Let ABCD be a trapezium whose vertices lie on the parabola y2 = 4x. Let the sides AD and BC of the trapezium be…

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Q2 jee_main_2026_24_january_evening Standard Equation of an Ellipse

Let the length of the latus rectum of an ellipse

x²a²+ y²b²=1

,

(a>b)

, be

30

. If its eccentricity is the maximum value of the function

f(t)=-(3)/(4)+2t-t²

, then

(a²+b²)

is equal to

A. $516$ $516$
B. $256$ $256$
C. $496$ $496$
D. $276$ $276$

Solution

Related Formula

Latus Rectum = 2b²a

e² = 1 - b²a² = a²-b²a²

Core Logic

First, find the maximum value of $f(t) = -(3)/(4) + 2t - t²$ $f(t) = -\frac{3}{4} + 2t - t^{2}$.

Completing the square or differentiating:

f(t) = -(t² - 2t + (3)/(4)) = -((t-1)² - 1 + (3)/(4)) = (1)/(4) - (t-1)²

The maximum value is $(1)/(4)$ $\frac{1}{4}$. Thus, eccentricity $e = (1)/(4)$ $e = \frac{1}{4}$.

e² = (1)/(16) a² - b²a² = (1)/(16) (1)

Step 1: Relating a and b

Given latus rectum is $30$ $30$:

2b²a = 30 b² = 15a (2)

Step 2: Solving for a and b

Substitute (2) into (1):

16(a² - 15a) = a²

15a² - 240a = 0

Since $a \neq 0$ $a \neq 0$, we have $15a - 240 = 0 a = 16$ $15a - 240 = 0 \implies a = 16$.

Then, $b² = 15(16) = 240$ $b^{2} = 15(16) = 240$. So $a² = 256$ $a^{2} = 256$.

Step 3: Final Calculation

We need to find $a² + b²$ $a^{2} + b^{2}$:

a² + b² = 256 + 240 = 496

Pattern Recognition

Whenever an ellipse's latus rectum and eccentricity are provided, it generates a standard system of two equations linking $a$ $a$ and $b²$ $b^{2}$. Solve for $a$ $a$ first since $b²$ $b^{2}$ is linear with respect to $a$ $a$ via latus rectum.

Chapter Mix

Class 11 Maths: Ellipse Class 12 Maths: Application of Derivatives

Q6 jee_main_2026_24_january_evening Reflection of a Parabola

Let the image of parabola

x²=4y

, in the line

x-y = 1

be

(y+α)²=b(x-c)

,

a, b, c in N

. Then

a+b+c

is equal to

A. $12$ $12$
B. $4$ $4$
C. $6$ $6$
D. $8$ $8$

Solution

Related Formula

Image of point (x₁, y₁) in line ax + by + c = 0 is given by:

(x - x₁)/(a) = (y - y₁)/(b) = -2 (ax₁ + by₁ + c)/(a² + b²)

Core Logic

Take a general parametric point $P$ $P$ on the parabola $x² = 4y$ $x^2 = 4y$, which is $P(2t, t²)$ $P(2t, t^2)$.

We find the mirror image $Q(h, k)$ $Q(h, k)$ of $P$ $P$ with respect to the line $x - y - 1 = 0$ $x - y - 1 = 0$.

Step 1: Finding the Image Coordinates

(h - 2t)/(1) = (k - t²)/(-1) = -2 (2t - t² - 1)/(1² + (-1)²)

(h - 2t)/(1) = (k - t²)/(-1) = -(2t - t² - 1) = t² - 2t + 1

Solving for $h$ $h$:

h - 2t = t² - 2t + 1 h = t² + 1

Solving for $k$ $k$:

k - t² = -(t² - 2t + 1) = -t² + 2t - 1 k = 2t - 1

Step 2: Eliminating the Parameter

From $k = 2t - 1$ $k = 2t - 1$, we get $t = (k + 1)/(2)$ $t = \frac{k + 1}{2}$.

Substitute $t$ $t$ into $h$ $h$:

h = ((k + 1)/(2))² + 1

h - 1 = ((k + 1)²)/(4)

(k + 1)² = 4(h - 1)

Step 3: Finding Target Values

Replacing $(h, k)$ $(h, k)$ with $(x, y)$ $(x, y)$, the image parabola is:

(y + 1)² = 4(x - 1)

Comparing this with $(y + α)² = b(x - c)$ $(y + \alpha)^2 = b(x - c)$: $α = 1, b = 4, c = 1$ $\alpha = 1, b = 4, c = 1$

a + b + c = 1 + 4 + 1 = 6

Pattern Recognition

To find the image of a conic section across a linear axis, it is almost always computationally cleaner to reflect its general parametric point rather than manipulating the implicit Cartesian equation through coordinate transformations.

Chapter Mix

Class 11 Maths: Parabola Class 11 Maths: Straight Lines

Q22 jee_main_2026_24_january_evening Locus of a Point

Let

(h, k)

lie on the circle

C : x² + y² = 4

and the point

(2h + 1, 3k + 2)

lie on an ellipse with eccentricity

e

. Then the value of

5e²

is equal to

Numerical Answer. Answer: 9 to 9

Solution

Related Formula

Parametric form of a circle x² + y² = r²: (r θ, r θ)

Eccentricity of an ellipse: e² = 1 - (b²)/(a²) (where a > b)

Core Logic

Let the point $P(h, k)$ $P(h, k)$ lie on $x² + y² = 4$ $x^2 + y^2 = 4$. Using parametric coordinates:

h = 2 θ, k = 2 θ

Let the target point be $Q(x, y)$ $Q(x, y)$:

x = 2h + 1 = 2(2 θ) + 1 = 4 θ + 1

$y = 3k + 2 = 3(2 θ) + 2 = 6 θ + 2$ (Wait, PDF states $3k+2$ $3k+2$, parametric solution in PDF says $6 θ + 3$ $6\sin\theta + 3$. Checking exact text: " $(2h + 1, 3k + 2)$ $(2h + 1, 3k + 2)$ lie on an ellipse..." but solution uses " $6 θ + 3$ $6\sin\theta + 3$". If the question meant $3k+3$ $3k+3$, that would be a typo in the question paper. However, $3(2 θ)+2 = 6 θ+2$ $3(2\sin\theta)+2 = 6\sin\theta+2$. This implies $(y-2)/(6) = θ$ $\frac{y-2}{6} = \sin\theta$. Either way, the denominators $a$ $a$ and $b$ $b$ of the resulting ellipse remain 4 and 6, so eccentricity is invariant to the constant offset).

Step 1: Finding the Locus

Isolating $θ$ $\cos\theta$ and $θ$ $\sin\theta$:

θ = (x - 1)/(4)

θ = (y - 2)/(6) (or (y-3)/(6) per the solution)

Using $²θ + ²θ = 1$ $\sin^2\theta + \cos^2\theta = 1$:

( (x - 1)/(4) )² + ( (y - 2)/(6) )² = 1

This is the equation of an ellipse where $a = 4$ $a = 4$ and $b = 6$ $b = 6$ (since $b > a$ $b > a$, it's a vertical ellipse).

Step 2: Calculating Eccentricity

For a vertical ellipse ( $b > a$ $b > a$), eccentricity is:

e² = 1 - (a²)/(b²) = 1 - (4²)/(6²) = 1 - (16)/(36)

e² = (36 - 16)/(36) = (20)/(36) = (5)/(9)

Step 3: Finding Final Target

We need the value of $(5)/(e²)$ $\frac{5}{e^2}$:

(5)/(e²) = (5)/((5)/(9)) = 9

Pattern Recognition

Affine transformations $(ax+b, cy+d)$ $(ax+b, cy+d)$ applied to a circle's locus purely stretch its semi-axes to the respective scaling constants ( $a$ $a$ and $c$ $c$). Translational constants ( $b$ $b$ and $d$ $d$) shift the center but do not affect the eccentricity.

Chapter Mix

Class 11 Maths: Ellipse Class 11 Maths: Circles

Q22 jee_main_2026_28_january_morning Ellipse and Hyperbola

For some

θ in (0, (π)/(2))

, let the eccentricity and the length of the latus rectum of the hyperbola

x² - y² ² θ = 8

be

e₁

and

₁

, respectively, and let the eccentricity and the length of the latus rectum of the ellipse

x² ² θ + y² = 6

be

e₂

and

₂

, respectively. If

e₁² = e₂² ( ² θ + 1)

, then

(( ₁ ₂)/(e₁ e₂)) ² θ

is equal to ____.

Numerical Answer. Answer: 8 to 8

Solution

Core Logic

For the hyperbola $(x²)/(8) - (y²)/(8 ²θ) = 1$ : $a² = 8, b² = 8 ²θ$ $a^2 = 8, b^2 = 8\cos^2\theta$ Eccentricity $e₁ = \sqrt(1 + (b²)/(a²)) = \sqrt(1 + (8 ²θ)/(8)) = \sqrt(1 + ²θ)$ . Latus rectum $₁ = (2b²)/(a) = 2(8 ²θ)2\sqrt(2) = 4\sqrt(2) ²θ$ .

For the ellipse $(x²)/(6 ²θ) + (y²)/(6) = 1$ : Here $a² = 6 ²θ, b² = 6$ $a^2 = 6\cos^2\theta, b^2 = 6$. Since $θ in (0, π/2)$ $\theta \in (0, \pi/2)$, $6 > 6 ²θ$ $6 > 6\cos^2\theta$, so the major axis is along the y-axis. Eccentricity $e₂ = \sqrt(1 - (a²)/(b²)) = \sqrt(1 - (6 ²θ)/(6)) = \sqrt(1 - ²θ) = θ$ . Latus rectum $₂ = (2a²)/(b) = 2(6 ²θ)\sqrt(6) = 2\sqrt(6) ²θ$ .

Step 1: Solve for Theta

Given relation: $e₁² = e₂² ( ²θ + 1)$ $e_1^2 = e_2^2 (\sec^2\theta + 1)$

1 + ²θ = ²θ (1 + (1)/( ²θ))

1 + ²θ = ²θ + ²θ

Replace $²θ$ $\sin^2\theta$ with $1 - ²θ$ $1 - \cos^2\theta$:

1 + ²θ = 1 - ²θ + ²θ

2 ²θ = ²θ = ( ²θ)/( ²θ)

2 ⁴θ = 1 - ²θ 2 ⁴θ + ²θ - 1 = 0

Factorizing:

(2 ²θ - 1)( ²θ + 1) = 0

Since $²θ + 1 \neq 0$ $\cos^2\theta + 1 \neq 0$, we have $2 ²θ = 1 ²θ = (1)/(2)$ . Since $θ in (0, π/2)$ $\theta \in (0, \pi/2)$, $θ = (π)/(4)$ $\theta = \frac{\pi}{4}$.

Step 2: Evaluate All Variables

For $θ = π/4$ $\theta = \pi/4$: $e₁ = \sqrt(1 + 1/2) = \sqrt(3/2)$ $e_1 = \sqrt{1 + 1/2} = \sqrt{3/2}$ $e₂ = (π/4) = 1/\sqrt(2)$ $e_2 = \sin(\pi/4) = 1/\sqrt{2}$ $₁ = 4\sqrt(2)(1/2) = 2\sqrt(2)$ $\ell_1 = 4\sqrt{2}(1/2) = 2\sqrt{2}$ $₂ = 2\sqrt(6)(1/2) = \sqrt(6)$ $\ell_2 = 2\sqrt{6}(1/2) = \sqrt{6}$ $²θ = 1$ $\tan^2\theta = 1$

Step 3: Final Calculation

Evaluate the target expression:

(( ₁ ₂)/(e₁ e₂)) ² θ = 2\sqrt(2) \cdot \sqrt(6)\sqrt(3/2) \cdot 1/\sqrt(2) \cdot 1

Numerator: $2\sqrt(12) = 4\sqrt(3)$ $2\sqrt{12} = 4\sqrt{3}$ Denominator: $\sqrt(3/4) = \sqrt(3)2$ $\sqrt{3/4} = \frac{\sqrt{3}}{2}$ Result = $4\sqrt(3) \sqrt(3)2 = 8$

Chapter Mix

Class 11 Mathematics: Conic Sections

Q3 jee_main_2026_28_january_evening Parabola and Triangles

Let

A

be the focus of the parabola

y² = 8x

. Let the line

y = mx + c

intersect the parabola at two distinct points

B

and

C

. If the centroid of the triangle

ABC

is

((7)/(3), (4)/(3))

, then

(BC)²

is equal to :

A. $41$ $41$
B. $80$ $80$
C. $89$ $89$
D. $32$ $32$

Solution

Related Formula

Centroid = ((x₁+x₂+x₃)/(3), (y₁+y₂+y₃)/(3))

D² = (x₂-x₁)² + (y₂-y₁)²

Core Logic

Focus of $y² = 8x$ $y^2 = 8x$ is $A(2, 0)$ $A(2, 0)$ since $4a = 8 \Rightarrow a=2$ $4a = 8 \Rightarrow a=2$. Let points on parabola be $B(2t₁², 4t₁)$ $B(2t_1^2, 4t_1)$ and $C(2t₂², 4t₂)$ $C(2t_2^2, 4t_2)$. The centroid of $Δ ABC$ $\Delta ABC$ is given as $((7)/(3), (4)/(3))$ $\left(\frac{7}{3}, \frac{4}{3}\right)$. Equating coordinates:

(2 + 2t₁² + 2t₂²)/(3) = (7)/(3) \Rightarrow t₁² + t₂² = (5)/(2)

(0 + 4t₁ + 4t₂)/(3) = (4)/(3) \Rightarrow t₁ + t₂ = 1

Parabola points and centroid configuration

Execution

Square the sum equation:

(t₁ + t₂)² = t₁² + t₂² + 2t₁t₂

1 = (5)/(2) + 2t₁t₂ \Rightarrow 2t₁t₂ = -(3)/(2) \Rightarrow t₁t₂ = -(3)/(4)

Find $(t₁ - t₂)²$ $(t_1 - t_2)^2$:

(t₁ - t₂)² = (t₁ + t₂)² - 4t₁t₂ = 1 - 4(-(3)/(4)) = 4

Calculate distance squared for $BC$ $BC$:

(BC)² = (2t₁² - 2t₂²)² + (4t₁ - 4t₂)²

(BC)² = 4(t₁ - t₂)²(t₁ + t₂)² + 16(t₁ - t₂)²

(BC)² = 4(4)(1) + 16(4) = 16 + 64 = 80

Pattern Recognition

Using parametric coordinates $(at², 2at)$ $(at^2, 2at)$ systematically reduces algebraic complexity when determining intersections or triangle properties on a parabola.

Chapter Mix

Class 11 Maths: Conic Sections

Let ABCD be a trapezium whose vertices lie on the parabola y² = 4x$y^2 = 4x$. Let the sides AD and BC of the trapezium be parallel to y-axis. If the diagonal AC is of length (25)/(4)$\frac{25}{4}$ and it passes through the point (1,0)$(1,0)$, then the area of ABCD is:

Solution & Explanation

Related Formula

Core Logic

Step 1: Using Diagonal Properties

Step 2: Finding Coordinates and Area

Pattern Recognition

Chapter Mix

More Conic Sections Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Relating a and b

Step 2: Solving for a and b

Step 3: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding the Image Coordinates

Step 2: Eliminating the Parameter

Step 3: Finding Target Values

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding the Locus

Step 2: Calculating Eccentricity

Step 3: Finding Final Target

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Solve for Theta

Step 2: Evaluate All Variables

Step 3: Final Calculation

Chapter Mix

Solution

Related Formula

Core Logic

Execution

Pattern Recognition

Chapter Mix

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Let ABCD be a trapezium whose vertices lie on the parabola $y² = 4x$ $y^2 = 4x$. Let the sides AD and BC of the trapezium be parallel to y-axis. If the diagonal AC is of length $(25)/(4)$ $\frac{25}{4}$ and it passes through the point $(1,0)$ $(1,0)$, then the area of ABCD is: