Let E1: (x2/9) + (y2/4) = 1 be an ellipse. Ellipses Ei 's are constructed such that their centres and eccentricities…

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More Conic Sections Previous-Year Questions — Page 8

Q59 jee_main_2025_29_jan_evening Chord with a Given Midpoint

α x + β y = 109

is the equation of the chord of the ellipse

(x²)/(9) +(y²)/(4) = 1

, whose mid point is

((5)/(2),(1)/(2))

, then

α +β

is equal to

A. $37$ $37$
B. $46$ $46$
C. $58$ $58$
D. $72$ $72$

Solution

Related Formula

Equation of a chord of a conic section with a given midpoint $(x₁, y₁)$ $(x_1, y_1)$ is:

$T = S₁$ $T = S_1$

Core Logic

Given midpoint $M((5)/(2), (1)/(2))$ and ellipse $(x²)/(9) + (y²)/(4) = 1$ $\frac{x^2}{9} + \frac{y^2}{4} = 1$.

Chord with a Given Midpoint diagram for Q59 - JEE Main 2025 Evening

Write $T$ $T$ and $S₁$ $S_1$ terms:

T: (x((5)/(2)))/(9) + (y((1)/(2)))/(4)

S₁: (((5)/(2))²)/(9) + (((1)/(2))²)/(4)

Equating both sides:

(5x)/(18) + (y)/(8) = (25)/(36) + (1)/(16)

Step 1: Simplify to Standard Form

Multiply the entire equation by 144 to eliminate fractions:

144((5x)/(18)) + 144((y)/(8)) = 144((25)/(36)) + 144((1)/(16))

40x + 18y = 4(25) + 9(1)

$40x + 18y = 109$ $40x + 18y = 109$

Comparing this directly with $α x + β y = 109$ $\alpha x + \beta y = 109$ provides:

α = 40, β = 18

α + β = 40 + 18 = 58

Pattern Recognition

Whenever you see 'chord whose midpoint is given', write $T = S₁$ $T = S_1$ automatically. Match coefficients directly at the final step after equating constant integers.

Chapter Mix

Class 11 Mathematics: Conic Sections

Q75 jee_main_2025_29_jan_evening Properties of Focal Chords

Let

y² = 12x

the parabola and

S

be its focus. Let

PQ

be a focal chord of the parabola such that (SP)

(SQ) = (147)/(4)

. Let

C

be the circle described taking

PQ

as a diameter. If the equation of a circle

C

64x² + 64y² - α x - 64\sqrt(3)y = β

, then \beta - \alpha is equal to

Numerical Answer. Answer: 1328 to 1328

Solution

Related Formula

Properties of focal chord parameter metrics in parabolas $y² = 4ax$ $y^2 = 4ax$:

t₁ \cdot t₂ = -1

Distance to the directrix property:

SP = a(1 + t²), SQ = a(1 + (1)/(t²))

Core Logic

Given parabola $y² = 12x a = 3$ $y^2 = 12x \implies a = 3$. Focus $S = (3, 0)$ $S = (3, 0)$. Set up focal segments product equation:

SP \cdot SQ = 3(1+t²) \cdot 3(1+(1)/(t²)) = (147)/(4)

9 \cdot ((1+t²)²)/(t²) = (147)/(4) ((1+t²)²)/(t²) = (49)/(12)

Solving for $t²$ $t^2$:

12t⁴ - 25t² + 12 = 0 t² = (3)/(4) or (4)/(3)

Step 1: Compute Endpoint Coordinate Bounds

Choosing $t = - \sqrt(3)2$ $t = -\frac{\sqrt{3}}{2}$ allows defining both chord coordinates symmetrically:

P(3t², 6t) P((9)/(4), -3\sqrt(3))

Q((3)/(t²), -(6)/(t)) Q(4, 4\sqrt(3))

Step 2: Derive Circle Equation

Write the diameter circle form equation:

(x - 4)(x - (9)/(4)) + (y - 4\sqrt(3))(y + 3\sqrt(3)) = 0

x² + y² - (25)/(4)x - \sqrt(3)y - 27 = 0

Multiply by 64 to clear the fractions and match the given equation template structure:

64x² + 64y² - 400x - 64\sqrt(3)y - 1728 = 0

Comparing directly with $64x² + 64y² - α x - 64\sqrt(3)y = β$ yields:

α = 400, β = 1728

β - α = 1728 - 400 = 1328

Pattern Recognition

The distance from focal chord endpoints to the focus equals their perpendicular distance to the directrix. This property connects parameter metrics to geometric lengths cleanly.

Chapter Mix

Class 11 Mathematics: Conic Sections Class 11 Mathematics: Circles

Q jee_main_2025_28_jan_morning Parabola and Trapezium Properties

Let ABCD be a trapezium whose vertices lie on the parabola

y² = 4x

. Let the sides AD and BC of the trapezium be parallel to y-axis. If the diagonal AC is of length

(25)/(4)

and it passes through the point

(1,0)

, then the area of ABCD is:

A. $(75)/(4)$ $\frac{75}{4}$
B. $(25)/(2)$ $\frac{25}{2}$
C. $(125)/(8)$ $\frac{125}{8}$
D. $(75)/(8)$ $\frac{75}{8}$

Solution

Related Formula

Area of a trapezium is given by:

Area = (1)/(2) \times (sum of parallel sides) \times (distance between them)

Core Logic

Let the coordinates of the vertices be parameterized on the parabola $y² = 4x$ $y^2 = 4x$. Since $AD$ $AD$ and $BC$ $BC$ are parallel to the y-axis, the coordinates take the form: $A(at₁², 2at₁)$ $A(at_1^2, 2at_1)$ and $D(at₁², -2at₁)$ $D(at_1^2, -2at_1)$ $B(at₂², 2at₂)$ $B(at_2^2, 2at_2)$ and $C(at₂², -2at₂)$ $C(at_2^2, -2at_2)$

Given $a=1$ $a=1$, the points simplify accordingly.

Parabola and Trapezium Properties diagram for Q52 - JEE Main 2025 Morning

Step 1: Using Diagonal Properties

The length of diagonal $AC$ $AC$ passing through focal point $(1,0)$ $(1,0)$ implies focal chord properties:

Length AC = a(t₁ + (1)/(t₁))² = (25)/(4)

t₁ + (1)/(t₁) = \pm(5)/(2) t₁ = 2 or (1)/(2)

Step 2: Finding Coordinates and Area

Substituting $t₁ = 2$ $t_1 = 2$, we get: $A((1)/(2), 1), D((1)/(4), -1), B(4, 4), C(4, -4)$

Evaluating the area formula:

Area = (1)/(2) \times (8 + 2) \times (4 - (1)/(4)) = (75)/(4)

Pattern Recognition

Focal chords of parabolas always satisfy $t₁ t₂ = -1$ $t_1 t_2 = -1$. Recognizing the passage through $(1,0)$ $(1,0)$ unlocks quick parametric simplifications.

Chapter Mix

Class 11 Maths: Conic Sections

Q jee_main_2025_03_april_morning Ellipse and Line Properties

A line passing through the point

P(\sqrt(5), \sqrt(5))

intersects the ellipse

(x²)/(36) + (y²)/(25) = 1

A

and

B

[cite: 567] such that

(PA) \cdot (PB)

is maximum. Then

5(PA² + PB²)

is equal to

A. 218
B. 377
C. 290
D. 338

Solution

Related Formula

Parametric line equation relative to an offset point $P(x₀, y₀)$ $P(x_0, y_0)$:

x = x₀ + r θ, y = y₀ + r θ

Ellipse and Line Properties diagram for Q56 - JEE Main 2025 Morning

Core Logic

Assume any line through $P(\sqrt(5), \sqrt(5))$ $P(\sqrt{5}, \sqrt{5})$ can be represented parametrically by:

Q(\sqrt(5) + r θ, \sqrt(5) + r θ)

Substitute coordinates into the standard ellipse equation $25x² + 36y² = 900$ $25x^2 + 36y^2 = 900$:

25(\sqrt(5) + r θ)² + 36(\sqrt(5) + r θ)² = 900

Expanding and gathering powers of $r$ $r$ yields:

r²(25 ²θ + 36 ²θ) + 2\sqrt(5)r(25 θ + 36 θ) - 595 = 0

The product of roots corresponds to the distance product:

PA \cdot PB = |r₁ r₂| = (595)/(25 ²θ + 36 ²θ) = (595)/(25 + 11 ²θ)

Step 1: Maximization Condition

To maximize $PA \cdot PB$ $PA \cdot PB$, the denominator must be minimized:

²θ = 0 θ = 0

This implies the chord line $AB$ $AB$ must run parallel to the x-axis:

yA = yB = \sqrt(5)

Substitute $y = \sqrt(5)$ $y = \sqrt{5}$ back into the ellipse equation to calculate x-coordinates:

(x²)/(36) + (5)/(25) = 1 (x²)/(36) = (4)/(5) x² = (144)/(5)

Therefore, the coordinates are $x = \pm 12\sqrt(5)$ $x = \pm \frac{12}{\sqrt{5}}$.

Step 2: Distance Value Summation

Compute $PA² + PB²$ $PA^2 + PB^2$ using coordinates directly:

PA² + PB² = (\sqrt(5) - 12\sqrt(5))² + (\sqrt(5) + 12\sqrt(5))²

= 2(5 + (144)/(5)) = (338)/(5)

Multiplying by 5 gives the required value:

5(PA² + PB²) = 338

Pattern Recognition

Shortcut: Represent lines through an arbitrary point parametrically in conic intersection problems. The product of distances $|r₁ r₂|$ $|r_1 r_2|$ directly falls out of the constant term over the leading coefficient, making trigonometric optimization straightforward.

Evaluation Rubric / Model Answer

338

Chapter Mix

Class 11 Mathematics: Conic Sections (Ellipse)

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding the Area Sequence Terms

Step 2: Summing the Infinite Geometric Series

Pattern Recognition

Chapter Mix

Reference Study Guides

More Conic Sections Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Simplify to Standard Form

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Endpoint Coordinate Bounds

Step 2: Derive Circle Equation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Using Diagonal Properties

Step 2: Finding Coordinates and Area

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Maximization Condition

Step 2: Distance Value Summation

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

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