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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Q9 jee_main_2026_28_january_evening Ellipse Parameters and Latus Rectum

An ellipse has its center at

(1,-2)

, one focus at

(3,-2)

and one vertex at

(5, - 2)

. Then the length of its latus rectum is :

A. $16\sqrt(3)$ $\frac{16}{\sqrt{3}}$
B. $6$ $6$
C. $4\sqrt(3)$ $4\sqrt{3}$
D. $6\sqrt(3)$ $6\sqrt{3}$

Solution

Related Formula

Latus Rectum (LR) = (2b²)/(a) = 2a(1-e²)

Core Logic

From the given coordinates on the major axis ( $y = -2$ $y = -2$): Center $C(1, -2)$ $C(1, -2)$, Focus $F₁(3, -2)$ $F_1(3, -2)$, Vertex $A₁(5, -2)$ $A_1(5, -2)$. Distance from center to vertex, $CA₁ = a = 5 - 1 = 4$ $CA_1 = a = 5 - 1 = 4$. Distance from center to focus, $CF₁ = ae = 3 - 1 = 2$ $CF_1 = ae = 3 - 1 = 2$.

Ellipse dimensions mapped to coordinates

Execution

Calculate eccentricity $e$ $e$:

ae = 2 \Rightarrow 4e = 2 \Rightarrow e = (1)/(2)

Use alternate formula for Latus Rectum:

LR = 2e((a)/(e) - ae) or directly 2a(1-e²)

LR = 2(4)(1 - (1)/(4)) = 8 \times (3)/(4) = 6

Pattern Recognition

Aligning focus, center, and vertex along a constant y-axis implies a standard shifted ellipse where absolute differences in x-coordinates yield standard parameters ( $a$ $a$ and $ae$ $ae$) directly.

Chapter Mix

Class 11 Maths: Conic Sections

Q10 jee_main_2026_28_january_evening Confocal Ellipse and Hyperbola

Let the ellipse

E: x²144 + y²169 = 1

and the hyperbola

H: x²16 - y²λ² = -1

have the same foci. If

e

and

L

respectively denote the eccentricity and the length of the latus rectum of

H

, then the value of

24(e + L)

is:

A. $296$ $296$
B. $126$ $126$
C. $148$ $148$
D. $67$ $67$

Solution

Related Formula

e = \sqrt(1 - (a²)/(b²)) (for vertical ellipse)

e = \sqrt(1 + (a²)/(b²)) (for conjugate hyperbola)

Core Logic

For Ellipse $E: (x²)/(144) + (y²)/(169) = 1$ $a² = 144, b² = 169$ $a^2 = 144, b^2 = 169$. Since $b > a$ $b > a$, the major axis is along the y-axis. Eccentricity $e' = \sqrt(1 - (144)/(169)) = \sqrt((25)/(169)) = (5)/(13)$ . Foci of ellipse = $(0, \pm be') = (0, \pm 13 \times (5)/(13)) = (0, \pm 5)$ .

Execution

For Hyperbola $H: (y²)/(λ²) - (x²)/(16) = 1$ Foci of conjugate hyperbola are $(0, \pm λ e)$ $(0, \pm \lambda e)$. Equating foci: $λ e = 5$ $\lambda e = 5$.

e = \sqrt(1 + (16)/(λ²))

λ \sqrt(1 + (16)/(λ²)) = 5 \Rightarrow λ² + 16 = 25 \Rightarrow λ² = 9 \Rightarrow λ = 3

Eccentricity of hyperbola, $e = (5)/(3)$ $e = \frac{5}{3}$. Length of latus rectum of hyperbola, $L = (2(16))/(λ) = (32)/(3)$ .

Calculate $24(e + L)$ $24(e + L)$:

24(e + L) = 24[(5)/(3) + (32)/(3)] = 24((37)/(3)) = 8 \times 37 = 296

Pattern Recognition

Confocal conics usually align along the same major axis. Notice the $-1$ $-1$ on the RHS of the hyperbola equation indicates a conjugate hyperbola orienting it vertically to match the $b>a$ $b>a$ ellipse.

Chapter Mix

Class 11 Maths: Conic Sections

Q55 jee_main_2025_02_april_evening Ellipse

If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is :

A. $4\sqrt(17)$ $\frac{4}{\sqrt{17}}$
B. $\sqrt(3)16$ $\frac{\sqrt{3}}{16}$
C. $3\sqrt(19)$ $\frac{3}{\sqrt{19}}$
D. $\sqrt(5)7$ $\frac{\sqrt{5}}{7}$

Solution

Related Formula

Length of minor axis = 2b

Distance between foci = 2ae

Eccentricity: e = \sqrt(1 - (b²)/(a²))

Core Logic

We set up an algebraic equation relating $b$ $b$, $a$ $a$, and $e$ $e$ from the given geometric condition, then substitute it into the eccentricity identity.

Step 1: Set up the geometric relation

Given that $2b = (1)/(4) (2ae)$ $2b = \frac{1}{4} (2ae)$:

b = (ae)/(4) (b)/(a) = (e)/(4)

Square both sides:

(b²)/(a²) = (e²)/(16)

Step 2: Solve for eccentricity

Using the eccentricity relation:

e² = 1 - (b²)/(a²)

e² = 1 - (e²)/(16)

e² (1 + (1)/(16)) = 1

(17)/(16) e² = 1 e² = (16)/(17) e = 4\sqrt(17)

Pattern Recognition

Standard Ellipse relations: For standard ellipses, the ratio of axes and the eccentricity are coupled quadratic equations. Expressing $b/a$ $b/a$ as a function of $e$ $e$ allows direct solving of the eccentricity.

Chapter Mix

Class 11 Mathematics: Conic Sections

Q67 jee_main_2025_02_april_evening Parabola

Let the point

P

of the focal chord

PQ

of the parabola

y² = 16x

be

(1, -4)

. If the focus of the parabola divides the chord

PQ

in the ratio

m : n

,

(m, n) = 1

, then

m² + n²

is equal to:

A. 17
B. 10
C. 37
D. 26

Solution

Related Formula

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

Focal Chord relation: t₁ t₂ = -1

Section Formula: (xc, yc) = ( (m x₂ + n x₁)/(m+n), (m y₂ + n y₁)/(m+n) )

Core Logic

We find the parametric parameters of coordinates $P$ $P$ and $Q$ $Q$, obtain their Cartesian values, and then apply the section formula with the focus $S$ $S$ to calculate the splitting ratio.

Step 1: Find coordinates of P and Q

For parabola $y² = 16x$ $y^2 = 16x$, the focal parameter is $a = 4$ $a = 4$. Focus is $S(4, 0)$ $S(4, 0)$. Let $P$ $P$ be $(a t₁², 2a t₁) = (1, -4)$ $(a t_1^2, 2a t_1) = (1, -4)$:

2a t₁ = -4 2(4) t₁ = -4 t₁ = -(1)/(2)

Since $PQ$ $PQ$ is a focal chord, the parametric points are coupled:

t₁ t₂ = -1 t₂ = 2

Now, calculate the coordinates of $Q$ $Q$:

Q \equiv (a t₂², 2 a t₂) = (4(4), 2(4)(2)) = (16, 16)

Step 2: Solve for the dividing ratio

Let the focus $S(4, 0)$ $S(4, 0)$ divide the line segment $PQ$ $PQ$ internally in the ratio $λ : 1$ $\lambda : 1$. Using the $y$ $y$-coordinate of the section formula:

yₛ = (λ yq + 1 yₚ)/(λ + 1)

0 = (λ(16) + 1(-4))/(λ + 1) 16λ - 4 = 0 λ = (1)/(4)

Thus, the focus $S$ $S$ divides the chord internally in the ratio $1:4$ $1:4$. Since $(1, 4) = 1$ $\gcd(1, 4) = 1$, we have $m = 1$ $m = 1$ and $n = 4$ $n = 4$:

m² + n² = 1² + 4² = 1 + 16 = 17

Pattern Recognition

Harmonic Mean Shortcut: In any parabola, the focus divides a focal chord internally into segments of lengths $SP$ $SP$ and $SQ$ $SQ$ such that the semi-latus rectum $2a$ $2a$ is the harmonic mean of these segments: $(1)/(SP) + (1)/(SQ) = (1)/(a)$ .

Chapter Mix

Class 11 Mathematics: Conic Sections

Q jee_main_2025_02_april_morning Properties of Hyperbola

Let one focus of the hyperbola

H: (x²)/(a²) - (y²)/(b²) = 1

be at

(\sqrt(10), 0)

and the corresponding directrix be

x = 9\sqrt(10)

. If

e

and

l

respectively are the eccentricity and the length of the latus rectum of

H

, then

9(e² + l)

is equal to:

A. $14$ $14$
B. $15$ $15$
C. $16$ $16$
D. $12$ $12$

Solution

Related Formula

For a standard hyperbola: Focus: $(\pm ae, 0)$ $(\pm ae, 0)$ Directrix: $x = \pm (a)/(e)$ $x = \pm \frac{a}{e}$ Eccentricity relation: $(ae)² = a² + b²$ $(ae)^2 = a^2 + b^2$ Length of latus rectum: $l = (2b²)/(a)$ $l = \frac{2b^2}{a}$

Core Logic

Given $ae = \sqrt(10)$ $ae = \sqrt{10}$ and $(a)/(e) = 9\sqrt(10)$ $\frac{a}{e} = \frac{9}{\sqrt{10}}$. Multiplying these gives $a²$ $a^2$, which determines both parameters.

Step 1: Find a and e

a² = (ae) \cdot ((a)/(e)) = \sqrt(10) \cdot 9\sqrt(10) = 9 a = 3

Substitute $a = 3$ $a = 3$ into $ae = \sqrt(10)$ $ae = \sqrt{10}$:

e = \sqrt(10)3 e² = (10)/(9)

Step 2: Find b and l

Using $(ae)² = a² + b²$ $(ae)^2 = a^2 + b^2$:

10 = 9 + b² b² = 1

Then the length of latus rectum $l$ $l$ is:

l = (2b²)/(a) = (2(1))/(3) = (2)/(3)

Step 3: Evaluate Final Expression

Calculate $9(e² + l)$ $9(e^2 + l)$:

9((10)/(9) + (2)/(3)) = 10 + 6 = 16

Pattern Recognition

Multiplying focus location by directrix location immediately eliminates $e$ $e$, giving $a²$ $a^2$ directly. Once $a²$ $a^2$ is known, $b²$ $b^2$ follow seamlessly via $(ae)² = a²+b²$ $(ae)^2 = a^2+b^2$.

Chapter Mix

Class 11 Mathematics: 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 5

Solution

Related Formula

Core Logic

Execution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Execution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set up the geometric relation

Step 2: Solve for eccentricity

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Find coordinates of P and Q

Step 2: Solve for the dividing ratio

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Find a and e

Step 2: Find b and l

Step 3: Evaluate Final Expression

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: