If alpha x + beta y = 109 is the equation of the chord of the ellipse (x2/9) +(y2/4) = 1, whose mid point is…

Year	2026	2025	2024	Total
Questions	22	38	16	76

Q10 jee_main_2026_22_january_morning Hyperbola and Line Intersection

If the line

α x + 2y = 1

, where

α in R

, does not meet the hyperbola

x² - 9y² = 9

, then a possible value of

α

is:

A. $0.6$ $0.6$
B. $0.8$ $0.8$
C. $0.5$ $0.5$
D. $0.7$ $0.7$

Solution

Related Formula

For a line y = mx + c and hyperbola (x²)/(a²) - (y²)/(b²) = 1:

If they do not intersect, the quadratic in x formed by substituting y has D < 0.

Core Logic

Given line: $α x + 2y = 1 y = (1 - α x)/(2)$ . Given hyperbola: $x² - 9y² = 9$ $x^2 - 9y^2 = 9$.

Substitute the expression for $y$ $y$ into the hyperbola's equation:

x² - 9((1 - α x)/(2))² = 9

Step 1: Solving for Discriminant

x² - (9(1 - 2α x + α² x²))/(4) = 9

Multiply by 4:

4x² - 9(1 - 2α x + α² x²) = 36

4x² - 9 + 18α x - 9α² x² - 36 = 0

(4 - 9α²)x² + 18α x - 45 = 0

For the line to NOT intersect the hyperbola, the quadratic must yield non-real roots, meaning Discriminant $D < 0$ $D < 0$.

D = (18α)² - 4(4 - 9α²)(-45) < 0

324α² + 180(4 - 9α²) < 0

324α² + 720 - 1620α² < 0

-1296α² + 720 < 0

1296α² > 720 α² > (720)/(1296) = (5)/(9)

Step 2: Finding Alpha Interval

α² - (5)/(9) > 0

α in (-\infty, - \sqrt(5)3) ( \sqrt(5)3, \infty)

Since $\sqrt(5) \approx 2.236$ $\sqrt{5} \approx 2.236$, we have $\sqrt(5)3 \approx 0.745$ $\frac{\sqrt{5}}{3} \approx 0.745$. So $α$ $\alpha$ must be strictly greater than $0.745$ $0.745$ (or less than $-0.745$ $-0.745$).

Checking the given options: (1) 0.6 (No) (2) 0.8 (Yes, $0.8 > 0.745$ $0.8 > 0.745$) (3) 0.5 (No) (4) 0.7 (No)

Pattern Recognition

Geometrically, for a line not to meet a hyperbola, its slope must lie within a specific range determined by the asymptotes ( $m = \pm b/a$ $m = \pm b/a$), and its $c²$ $c^2$ must satisfy $c² < a²m² - b²$ $c^2 < a^2m^2 - b^2$. Direct substitution to enforce $D < 0$ $D < 0$ is purely mechanical and robust.

Chapter Mix

Class 11 Maths: Conic Sections

Q17 jee_main_2026_22_january_morning Properties of Parabola

If the chord joining the points

P₁(x₁, y₁)

and

P₂(x₂, y₂)

on the parabola

y² = 12x

subtends a right angle at the vertex of the parabola, then

x₁x₂ - y₁y₂

is equal to

A. $288$ $288$
B. $280$ $280$
C. $284$ $284$
D. $292$ $292$

Solution

Related Formula

If a chord joining t₁ and t₂ subtends a right angle at the vertex (0,0), then t₁ t₂ = -4.

Parametric coordinates for y² = 4ax: (at², 2at)

Core Logic

Given parabola $y² = 12x 4a = 12 a = 3$ .

Let the points be $P₁(x₁, y₁) = (3t₁², 6t₁)$ $P_1(x_1, y_1) = (3t_1^2, 6t_1)$ and $P₂(x₂, y₂) = (3t₂², 6t₂)$ $P_2(x_2, y_2) = (3t_2^2, 6t_2)$.

Since the chord subtends a right angle at the vertex (origin),

mOP₁ \cdot mOP₂ = -1

(6t₁)/(3t₁²) \cdot (6t₂)/(3t₂²) = -1

(2)/(t₁) \cdot (2)/(t₂) = -1 t₁ t₂ = -4

Step 1: Calculating the Expression

We need to evaluate $x₁ x₂ - y₁ y₂$ $x_1 x_2 - y_1 y_2$:

x₁ x₂ = (3t₁²)(3t₂²) = 9(t₁ t₂)²

y₁ y₂ = (6t₁)(6t₂) = 36(t₁ t₂)

Substitute $t₁ t₂ = -4$ $t_1 t_2 = -4$:

x₁ x₂ - y₁ y₂ = 9(-4)² - 36(-4)

$= 9(16) + 144$ $= 9(16) + 144$

= 144 + 144 = 288

Pattern Recognition

Right angles subtended at the vertex by a chord on $y² = 4ax$ $y^2 = 4ax$ instantly lock the parameter product to $t₁ t₂ = -4$ $t_1 t_2 = -4$. Substitute this directly into any coordinate products required by the problem.

Chapter Mix

Class 11 Maths: Conic Sections

Q11 jee_main_2026_22_january_evening Hyperbola Properties and Area of Triangle

Let

P(10, 2\sqrt(15))

be a point on the hyperbola

(x²)/(a²) - (y²)/(b²) = 1

, whose foci are

S

and

S'

. If the length of its latus rectum is

8

, then the square of the area of

Δ PSS'

is equal to:

A. 4200
B. 900
C. 1462
D. 2700

Solution

Related Formula

Latus rectum length $= (2b²)/(a) = 8 b² = 4a$ . Focal length $= 2ae = 2\sqrt(a² + b²)$ $= 2ae = 2\sqrt{a^2 + b^2}$.

Core Logic

Substitute $P(10, 2\sqrt(15))$ $P(10, 2\sqrt{15})$ and $b² = 4a$ $b^2 = 4a$ into hyperbola equation:

(100)/(a²) - (60)/(4a) = 1 a² + 15a - 100 = 0

(a + 20)(a - 5) = 0 a = 5 (a > 0)

Thus, $b² = 20 b = \sqrt(20)$ $b^2 = 20 \implies b = \sqrt{20}$.

Step 1: Calculate Focal Distance and Area

Focal distance $SS' = 2ae = 2 \sqrt(a² + b²) = 2 \sqrt(25 + 20) = 6\sqrt(5)$ . Area of $Δ PSS' = (1)/(2) \times base \times height = (1)/(2) (6\sqrt(5)) (2\sqrt(15)) = 30\sqrt(3) = A$ .

Step 2: Square of Area

A² = (30\sqrt(3))² = 900 \times 3 = 2700

Pattern Recognition

Use latus rectum relation to reduce hyperbola parameter to single variable quadratic.

Chapter Mix

Class 11 Maths: Conic Sections

Q16 jee_main_2026_22_january_evening Ellipse Focal Distances

Let

S

and

S'

be the foci of the ellipse

(x²)/(25) + (y²)/(9) = 1

and

P(α, β)

be a point on the ellipse in the first quadrant. If

(SP)² + (S'P)² - SP \cdot S'P = 37

, then

α² + β²

is equal to:

A. 15
B. 11
C. 17
D. 13

Solution

Related Formula

For ellipse $(x²)/(a²) + (y²)/(b²) = 1$ $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: $SP + S'P = 2a = 10$ $SP + S'P = 2a = 10$, $e = \sqrt(1 - b²/a²) = \sqrt(1 - 9/25) = 4/5$ . Focal distances: $SP = a - eα = 5 - (4)/(5)α$ , $S'P = a + eα = 5 + (4)/(5)α$ .

Core Logic

Using algebraic identity:

(SP + S'P)² - 3 SP \cdot S'P = 37 100 - 3 SP \cdot S'P = 37 SP \cdot S'P = 21

Substitute focal distance formula:

25 - (16)/(25)α² = 21 (16)/(25)α² = 4 α² = (25)/(4)

Step 1: Calculate beta^2 and Sum

Substitute $α²$ $\alpha^2$ into ellipse equation $(α²)/(25) + (β²)/(9) = 1$ :

(1)/(4) + (β²)/(9) = 1 β² = (27)/(4)

α² + β² = (25)/(4) + (27)/(4) = (52)/(4) = 13

Pattern Recognition

Express $(SP)² + (S'P)² - SP \cdot S'P$ $(SP)^2 + (S'P)^2 - SP \cdot S'P$ in terms of $(SP+S'P)$ $(SP+S'P)$ to determine $SP \cdot S'P$ $SP \cdot S'P$ instantly.

Chapter Mix

Class 11 Maths: Conic Sections

Q17 jee_main_2026_22_january_evening Locus of Midpoint of Chord

Let the locus of the mid-point of the chord through the origin

O

of the parabola

y² = 4x

be the curve

S

. Let

P

be any point on

S

. Then the locus of the point, which internally divides

OP

in the ratio

3:1

, is:

A. $3y² = 2x$ $3y^2 = 2x$
B. $2y² = 3x$ $2y^2 = 3x$
C. $3x² = 2y$ $3x^2 = 2y$
D. $2x² = 3y$ $2x^2 = 3y$

Solution

Related Formula

Section formula for internal division in ratio $m:n$ $m:n$:

R(h,k) = ( (m x₂ + n x₁)/(m+n), (m y₂ + n y₁)/(m+n) )

Core Logic

Locus of midpoint diagram for Q17 - JEE Main 2026 Evening

Let chord endpoint be $Q(t², 2t)$ $Q(t^2, 2t)$. Midpoint $M(h,k)$ $M(h,k)$ of $OQ$ $OQ$:

h = (t²)/(2), k = t k² = 2h

So curve $S$ $S$ is $y² = 2x$ $y^2 = 2x$.

Now $P$ $P$ lies on $S: y² = 2x$ $S: y^2 = 2x$, so $P = ((t²)/(2), t)$ $P = \left(\frac{t^2}{2}, t\right)$. Point $R(h,k)$ $R(h,k)$ divides $OP$ $OP$ in ratio $3:1$ $3:1$:

Step 1: Section Formula Application

h = (3(t²/2) + 0)/(4) = (3t²)/(8), k = (3(t) + 0)/(4) = (3t)/(4)

From $k = (3t)/(4) t = (4k)/(3)$ . Substitute into $h$ $h$:

h = (3)/(8) ((4k)/(3))² = (3)/(8) \cdot (16k²)/(9) = (2k²)/(3)

2k² = 3h 2y² = 3x

Pattern Recognition

Parametrize midpoint curve $S$ $S$, then re-apply section ratio to derive final locus equation.

Chapter Mix

Class 11 Maths: Conic Sections

If α x + β y = 109$\alpha x + \beta y = 109$ is the equation of the chord of the ellipse (x²)/(9) +(y²)/(4) = 1$\frac{x^2}{9} +\frac{y^2}{4} = 1$, whose mid point is ((5)/(2),(1)/(2))$\left(\frac{5}{2},\frac{1}{2}\right)$, then α +β$\alpha +\beta$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Simplify to Standard Form

Pattern Recognition

Chapter Mix

Reference Study Guides

More Conic Sections Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Solving for Discriminant

Step 2: Finding Alpha Interval

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating the Expression

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Focal Distance and Area

Step 2: Square of Area

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate beta^2 and Sum

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Section Formula Application

Pattern Recognition

Chapter Mix

More Conic Sections Questions — jee_main_2025_29_jan_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

If $α x + β y = 109$ $\alpha x + \beta y = 109$ is the equation of the chord of the ellipse $(x²)/(9) +(y²)/(4) = 1$ $\frac{x^2}{9} +\frac{y^2}{4} = 1$, whose mid point is $((5)/(2),(1)/(2))$ $\left(\frac{5}{2},\frac{1}{2}\right)$, then $α +β$ $\alpha +\beta$ is equal to