Eliminating t: k² = 16a²t² = 8ah, a parabola with latus rectum 8a = 2l.

Teacher's Note:
1. Use the tangent slope and common point to link h and k.

Answer: (a) \( d^2+(2b+3c)^2=0 \)

Explanation:
1. The parabolas intersect at (0, 0) and (4a, 4a).
2. Through (0, 0): 4d = 0, so d = 0. Through (4a, 4a): 8ab + 12ac + 4d = 0, so 2b + 3c = 0 (as a ≠ 0, d = 0).
3. Both conditions together give d² + (2b + 3c)² = 0.

Teacher's Note:
1. A sum of two squares vanishes only if both terms vanish.

Answer: (a) \( t_2=-t_1-\dfrac2{t_1} \)

Explanation:
1. The parabola is y² = 4bx. The slope of the normal at t₁ is −t₁, and the chord joining t₁ and t₂ has slope 2/(t₁ + t₂).
2. Equating: 2/(t₁ + t₂) = −t₁, so t₁ + t₂ = −2/t₁.
3. t₂ = −t₁ − 2/t₁.

Teacher's Note:
1. The normal at t₁ is a chord whose slope is −t₁.

Answer: (c) y² = x

Explanation:
1. P = (x/4, y/4) = (X, Y), so x = 4X and y = 4Y.
2. 16Y² = 4(4X), so Y² = X.
3. The locus is y² = x.

Teacher's Note:
1. Substitute x = 4X and y = 4Y in the parabola.

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

Explanation:
1. The vertex is (1, 1) and the focus (2, 2), so the focal length is a = √2. The directrix is perpendicular to the axis, through the point at distance a from the vertex on the far side from the focus: x + y = 0.
2. Focus-distance = directrix-distance: (x − 2)² + (y − 2)² = (x + y)²/2.
3. Simplifying: x² − 2xy + y² − 8x − 8y + 16 = 0, i.e. (x − y)² = 8(x + y − 2).

Teacher's Note:
1. Use the focus-directrix definition with the focus (2, 2) and the directrix x + y = 0.

Answer: (a) 20

Explanation:
1. Let B = (t², 2t). AB² = t⁴ + 4t² = 5 gives t² = 1, so take B = (1, 2) (t = 1).
2. Perpendicular chords from the vertex have parameters with t t' = −4, so t' = −4 and C = (16, −8).
3. AC = √(256 + 64) = 8√5, and the area is ½ · √5 · 8√5 = 20.

Teacher's Note:
1. For perpendicular chords from the vertex, t₁t₂ = −4.

Answer: (a) \( xy^2+2a(2a^2+y^2)=0 \)

Explanation:
1. If the normal at t₁ meets the curve again at t₂ = −t₁ − 2/t₁, the tangents at the ends meet at (x, y) = (at₁t₂, a(t₁ + t₂)).
2. x = a t₁(−t₁ − 2/t₁) = −a(t₁² + 2) and y = −2a/t₁, so t₁ = −2a/y.
3. x = −a(4a²/y² + 2), so xy² = −a(4a² + 2y²) = −2a(2a² + y²), i.e. xy² + 2a(2a² + y²) = 0.

Teacher's Note:
1. Use t₂ = −t₁ − 2/t₁ and the formula for the pole of the chord.

Answer: (a) \( y^2=a(x-3a) \)

Explanation:
1. The normal with slope m through (h, k): am³ + (2a − h)m + k = 0, with roots m₁, m₂, m₃ and m₁ + m₂ + m₃ = 0.
2. With m₁m₂ = −1 and m₁m₂m₃ = −k/a, we get m₃ = k/a. Then m₁m₂ + m₃(m₁ + m₂) = (2a − h)/a gives −1 − m₃² = (2a − h)/a.
3. So −a − k²/a = 2a − h, i.e. k² = a(h − 3a). The locus is y² = a(x − 3a).

Teacher's Note:
1. Use the sums and products of the roots of the cubic in m.

Answer: (a) \( (4a,0) \)

Explanation:
1. P = (at², 2at) and Q has parameter −4/t. The mid-point is (x, y) with x = (a/2)(t² + 16/t²) and y = a(t − 4/t).
2. y² = a²(t² − 8 + 16/t²), so t² + 16/t² = y²/a² + 8 and x = y²/(2a) + 4a.
3. y² = 2a(x − 4a): the vertex is (4a, 0).

Teacher's Note:
1. Eliminate t between the mid-point coordinates.

Answer: (a) \( \left(\dfrac p2,p\right) \) or \( \left(\dfrac p2,-p\right) \)

Explanation:
1. The focus is (p/2, 0) and the directrix x = −p/2, so the radius is p: (x − p/2)² + y² = p².
2. Substituting y² = 2px: x² + px − 3p²/4 = 0, so x = p/2 (the root −3p/2 gives no point).
3. Then y² = p², so y = ±p.

Teacher's Note:
1. The radius equals the distance from the focus to the directrix.