JEE Mathematics Ellipse MCQs Set 04

Answer: (c) \( \dfrac{1}{2x^2}+\dfrac{1}{4y^2}=1 \)

Explanation:
1. The ellipse is x²/2 + y² = 1. The tangent at (√2 cos θ, sin θ) is x cos θ/√2 + y sin θ = 1, with intercepts √2/cos θ and 1/sin θ.
2. The mid-point (x, y) = (1/(√2 cos θ), 1/(2 sin θ)), so cos²θ = 1/(2x²) and sin²θ = 1/(4y²).
3. Adding: 1/(2x²) + 1/(4y²) = 1.

Teacher's Note:
1. Write the tangent in parametric form, find the intercepts and eliminate θ.

Answer: (d) \( (4\sqrt3,2\sqrt2) \)

Explanation:
1. 2ae = 2b gives b = ae, so a² = b² + a²e² = 2b². The latus rectum 2b²/a = 8 gives b² = 4a.
2. Then a² = 2b² = 8a, so a = 8 and b² = 32: x²/64 + y²/32 = 1.
3. (4√3, 2√2): 48/64 + 8/32 = 3/4 + 1/4 = 1. The other points give 7/8, 7/8 and 9/8.

Teacher's Note:
1. Use b = ae to link a and b, then test the points.

Answer: (a) a < 4

Explanation:
1. Both denominators must be positive: 10 − a > 0 and 4 − a > 0, so a < 10 and a < 4.
2. The denominators are never equal (they differ by 6), so the ellipse is never a circle.
3. Hence a < 4.

Teacher's Note:
1. An ellipse needs both denominators positive.

Answer: (c) 4

Explanation:
1. The right angle at B means b = c (the semi-focal distance), so a = √2 b.
2. Area = ½ · 2c · b = b² = 8.
3. Latus rectum = 2b²/a = 16/(√2 · 2√2) = 16/4 = 4.

Teacher's Note:
1. A right angle at the end of the minor axis gives b = ae.

Answer: (a) 4x − 2y = 1

Explanation:
1. a/e = 4 with e = 1/2 gives a = 2, and b² = a²(1 − e²) = 3: x²/4 + y²/3 = 1.
2. The normal at (x₁, y₁) is a²x/x₁ − b²y/y₁ = a² − b²: 4x/1 − 3y/(3/2) = 1.
3. That is 4x − 2y = 1.

Teacher's Note:
1. Use the normal formula a²x/x₁ − b²y/y₁ = a² − b².

Answer: (d) 27

Explanation:
1. a = 3, b² = 5, e = 2/3, so the ends of the latera recta are (±2, ±5/3).
2. The tangent at (2, 5/3) is 2x/9 + y/3 = 1, meeting the axes at (9/2, 0) and (0, 3). By symmetry the quadrilateral is a rhombus with vertices (±9/2, 0), (0, ±3).
3. Area = ½ · 9 · 6 = 27.

Teacher's Note:
1. The four tangents form a rhombus with diagonals along the axes.

Answer: (c) \( \dfrac92 \)

Explanation:
1. At a common point, the slopes are 3/y (parabola) and −9x/(by) (ellipse).
2. Perpendicular: (3/y)(−9x/(by)) = −1, so 27x = by² = 6bx (using y² = 6x).
3. Hence b = 27/6 = 9/2.

Teacher's Note:
1. Set the product of the slopes at the common point equal to −1.

Answer: (c) \( (x^2+y^2)^2=6x^2+2y^2 \)

Explanation:
1. The ellipse is x²/6 + y²/2 = 1, so a² = 6 and b² = 2.
2. For the ellipse the foot of the perpendicular from the centre on a tangent lies on (x² + y²)² = a²x² + b²y².
3. So (x² + y²)² = 6x² + 2y².

Teacher's Note:
1. The locus is the pedal curve (x² + y²)² = a²x² + b²y².

Answer: (d) \( x^2+4y^2=16 \)

Explanation:
1. The first diameter is 2, so the semi-minor axis is 2. The second diameter is 4, so the semi-major axis is 4.
2. With the major axis along the x-axis: x²/16 + y²/4 = 1.
3. That is x² + 4y² = 16.

Teacher's Note:
1. The diameters of the circles are 2 and 4.

Answer: (d) \( 3x^2+5y^2-32=0 \)

Explanation:
1. e² = 1 − b²/a² = 2/5 gives b² = 3a²/5.
2. Passing through (−3, 1): 9/a² + 5/(3a²) = 1, so 32/(3a²) = 1 and a² = 32/3, b² = 32/5.
3. The ellipse is 3x²/32 + 5y²/32 = 1, i.e. 3x² + 5y² − 32 = 0.

Teacher's Note:
1. Express b² through a² using e, then substitute the point.

Answer: (a) \( x^2+12y^2=16 \)

Explanation:
1. The rectangle has vertices (±2, ±1).
2. The outer ellipse x²/16 + y²/b² = 1 (it passes through (4, 0)) contains (2, 1): 4/16 + 1/b² = 1, so b² = 4/3.
3. x²/16 + 3y²/4 = 1, i.e. x² + 12y² = 16.

Teacher's Note:
1. The corners of the rectangle must lie on the outer ellipse.

Answer: (b) \( \dfrac83 \)

Explanation:
1. For a point (x, 0) on the axis: |x| = ½|x − 4|.
2. x = ½(4 − x) gives x = 4/3, and −x = ½(4 − x) gives x = −4.
3. The vertices are 4/3 and −4, so 2a = 16/3 and a = 8/3.

Teacher's Note:
1. Find the two vertices on the axis using the focus-directrix property.

Answer: (d) \( \dfrac35 \)

Explanation:
1. 2c = 6 gives c = 3, and 2b = 8 gives b = 4.
2. a² = b² + c² = 25, so a = 5.
3. e = c/a = 3/5.

Teacher's Note:
1. Find a from a² = b² + c².

Answer: (d) 2ab sq. unit

Explanation:
1. Take the vertex (a cos θ, b sin θ). The rectangle has area 4ab sin θ cos θ = 2ab sin 2θ.
2. The maximum of sin 2θ is 1, so the area is 2ab.

Teacher's Note:
1. Use the parametric form of a vertex of the rectangle.

Answer: (d) \( \dfrac{1}{\sqrt2} \)

Explanation:
1. Angle FBF′ = 90° means OB = OF, so b = ae.
2. Then a² = b² + a²e² = 2a²e², so e² = 1/2.
3. e = 1/√2.

Teacher's Note:
1. In the right isosceles triangle FBF′, the median OB equals OF.

Answer: (c) a point, if k = 0

Explanation:
1. Completing the squares: 2x² − 8x = 2(x − 2)² − 8 and 3y² − 18y = 3(y − 3)² − 27, so the left side is 2(x − 2)² + 3(y − 3)² − 35 + 35 = 2(x − 2)² + 3(y − 3)². The equation becomes 2(x − 2)² + 3(y − 3)² = k.
2. For k > 0 this is an ellipse, for k = 0 it is the single point (2, 3), and for k < 0 there is no locus.
3. Only the statement 'a point, if k = 0' is true.

Teacher's Note:
1. A sum of positive multiples of squares equals k.

Answer: (c) 13

Explanation:
1. F₁F₂ is a diameter of the circle, so angle F₁PF₂ = 90°.
2. With PF₁ + PF₂ = 17 and PF₁² + PF₂² = (2c)²: the area ½ PF₁ · PF₂ = 30 gives PF₁ · PF₂ = 60. Then 4c² = 289 − 120 = 169.
3. 2c = 13.

Teacher's Note:
1. A right angle at P lets you use (PF₁ + PF₂)² = PF₁² + PF₂² + 2 PF₁ · PF₂.

Answer: (c) (0, 4) and (0, 1)

Explanation:
1. For the mid-point (h, k), T = S₁: hx/16 + ky/25 − 1 = h²/16 + k²/25 − 1. Through (0, 5): k/5 = h²/16 + k²/25.
2. So x²/16 + (y² − 5y)/25 = 0, i.e. x²/4 + (y − 5/2)²/(25/4) = 1: centre (0, 5/2), semi-axes 2 and 5/2.
3. c² = 25/4 − 4 = 9/4, so the foci are (0, 5/2 ± 3/2) = (0, 4) and (0, 1).

Teacher's Note:
1. Use T = S₁ for the mid-point of a chord.

Answer: (b) \( a^2l^2+b^2m^2=2n^2 \)

Explanation:
1. The points are (a cos θ, b sin θ) and (−a sin θ, b cos θ).
2. They lie on the line: la cos θ + mb sin θ = −n and −la sin θ + mb cos θ = −n.
3. Squaring and adding: a²l² + b²m² = 2n².

Teacher's Note:
1. Square and add the two conditions to eliminate θ.

Answer: (c) \( b^2 \)

Explanation:
1. The tangent at θ is bx cos θ + ay sin θ − ab = 0. With D = √(b²cos²θ + a²sin²θ), the perpendiculars from (ae, 0) and (−ae, 0) are ab(1 − e cos θ)/D and ab(1 + e cos θ)/D.
2. Their product is a²b²(1 − e²cos²θ)/D². Since D² = a² − (a² − b²)cos²θ = a²(1 − e²cos²θ), the product is b².
3. It is b² for every tangent.

Teacher's Note:
1. This product is the standard constant b².