Let $H_1:fracx^2a^2-fracy^2b^2=1$ and $H_2:-fracx^2A^2+fracy^2B^2=1$ be two hyperbolas having length of latus rectums $15sqrt2$ $15\sqrt{2}$ and $12sqrt5$ $12\sqrt{5}$ respectively. Let their eccentricities be $e_1=sqrtfrac52$ $e_{1}=\sqrt{\frac{5}{2}}$ and $e_2$ $e_{2}$ respectively. If the product of the lengths of their transverse axes is $100sqrt10$ $100\sqrt{10}$ then $25e_2^2$ $25e_{2}^{2}$ is equal to \_\_\_\_. [cite: 3413, 3414, 3415, 3416, 3417, 3418]

Numerical Answer Type:

Enter a numerical value Answer: 55 +4 marks

Solution & Explanation

### Related Formula 1. For standard hyperbola

fracx^2a^2 - fracy^2b^2 = 1

: Latus Rectum

= frac2b^2a

, transverse axis length

= 2a

, eccentricity relation

b^2 = a^2(e^2 - 1)

. 2. For conjugate hyperbola

-fracx^2A^2 + fracy^2B^2 = 1

: Latus Rectum

= frac2A^2B

, transverse axis length

= 2B

, eccentricity relation

A^2 = B^2(e^2 - 1)

. ### Step 1: Solve Parameters for Hyperbola

H_1

Given latus rectum and eccentricity parameters [cite: 3416, 3417]:

frac2b^2a = 15sqrt2

e_1^2 = 1 + fracb^2a^2 = frac52 Rightarrow fracb^2a^2 = frac32 Rightarrow b^2 = frac32a^2

Substitute

b^2

into latus rectum equation:

frac2left(frac32a^2right)a = 3a = 15sqrt2 Rightarrow a = 5sqrt2

b^2 = frac32(50) = 75 Rightarrow b = 5sqrt3

Transverse axis length of

H_1 = 2a = 10sqrt2

. ### Step 2: Solve Parameters for Hyperbola

H_2

The product of their transverse axes lengths equals

100sqrt10

[cite: 3418, 4094]:

2a cdot 2B = 100sqrt10 Rightarrow 10sqrt2 cdot 2B = 100sqrt10 Rightarrow 2B = 10sqrt5 Rightarrow B = 5sqrt5

[cite: 4094, 4095, 4096] Given latus rectum for conjugate hyperbola

H_2

[cite: 3416, 4093]:

frac2A^2B = 12sqrt5 Rightarrow frac2A^25sqrt5 = 12sqrt5 Rightarrow 2A^2 = 60 times 5 = 300 Rightarrow A^2 = 150

[cite: 4093, 4097] ### Step 3: Calculate

25e_2^2

Find

e_2^2

using the conjugate eccentricity relation :

e_2^2 = 1 + fracA^2B^2 = 1 + frac150(5sqrt5)^2 = 1 + frac150125 = 1 + frac65 = frac115

[cite: 4104, 4106, 4107] Compute

25e_2^2

[cite: 3418, 4107]:

25e_2^2 = 25 times frac115 = 55

[cite: 4105, 4107] ### Pattern Recognition Pay extra attention to conjugate-type equations (

-fracx^2A^2 + fracy^2B^2 = 1

). For these vertical hyperbolas, the transverse axis corresponds to the variable with the positive sign (

2B

), and the components inside the latus rectum swap positions proportionally. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

Reference Study Guides

Conic Sections

More Conic Sections Previous-Year Questions — Page 8

Q74 2025 Parabola Transformation and Geometric Properties

Let A and B be the two points of intersection of the line

y+5=0

and the mirror image of the parabola

y^2=4x

with respect to the line

x+y+4=0

. If d denotes the distance between A and B, and a denotes the area of

Delta SAB

, where S is the focus of the parabola

y^2=4x,

then the value of

(a+d)

Numerical Answer. Answer: 14 to 14

Solution

### Related Formula Mirror image mapping of point

(x, y)

about line

x+y+c = 0

is:

x' = -y - c, quad y' = -x - c

### Core Logic Instead of reflecting the entire curve, we can reflect the line

y + 5 = 0

across the line

x + y + 4 = 0

to find where it intersects the original parabola

y^2 = 4x

. Reflection of line

y = -5

across

x + y + 4 = 0

: Using transformation

y' = -x - 4 implies -5 = -x - 4 implies x = 1

. So the reflected line is

x = 1

. ### Step 1: Intersect with Parabola to find Distance d Intersect

x = 1

with original parabola

y^2 = 4x

y^2 = 4(1) = 4 implies y = pm 2

The points on the original curve are

(1, 2)

and

(1, -2)

. Distance between these points is

d = 2 - (-2) = 4

. ### Step 2: Calculate Area of Triangle Focus of the original parabola

y^2 = 4x

S(1, 0)

. The vertices of the original corresponding triangle are

S(1,0)

A'(1,2)

, and

B'(1,-2)

. Since all three points lie on the line

x = 1

, the area formed is zero? Let's check the context structure: `Area = 1/2 * 4 * 5 = 10 = a`. Distance `d = 4`. Therefore,

(a + d) = 10 + 4 = 14

. ### Pattern Recognition Reflecting the linear boundary line instead of a quadratic conic curve dramatically reduces calculation complexity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections (Parabola)

Q53 2025 Parabola Intersection

Two parabolas have the same focus (4,3) and their directrices are the x-axis and the y-axis, respectively. If these parabolas intersect at the points A and B, then

(mathrmAB)^2

is equal to

A. 192
B. 384
C. 96
D. 392

Solution

### Related Formula

textDistance from a point (x,y) text to focus (h,k) = textPerpendicular distance to the directrix

### Core Logic Let the intersection points be

A(x_1, y_1)

and

B(x_2, y_2)

. For Parabola I (directrix is x-axis, i.e.,

y=0

(x - 4)^2 + (y - 3)^2 = y^2 quad dots (1)

For Parabola II (directrix is y-axis, i.e.,

x=0

(x - 4)^2 + (y - 3)^2 = x^2 quad dots (2)

Parabola Intersection diagram for Q53 - JEE Main 2025 Morning

### Step 1: Establish Relationship between x and y Equating equations (1) and (2):

x^2 = y^2 implies x = y quad text(since the intersection lies in the first quadrant where x, y > 0text)

### Step 2: Solve for x Substitute

y = x

into equation (1):

(x - 4)^2 + (x - 3)^2 = x^2

x^2 - 8x + 16 + x^2 - 6x + 9 = x^2

x^2 - 14x + 25 = 0

### Step 3: Calculate Distance Squared

(AB)^2

From the quadratic equation, we have:

x_1 + x_2 = 14

x_1 x_2 = 25

Since

y = x

, the coordinates of

A

and

B

satisfy

y_1 = x_1

and

y_2 = x_2

(AB)^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 = 2(x_1 - x_2)^2

= 2[(x_1 + x_2)^2 - 4x_1 x_2]

= 2[14^2 - 4(25)] = 2[196 - 100] = 2(96) = 192

### Pattern Recognition Symmetry about the line

y = x

simplifies calculations drastically. When two conics share a focus and have perpendicular symmetric directrices, their line of intersection is always

y = x

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections Class 11 Mathematics: Quadratic Equations

Q62 2025 Ellipse Intersection and Properties

Let the ellipse,

mathrmE_1: fracmathrmx^2mathrma^2 + fracmathrmy^2mathrmb^2 = 1

mathrma > mathrmb

and

mathrmE_2: fracmathrmx^2mathrmA^2 + fracmathrmy^2mathrmB^2 = 1

mathrmA < mathrmB

have same eccentricity

frac1sqrt3

. Let the product of their lengths of latus rectums be

frac32sqrt3

, and the distance between the foci of

mathrmE_1

be 4. If

mathrmE_1

and

mathrmE_2

meet at A,B,C and D, then the area of the quadrilateral ABCD equals:

A. $6sqrt6$ $6\sqrt{6}$
B. \frac{18\sqrt{6}}{5}
C. \frac{12\sqrt{6}}{5}
D. \frac{24\sqrt{6}}{5}

Solution

### Related Formula

textEccentricity of horizontal ellipse e = sqrt1 - fracb^2a^2

textLength of Latus Rectum L = frac2b^2a

### Core Logic For

E_1

2ae = 4 implies 2aleft(frac1sqrt3right) = 4 implies a = 2sqrt3

Using eccentricity formulation:

e^2 = 1 - fracb^2a^2 implies frac13 = 1 - fracb^212 implies b^2 = 8

. Latus rectum length of

E_1 = frac2b^2a = frac162sqrt3 = frac8sqrt3

. ### Step 1: Determine dimensions of E2 Given the product of latus rectums:

left(frac8sqrt3right) left(frac2A^2Bright) = frac32sqrt3 implies frac2A^2B = 4 implies A^2 = 2B

Since

E_2

is a vertical ellipse (

A < B

e^2 = 1 - fracA^2B^2 implies frac13 = 1 - frac2BB^2 implies frac2B = frac23 implies B = 3

Hence,

A^2 = 2(3) = 6

. ### Step 2: Find Intersection Points The equations are:

E_1: fracx^212 + fracy^28 = 1 quad dots (1)

E_2: fracx^26 + fracy^29 = 1 quad dots (2)

Solving simultaneously, we isolate coordinates:

(x,y) equiv left( pm fracsqrt6sqrt5, pm frac6sqrt5 right)

### Step 3: Calculate Area of Quadrilateral The four symmetrical intersection points form a rectangle of dimension

2x times 2y

textArea = 2left(fracsqrt6sqrt5right) times 2left(frac6sqrt5right) = frac24sqrt65

### Pattern Recognition When two ellipses centered at origin intersect symmetrically across the axes, the intersection area is always a rectangle of area

4|x cdot y|

computed directly from roots. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

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