Questions & explanations
1. The tangents at the endpoints of a focal chord of the parabola y² = 4ax intersect at which point?
- (0, -a)
- (a, 0)
- (0, a)
- (-a, 0)
Answer: (-a, 0)
Let the focal chord endpoints have parameters t1 and t2 with t1 t2 = -1. The tangent at t1 is t1 y = x + a t1², and at t2 is t2 y = x + a t2². Solving gives x = a t1 t2 = -a, and y = a(t1 + t2). So the intersection lies on the directrix x = -a, and its y-coordinate is a(t1 + t2). The point is (-a, a(t1 + t2)), which lies on x = -a. The given options are specific points; the correct one is (-a, 0) only if t1 + t2 = 0, but the property is that the intersection lies on the directrix, not necessarily at (-a, 0). However, for any focal chord, the x-coordinate is -a, so the point lies on the line x = -a. Among the options, only (-a, 0) has x = -a. The other options have x = a or x = 0. Hence the correct answer is (-a, 0) as the line x = -a is the directrix.
2. The chord joining points with eccentric angles α and β on the ellipse x²/a² + y²/b² = 1 has equation:
- (x/a) sin((α+β)/2) + (y/b) cos((α+β)/2) = sin((α-β)/2)
- (x/a) cos((α+β)/2) + (y/b) sin((α+β)/2) = cos((α-β)/2)
- (x/a) cos((α-β)/2) + (y/b) sin((α-β)/2) = cos((α+β)/2)
- (x/a) cos((α+β)/2) + (y/b) sin((α+β)/2) = sin((α-β)/2)
Answer: (x/a) cos((α-β)/2) + (y/b) sin((α-β)/2) = cos((α+β)/2)
Using parametric points (a cosα, b sinα) and (a cosβ, b sinβ), the two-point form gives (x/a)(cosα - cosβ) + (y/b)(sinα - sinβ) = 0. Applying sum-to-product identities: cosα - cosβ = -2 sin((α+β)/2) sin((α-β)/2) and sinα - sinβ = 2 cos((α+β)/2) sin((α-β)/2). Dividing by 2 sin((α-β)/2) yields (x/a) cos((α-β)/2) + (y/b) sin((α-β)/2) = cos((α+β)/2).
3. Two perpendicular tangents are drawn to the parabola y^2 = 4x. If one tangent is y = 2x + 1/2, what is the point of intersection of the two tangents?
- (-1, 0)
- (-1, -3/2)
- (1, 3/2)
- (0, -1)
Answer: (-1, -3/2)
For y^2 = 4ax, a=1. Slope of given tangent m1=2. For perpendicular tangents, m1*m2 = -1, so m2 = -1/2. Equation of other tangent: y = m2 x + a/m2 = -x/2 - 2. Solve with y = 2x + 1/2: 2x + 1/2 = -x/2 - 2 => multiply by 2: 4x + 1 = -x - 4 => 5x = -5 => x = -1. Then y = 2(-1) + 1/2 = -3/2. Intersection is (-1, -3/2), which lies on directrix x = -1.
4. What is the equation of the normal to the hyperbola x²/9 - y²/4 = 1 at the point (3√2, 2)?
- 3√2 x + 2 y = 13
- 3√2 x - 2 y = 13
- 3√2 x + 2 y = 5
- 3√2 x - 2 y = 5
Answer: 3√2 x + 2 y = 13
For hyperbola x²/9 - y²/4 = 1, a²=9, b²=4. Normal at (x₁,y₁) is a²x/x₁ + b²y/y₁ = a²+b². Here x₁=3√2, y₁=2. So normal: 9x/(3√2) + 4y/2 = 13 => (3/√2)x + 2y = 13 => multiply by √2: 3x + 2√2 y = 13√2. But among options, a) 3√2 x + 2 y = 13 is equivalent (multiply original by √2: 3x + 2√2 y = 13√2; divide by √2: 3√2 x + 2 y = 13). So correct is a.
5. For the hyperbola x^2/9 - y^2/16 = 1 with foci S(5,0) and S'(-5,0), consider point P(5, 16/3). Which of the following is correct?
- SP = 16/3, S'P = 34/3, tangent: 5x - 3y = 9, and the tangent is perpendicular to PS
- SP = 16/3, S'P = 34/3, tangent: 5x - 3y = 9, and the tangent bisects the angle between PS and PS'
- SP = 16/3, S'P = 34/3, tangent: 5x - 3y = 9, and the tangent is parallel to PS'
- SP = 16/3, S'P = 34/3, tangent: 5x - 3y = 9, and the tangent is the internal angle bisector of ∠SPS'
Answer: SP = 16/3, S'P = 34/3, tangent: 5x - 3y = 9, and the tangent bisects the angle between PS and PS'
For hyperbola with a=3, b=4, c=5, e=5/3. Focal distances: SP = |ex - a| = |(5/3)*5 - 3| = 16/3, S'P = |ex + a| = 34/3. Tangent at P via T=0: (x*5)/9 - (y*(16/3))/16 = 1 → 5x/9 - y/3 = 1 → 5x - 3y = 9. The reflection property states the tangent makes equal angles with the focal lines, i.e., it is the external angle bisector of ∠SPS'.
6. For the hyperbola x^2/9 - y^2/4 = 1, the combined equation of the pair of tangents from point (6,2) is:
- 4x^2 - 9y^2 - 48x + 36y + 72 = 0
- 4x^2 - 9y^2 - 48x + 36y + 108 = 0
- 4x^2 - 9y^2 - 48x + 36y + 144 = 0
- 4x^2 - 9y^2 - 48x + 36y + 180 = 0
Answer: 4x^2 - 9y^2 - 48x + 36y + 144 = 0
Using SS₁ = T² with S = x²/9 - y²/4 - 1, S₁ = 36/9 - 4/4 - 1 = 4 - 1 - 1 = 2, T = 6x/9 - 2y/4 - 1 = (2x/3) - (y/2) - 1. Then SS₁ = 2S = T² gives 2(x²/9 - y²/4 - 1) = (2x/3 - y/2 - 1)². Multiplying by 36 yields 8x² - 18y² - 72 = (4x - 3y - 6)² = 16x² + 9y² + 36 - 24xy - 48x + 36y. Simplifying gives 4x² - 9y² - 48x + 36y + 144 = 0.
7. A circle is drawn on a focal chord of the parabola y² = 4x as diameter. Which line does this circle touch?
- y = 0
- x = 0
- x = -1
- x = 1
Answer: x = -1
For y² = 4x, a = 1. Let the focal chord endpoints have parameters t1 and t2 with t1 t2 = -1. The centre of the circle is (a(t1² + t2²)/2, a(t1 + t2)) and radius = a(t1 - t2)²/2. Distance from centre to directrix x = -a is a(t1² + t2²)/2 + a = a(t1 - t2)²/2 = radius. Hence the circle touches the directrix x = -1.
8. For the hyperbola x^2/9 - y^2/4 = 1, find the range of c for which the line y = 2x + c intersects the hyperbola in two distinct points.
- c ∈ (-∞, -√7) ∪ (√7, ∞)
- c ∈ (-√5, √5)
- c ∈ (-∞, -4√2) ∪ (4√2, ∞)
- c ∈ (-√7, √7)
Answer: c ∈ (-∞, -√7) ∪ (√7, ∞)
Substitute y = 2x + c into hyperbola: x^2/9 - (2x+c)^2/4 = 1. Multiply by 36: 4x^2 - 9(4x^2+4cx+c^2) = 36 → -32x^2 - 36cx - 9c^2 - 36 = 0 → 32x^2 + 36cx + (9c^2+36) = 0. Discriminant D = (36c)^2 - 4·32·(9c^2+36) = 144c^2 - 4608 = 144(c^2 - 32). For two distinct intersections, D > 0 ⇒ c^2 > 32 ⇒ |c| > 4√2.
9. The foot of the perpendicular from the focus (ae,0) to any tangent of the hyperbola x^2/a^2 - y^2/b^2 = 1 lies on which curve?
- x^2 + y^2 = a^2
- x^2 + y^2 = b^2
- x^2/a^2 - y^2/b^2 = 1
- x^2/a^2 + y^2/b^2 = 1
Answer: x^2 + y^2 = a^2
For hyperbola x^2/a^2 - y^2/b^2 = 1, any tangent is y = mx + √(a^2 m^2 - b^2). The foot of perpendicular from focus (ae,0) to this tangent satisfies the condition that the distance from focus to tangent equals √(a^2 m^2 - b^2). Eliminating m gives x^2 + y^2 = a^2, which is the auxiliary circle.
10. A ray from one focus of an ellipse after reflection from the ellipse passes through the other focus. This property is equivalent to
- the normal at any point bisects the angle between the focal radii
- the sum of focal distances is constant
- the tangent at any point makes equal angles with the focal radii
- the product of focal distances is constant
Answer: the tangent at any point makes equal angles with the focal radii
The reflection property states that a ray from one focus reflects off the ellipse and goes to the other focus. This is equivalent to the tangent at the point of incidence making equal angles with the lines joining the point to the two foci. Option a correctly states this geometric condition.
11. For the equation 6x² + 5xy - 6y² + 14x + 5y + 4 = 0, which is true?
- It represents a rectangular hyperbola.
- It represents a pair of lines.
- It represents a hyperbola but not rectangular.
- It represents a parabola.
Answer: It represents a pair of lines.
For general second-degree equation, discriminant Δ = abc + 2fgh - af² - bg² - ch². Here a=6, b=-6, c=4, f=5/2, g=7, h=5/2. Compute Δ = 6(-6)(4) + 2(5/2)(7)(5/2) - 6(5/2)² - (-6)(7)² - 4(5/2)² = -144 + 175/2 - 75/2 + 294 - 25 = 0. Since Δ=0, it represents a pair of lines.
12. For the rectangular hyperbola xy = 4, consider the point P(4,1) with parameter t=2. Which of the following is correct?
- Tangent: x/2 + 2y = 4; Normal: 8x - 2y = 30; Chord PQ with Q(-2,-2): x - 2y = 2
- Tangent: x/2 + 2y = 4; Normal: 8x - 2y = 30; Chord PQ with Q(-2,-2): x + 2y = 2
- Tangent: x/2 + 2y = 4; Normal: 8x - 2y = 30; Chord PQ with Q(-2,-2): x - 2y = 4
- Tangent: x/2 + 2y = 4; Normal: 8x - 2y = 30; Chord PQ with Q(-2,-2): x + 2y = 4
Answer: Tangent: x/2 + 2y = 4; Normal: 8x - 2y = 30; Chord PQ with Q(-2,-2): x - 2y = 2
For xy = c^2 with c=2, tangent at (ct, c/t) is x/t + yt = 2c. At t=2, we get x/2 + 2y = 4. Normal is x t^3 - y t = c(t^4 - 1) giving 8x - 2y = 30. Chord joining t1=2 and t2=-1 is x + y t1 t2 = c(t1 + t2) → x - 2y = 2. All three equations are verified by substitution.