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:

Solution & Explanation

### 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

More Conic Sections Previous-Year Questions — Page 12

Q27 jee_main_2024_31_jan_morning Ellipse and Hyperbola Properties
Let the foci and length of the latus rectum of an ellipse fracx^2a^2 + fracy^2b^2 = 1, a > b be (pm 5, 0) and sqrt50, respectively. Then, the square of the eccentricity of the hyperbola fracx^2b^2 - fracy^2a^2 b^2 = 1 equals
Numerical Answer. Answer: 51 to 51

Solution

### Core Logic For the ellipse, foci are at (pm 5, 0) implies ae = 5. Latus rectum = frac2b^2a = sqrt50 = 5sqrt2 implies b^2 = frac5sqrt2a2. ### Step 1: Solve for a and b Using b^2 = a^2(1 - e^2): a^2 - (ae)^2 = b^2 implies a^2 - 25 = frac5sqrt2a2 2a^2 - 5sqrt2a - 50 = 0 2a^2 - 10sqrt2a + 5sqrt2a - 50 = 0 2a(a - 5sqrt2) + 5sqrt2(a - 5sqrt2) = 0 a = 5sqrt2 (since a > 0). Now, b^2 = frac5sqrt2(5sqrt2)2 = 25 implies b = 5. ### Step 2: Hyperbola Eccentricity The hyperbola is fracx^2b^2 - fracy^2a^2b^2 = 1. Here, semi-major axis A = b and semi-minor axis B = ab. Using eccentricity formula for hyperbola e_H^2 = 1 + fracB^2A^2: e_H^2 = 1 + fraca^2 b^2b^2 = 1 + a^2 Since a = 5sqrt2, a^2 = 50. e_H^2 = 1 + 50 = 51 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections
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