JEE Mathematics Hyperbola MCQs Set 05

Answer: (a) \( (6,5\sqrt2) \)

Explanation:
1. a = 2 and c = 3, so b² = 5: x²/4 − y²/5 = 1.
2. (6, 5√2): 9 − 10 = −1 ≠ 1. The others give (2√6, 5): 6 − 5 = 1; (−6, 2√10): 9 − 8 = 1; (4, √15): 4 − 3 = 1.
3. So (6, 5√2) does not lie on it.

Teacher's Note:
1. Find a, b from the vertices and focus, then test the points.

Answer: (d) x − y + 1 = 0

Explanation:
1. The hyperbola is x²/5 − y²/4 = 1. For slope m = 1, c² = a²m² − b² = 5 − 4 = 1.
2. So the tangents are y = x ± 1, i.e. x − y ± 1 = 0.
3. Only x − y + 1 = 0 is among the options.

Teacher's Note:
1. Use c² = a²m² − b².

Answer: (d) \( \dfrac{2}{\sqrt3} \)

Explanation:
1. a = 2, so 16/4 − 4/b² = 1, giving b² = 4/3.
2. e² = 1 + b²/a² = 1 + 1/3 = 4/3.
3. e = 2/√3.

Teacher's Note:
1. Substitute the point to find b².

Answer: (c) \( \dfrac{13}{12} \)

Explanation:
1. 2b = 5 gives b = 5/2, and 2c = 13 gives c = 13/2.
2. a² = c² − b² = 169/4 − 25/4 = 36, so a = 6.
3. e = c/a = 13/12.

Teacher's Note:
1. Use c² = a² + b².

Answer: (a) \( (2\sqrt2,3\sqrt3) \)

Explanation:
1. With c = 2, a² + b² = 4. Through P: 2/a² − 3/b² = 1 gives a⁴ − 9a² + 8 = 0, so a² = 1 (a² = 8 is too large) and b² = 3: x² − y²/3 = 1.
2. The tangent at (√2, √3) is √2 x − (√3/3) y = 1.
3. For (2√2, 3√3): 4 − 3 = 1 ✓.

Teacher's Note:
1. Find the hyperbola first, then test each point on the tangent.

Answer: (c) \( \dfrac{2}{\sqrt3} \)

Explanation:
1. 2b = c (half of 2c), so c = 2b and a² = c² − b² = 3b².
2. Latus rectum 2b²/a = 8 gives b² = 4a, and a² = 3b² = 12a, so a = 12 and b² = 48.
3. e = c/a = 2b/a = 2√48/12 = 2/√3.

Teacher's Note:
1. Use c = 2b to relate a and b.

Answer: (d) \( 45\sqrt5 \)

Explanation:
1. The chord of contact PQ from T(0, 3) is 4x·0 − 3y = 36, i.e. y = −12.
2. On y = −12: 4x² = 36 + 144, so x² = 45 and x = ±3√5, giving PQ = 6√5.
3. The distance from T to the chord is 15, so the area is ½ · 6√5 · 15 = 45√5.

Teacher's Note:
1. Find the chord of contact first.

Answer: (c) y = 6/x

Explanation:
1. If the tangent at (x, y) meets the axes at (2x, 0) and (0, 2y), its slope is −2y/(2x) = −y/x.
2. So dy/dx = −y/x, giving xy = constant.
3. Through (2, 3): xy = 6, so y = 6/x.

Teacher's Note:
1. The intercepts are twice the coordinates of the point of contact.

Answer: (c) 3x² − y² = 3

Explanation:
1. c = 2 and e = 2, so a = 1 and b² = c² − a² = 3.
2. x² − y²/3 = 1, i.e. 3x² − y² = 3.

Teacher's Note:
1. a = c/e and b² = c² − a².

Answer: (d) Abscissa of foci

Explanation:
1. Here a² = cos²α and b² = sin²α, so c² = a² + b² = 1 and the foci are (±1, 0) for every α.
2. The eccentricity 1/cos α, the vertices (±cos α, 0) and the directrices x = ± cos²α all change with α.

Teacher's Note:
1. Compute c² = a² + b².

Answer: (a) a hyperbola

Explanation:
1. Tangency of y = mx + c needs c² = a²m² − b², so β² = a²α² − b².
2. The locus is a²x² − y² = b², which is a hyperbola.

Teacher's Note:
1. Apply c² = a²m² − b² with m = α, c = β.

Answer: (c) a hyperbola, if 8 < λ < 12

Explanation:
1. For λ < 8 both denominators are positive, so the curve is an ellipse (for example λ = 0).
2. For 8 < λ < 12 the first denominator is positive and the second negative, so it is a hyperbola.
3. Only the statement 'a hyperbola, if 8 < λ < 12' is true.

Teacher's Note:
1. A hyperbola needs denominators of opposite signs.

Answer: (c) 2xy − 4x + 4y + 1 = 0

Explanation:
1. (x − 1)² + (y + 1)² = 2 · (x − y + 1)²/2 = (x − y + 1)².
2. x² − 2x + y² + 2y + 2 = x² + y² + 1 − 2xy + 2x − 2y.
3. Simplifying: 2xy − 4x + 4y + 1 = 0.

Teacher's Note:
1. Use focus-directrix: distance to focus = e × distance to directrix.

Answer: (b) \( \dfrac{(x-5)^2}{25}-\dfrac{y^2}{144}=1 \)

Explanation:
1. The centre is (5, 0) and a = 5. The focus is at distance c = 13 from the centre, so b² = 169 − 25 = 144.
2. The equation is (x − 5)²/25 − y²/144 = 1.

Teacher's Note:
1. The centre is the mid-point of the vertices.

Answer: (d) a hyperbola

Explanation:
1. Multiplying the two equations: x²/a² − y²/b² = λ · (1/λ) = 1.
2. This is a hyperbola.

Teacher's Note:
1. Multiply the equations to eliminate λ.

Answer: (b) \( y=3\sqrt{\dfrac27}x+\dfrac{15}{\sqrt7} \)

Explanation:
1. The hyperbola is x²/16 − y²/9 = 1. A tangent y = mx + c needs c² = 16m² − 9 and, for the circle, c² = 9(1 + m²).
2. 16m² − 9 = 9 + 9m² gives m² = 18/7, so c² = 9 · 25/7 and c = 15/√7.
3. The tangent is y = 3√(2/7) x + 15/√7.

Teacher's Note:
1. Equate the two expressions for c².

Answer: (b) \( a^2\cos^2\alpha-b^2\sin^2\alpha=p^2 \)

Explanation:
1. Write the line as y = −cot α · x + p/sin α, so m = −cot α and c = p/sin α.
2. c² = a²m² − b²: p²/sin²α = a² cot²α − b².
3. Multiplying by sin²α: p² = a² cos²α − b² sin²α.

Teacher's Note:
1. Convert the line to slope form.