Let the image of parabola x2=4y, in the line x-y = 1 be (y+alpha)2=b(x-c), a, b, c N. Then a+b+c is equal to

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

Q1 jee_main_2026_23_january_morning Hyperbola

Let the domain of the function

f(x) = ₃ ₅ ₇(9x - x² - 13)

be the interval

(m, n)

. Let the hyperbola

(x²)/(a²) - (y²)/(b²) = 1

have eccentricity

(n)/(3)

and the length of the latus rectum

(8m)/(3)

. Then

b² - a²

is equal to:

A. $5$ $5$
B. $11$ $11$
C. $9$ $9$
D. $7$ $7$

Solution

Related Formula

e = \sqrt(1 + (b²)/(a²))

L.R. = (2b²)/(a)

Core Logic

For the domain of the given logarithmic function, the argument of the innermost logarithm must be strictly greater than 1 because of the nested logs:

₅( ₇(9x-x²-13))>0

\Rightarrow ₇(9x-x²-13) > 1

\Rightarrow 9x-x²-13 > 7

\Rightarrow x²-9x+20 < 0

\Rightarrow (x-4)(x-5) < 0

Thus, $4 < x < 5$ $4 < x < 5$. Therefore, the domain interval is $(4, 5)$ $(4, 5)$, yielding $m = 4$ $m = 4$ and $n = 5$ $n = 5$.

Step 1: Hyperbola Properties

Given eccentricity $e = (n)/(3) = (5)/(3)$ $e = \frac{n}{3} = \frac{5}{3}$:

e = \sqrt(1 + (b²)/(a²)) = (5)/(3) \Rightarrow (b²)/(a²) = (25)/(9) - 1 = (16)/(9)

\Rightarrow (b)/(a) = (4)/(3)

Given the length of the latus rectum is $(8m)/(3)$ $\frac{8m}{3}$:

(2b²)/(a) = (8(4))/(3) = (32)/(3)

\Rightarrow 2b((b)/(a)) = (32)/(3) \Rightarrow 2b((4)/(3)) = (32)/(3)

\Rightarrow 8b = 32 \Rightarrow b = 4

Since $(b)/(a) = (4)/(3)$ $\frac{b}{a} = \frac{4}{3}$, we get $a = 3$ $a = 3$.

Step 2: Final Calculation

We need to find $b² - a²$ $b^2 - a^2$:

b² - a² = (4)² - (3)² = 16 - 9 = 7

Pattern Recognition

Nested logarithmic domains require unpacking from the outside in: $ₐ(X) > 0 \Rightarrow X > 1$ $\log_a(X) > 0 \Rightarrow X > 1$. Linking function domains to coordinate geometry parameters is a standard JEE cross-topic pattern.

Chapter Mix

Class 11 Maths: Conic Sections Class 11 Maths: Relations and Functions

Q5 jee_main_2026_23_january_morning Ellipse

Let the line

y - x = 1

intersect the ellipse

(x²)/(2) + (y²)/(1) = 1

at the points

A

and

B

. Then the angle made by the line segment

AB

at the center of the ellipse is:

A. $π - ⁻¹((1)/(4))$
B. $(π)/(2) + ⁻¹((1)/(4))$
C. $(π)/(2) + 2 ⁻¹((1)/(4))$
D. $(π)/(2) - ⁻¹((1)/(4))$

Solution

Related Formula

θ = | (m₁ - m₂)/(1 + m₁ m₂) |

Core Logic

Find the intersection points of the line $y = x + 1$ $y = x + 1$ and the ellipse $(x²)/(2) + y² = 1$ $\frac{x^2}{2} + y^2 = 1$.

Ellipse diagram for Q5 - JEE Main 2026 Morning

Substitute

y = x + 1

into the ellipse equation:

(x²)/(2) + (x + 1)² = 1

x² + 2(x² + 2x + 1) = 2

3x² + 4x = 0 \Rightarrow x(3x + 4) = 0

This gives $x = 0$ $x = 0$ or $x = -(4)/(3)$ $x = -\frac{4}{3}$.

Step 1: Calculate Intersection Points

For $x = 0$ $x = 0$, $y = 1 \Rightarrow A(0, 1)$ $y = 1 \Rightarrow A(0, 1)$. For $x = -(4)/(3)$ $x = -\frac{4}{3}$, $y = -(4)/(3) + 1 = -(1)/(3) \Rightarrow B(-(4)/(3), -(1)/(3))$ .

Step 2: Find Angle at the Origin

Let $O(0,0)$ $O(0,0)$ be the center of the ellipse. The angle made by segment $AB$ $AB$ at $O$ $O$ is $∠ AOB$ $\angle AOB$. The slope of $OA$ $OA$ is $m₁ = (1 - 0)/(0 - 0) = \infty$ $m_1 = \frac{1 - 0}{0 - 0} = \infty$ (which means $OA$ $OA$ is along the y-axis, angle is $π/2$ $\pi/2$). The slope of $OB$ $OB$ is $m₂ = (-1/3 - 0)/(-4/3 - 0) = (1)/(4)$ . The angle of $OB$ $OB$ with the positive x-axis is $θ = ⁻¹((1)/(4))$ . The total angle $∠ AOB$ $\angle AOB$ is $(π)/(2) + θ = (π)/(2) + ⁻¹((1)/(4))$ .

Pattern Recognition

When solving line-conic intersection, explicit extraction of points $(A, B)$ $(A, B)$ is often simpler than using homogenization if the intersection yields simple rational or integer coordinates.

Chapter Mix

Class 11 Maths: Conic Sections Class 11 Maths: Straight Lines

Let the image of parabola x²=4y$x^{2}=4y$, in the line x-y = 1$x-y = 1$ be (y+α)²=b(x-c)$(y+\alpha)^{2}=b(x-c)$, a, b, c in N$a, b, c \in \mathbb{N}$. Then a+b+c$a+b+c$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding the Image Coordinates

Step 2: Eliminating the Parameter

Step 3: Finding Target Values

Pattern Recognition

Chapter Mix

Reference Study Guides

More Parabola Previous-Year Questions — Page 3

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

Solution

Related Formula

Core Logic

Step 1: Hyperbola Properties

Step 2: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Intersection Points

Step 2: Find Angle at the Origin

Pattern Recognition

Chapter Mix

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Let the image of parabola $x²=4y$ $x^{2}=4y$, in the line $x-y = 1$ $x-y = 1$ be $(y+α)²=b(x-c)$ $(y+\alpha)^{2}=b(x-c)$, $a, b, c in N$ $a, b, c \in \mathbb{N}$. Then $a+b+c$ $a+b+c$ is equal to