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A line passing through the point P(sqrt5, sqrt5) intersects the ellipse fracx^236 + fracy^225 = 1 at A and B [cite: 567] such that (PA) cdot (PB) is maximum. Then 5(PA^2 + PB^2) is equal to[cite: 570]:

Solution & Explanation

### Related Formula Parametric line equation relative to an offset point P(x_0, y_0): x = x_0 + rcostheta, quad y = y_0 + rsintheta
Ellipse and Line Properties diagram for Q56 - JEE Main 2025 Morning
Ellipse and Line Properties diagram for Q56 - JEE Main 2025 Morning
### Core Logic Assume any line through P can be represented parametrically by [cite: 1277]: Q(sqrt5 + rcostheta, sqrt5 + rsintheta) [cite: 1277] Substitute coordinates into the standard ellipse formula [cite: 1278]: 25(sqrt5 + rcostheta)^2 + 36(sqrt5 + rsintheta)^2 = 900 [cite: 1278] Expanding and gathering powers of r yields [cite: 1280]: r^2(25cos^2theta + 36sin^2theta) + 2sqrt5r(25costheta + 36sintheta) - 595 = 0 [cite: 1280] The product of roots corresponds to the distance product value[cite: 1281, 1283]: PA cdot PB = |r_1 r_2| = frac59525cos^2theta + 36sin^2theta = frac59525 + 11sin^2theta [cite: 1283] ### Step 1: Maximization condition To make PA cdot PB maximum, the denominator must be minimized [cite: 1284]: sin^2theta = 0 implies theta = 0 [cite: 1284] This implies the chord line AB must run parallel to the x-axis [cite: 1285]: y_A = y_B = sqrt5 [cite: 1285] Substitute y = sqrt5 back into the ellipse equation to calculate x-coordinates [cite: 1286]: fracx^236 + frac525 = 1 implies fracx^236 = frac45 implies x^2 = frac1445 [cite: 1287] Therefore, the coordinates are x = pm frac12sqrt5. ### Step 2: Distance value summation Compute PA^2 + PB^2 using coordinates directly [cite: 1289]: PA^2 + PB^2 = left(sqrt5 - frac12sqrt5right)^2 + left(sqrt5 + frac12sqrt5right)^2 [cite: 1289] = 2left(5 + frac1445right) = frac3385 [cite: 1290] Multiplying by 5 gives the target integer answer [cite: 1290]: 5(PA^2 + PB^2) = 338 [cite: 1290] ### Pattern Recognition Parametric distances from a point intersecting a conic configuration usually form a standard quadratic equation in r. The angle parameter immediately simplifies the boundary constraint optimization. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections (Ellipse)

Reference Study Guides

Conic Sections

More Conic Sections Previous-Year Questions — Page 7

Q70 2025 Parabola Equation with Given Vertex and Directrix
If the equation of the parabola with vertex Vleft(frac32,3right) and the directrix x+2y=0 is alpha x^2+beta y^2-gamma xy-30x-60y+225=0, then alpha+beta+gamma is equal to: [cite: 3348, 3354, 3355]
  • A. 6
  • B. 8
  • C. 7
  • D. 9

Solution

### Related Formula The locus definition of a parabola states that the squared distance from any point P(x,y) to the focus S(x_0, y_0) equals the squared perpendicular distance to the directrix line Ax + By + C = 0: (x - x_0)^2 + (y - y_0)^2 = frac(Ax + By + C)^2A^2 + By^2 ### Step 1: Determine the Focus coordinates The axis line of the parabola is perpendicular to the directrix x + 2y = 0 and passes through the vertex V(1.5, 3) [cite: 3354, 3355]. Slope of directrix = -0.5 Rightarrow Slope of axis = 2. Equation of axis : y - 3 = 2left(x - frac32right) Rightarrow y - 2x = 0 [cite: 4059, 4060] The intersection of the axis (y - 2x = 0) and directrix (x + 2y = 0) gives the foot of the directrix, which is (0, 0) [cite: 4057, 4060]. Since the vertex is the midpoint between the focus and the foot of the directrix : left(frac32, 3right) = left(fracx_f + 02, fracy_f + 02right) Rightarrow textFocus S = (3, 6) ### Step 2: Derive the Parabola Locus Equation Equate the distance equations from point P(x,y) : (x - 3)^2 + (y - 6)^2 = frac(x + 2y)^21^2 + 2^2 5left(x^2 - 6x + 9 + y^2 - 12y + 36right) = x^2 + 4xy + 4y^2 5x^2 - 30x + 45 + 5y^2 - 60y + 180 = x^2 + 4xy + 4y^2 4x^2 + y^2 - 4xy - 30x - 60y + 225 = 0 ### Step 3: Coefficient Extraction Compare with the equation template alpha x^2 + beta y^2 - gamma xy - 30x - 60y + 225 = 0 : alpha = 4, quad beta = 1, quad gamma = 4 alpha + beta + gamma = 4 + 1 + 4 = 9 ### Pattern Recognition The vertex is always exactly midway between the focus and the foot of the directrix line along the line of symmetry. Finding the origin (0,0) as the foot quickly reveals the focus coordinates via doubling. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections
Q74 2025 Hyperbola - Latus Rectum and Eccentricity
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 and 12sqrt5 respectively. Let their eccentricities be e_1=sqrtfrac52 and e_2 respectively. If the product of the lengths of their transverse axes is 100sqrt10 then 25e_2^2 is equal to \_\_\_\_. [cite: 3413, 3414, 3415, 3416, 3417, 3418]
Numerical Answer. Answer: 55

Solution

### 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
Q60 2025 Properties of Ellipse
Let the product of the focal distances of the point left(sqrt3,frac12right) on the ellipse fracx^2a^2 +fracy^2b^2 = 1 (a > b) be frac74 . Then the absolute difference of the eccentricities of two such ellipses is :
  • A. frac3 - 2sqrt23sqrt2
  • B. frac1 - sqrt3sqrt2
  • C. frac3 - 2sqrt22sqrt3
  • D. frac1 - 2sqrt2sqrt3

Solution

### Related Formula For a point P(x_1, y_1) on a standard horizontal ellipse, the focal distances are given by (a + ex_1) and (a - ex_1). Their algebraic product equals: textProduct = a^2 - e^2 x_1^2 ### Core Logic Substitute the given coordinate x_1 = sqrt3 into our focal distance product value: a^2 - e^2(sqrt3)^2 = frac74 implies a^2 - 3e^2 = frac74 4a^2 = 7 + 12e^2 quad dots (1) ### Step 1: Apply Point Ingestion into Conic Equation Since the point left(sqrt3, frac12right) lies directly on the ellipse perimeter: frac3a^2 + frac(1/2)^2b^2 = 1 implies frac3a^2 + frac14b^2 = 1 Using the standard eccentricity identity b^2 = a^2(1-e^2): frac3a^2 + frac14a^2(1-e^2) = 1 implies 12(1-e^2) + 1 = 4a^2(1-e^2) 13 - 12e^2 = 4a^2(1-e^2) quad dots (2) ### Step 2: Solve the Bi-quadratic Equation for Eccentricity Substitute 4a^2 from equation (1) directly into equation (2): 13 - 12e^2 = (7 + 12e^2)(1-e^2) 13 - 12e^2 = 7 - 7e^2 + 12e^2 - 12e^4 12e^4 - 17e^2 + 6 = 0 Factorize the quadratic form in terms of e^2: (4e^2 - 3)(3e^2 - 2) = 0 implies e^2 = frac34 text or e^2 = frac23 This yields two distinct valid eccentricity parameters: e_1 = fracsqrt32, quad e_2 = sqrtfrac23 ### Step 3: Evaluate Target Absolute Difference Find the difference between the two eccentricity parameters: textDifference = fracsqrt32 - fracsqrt2sqrt3 = frac3 - 2sqrt22sqrt3 ### Pattern Recognition Converting b^2 to a^2(1-e^2) early decouples the multi-variable polynomial down to a standard bi-quadratic format in eccentricity, which can be solved easily. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections
Q55 2025 Chord of Ellipse
If the midpoint of a chord of the ellipse fracx^29+fracy^24=1 is (sqrt2,4/3), and the length of the chord is frac2sqrtalpha3 , then alpha is:
  • A. 18
  • B. 22
  • C. 26
  • D. 20

Solution

### Related Formula Equation of a chord of an ellipse with given midpoint (x_1, y_1) is given by T = S_1: fracxx_1a^2 + fracyy_1b^2 = fracx_1^2a^2 + fracy_1^2b^2 ### Core Logic Given ellipse: fracx^29 + fracy^24 = 1 and midpoint (x_1, y_1) = (sqrt2, frac43). Applying T = S_1: fracxsqrt29 + fracy(4/3)4 = frac(sqrt2)^29 + frac(4/3)^24 fracsqrt2x9 + fracy3 = frac29 + frac1636 fracsqrt2x9 + fracy3 = frac29 + frac49 = frac69 Multiplying through by 9: sqrt2x + 3y = 6 implies 3y = 6 - sqrt2x ### Step 1: Find Intersection Points with Ellipse Substitute 3y = 6 - sqrt2x into the multiplied form of ellipse 4x^2 + 9y^2 = 36: 4x^2 + (3y)^2 = 36 4x^2 + (6 - sqrt2x)^2 = 36 4x^2 + 36 + 2x^2 - 12sqrt2x = 36 6x^2 - 12sqrt2x = 0 6x(x - 2sqrt2) = 0 Thus, x = 0 or x = 2sqrt2. ### Step 2: Find y-coordinates and Chord Length If x_1 = 0 implies 3y_1 = 6 implies y_1 = 2 If x_2 = 2sqrt2 implies 3y_2 = 6 - sqrt2(2sqrt2) = 6 - 4 = 2 implies y_2 = frac23 The end points of the chord are A(0, 2) and B(2sqrt2, frac23). textLength of chord AB = sqrt(2sqrt2 - 0)^2 + left(frac23 - 2right)^2 AB = sqrt8 + left(-frac43right)^2 = sqrt8 + frac169 = sqrtfrac889 = fracsqrt4 times 223 = frac2sqrt223 Comparing with frac2sqrtalpha3, we get alpha = 22. ### Pattern Recognition When the intersection equation results in a simple factoring like 6x^2 - 12sqrt2x = 0, calculating the explicit coordinates is incredibly fast compared to using general formula roots equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections (Ellipse)
Q69 2025 Intersection of Curves and Locus of Centroid
If A and B are the points of intersection of the circle x^2+y^2-8x=0 and the hyperbola fracx^29-fracy^24=1 and a point P moves on the line 2x-3y+4=0, then the centroid of Delta PAB lies on the line:
  • A. 4x-9y=12
  • B. x+9y=36
  • C. 9x-9y=32
  • D. 6x-9y=20

Solution

### Related Formula Centroid (h, k) of a triangle with vertices (x_1,y_1), (x_2,y_2), (x_3,y_3): h = fracx_1 + x_2 + x_33, quad k = fracy_1 + y_2 + y_33 ### Core Logic Given equations: 1) Circle: y^2 = 8x - x^2 2) Hyperbola: 4x^2 - 9y^2 = 36 Substitute circle's y^2 into hyperbola equation: 4x^2 - 9(8x - x^2) = 36 implies 4x^2 - 72x + 9x^2 = 36 13x^2 - 72x - 36 = 0 implies (13x + 6)(x - 6) = 0 If x = -6/13, y^2 < 0 (rejected). Thus, x = 6. Substituting x = 6 into circle: y^2 = 8(6) - 6^2 = 48 - 36 = 12 implies y = pm sqrt12. The intersection points are A(6, sqrt12) and B(6, -sqrt12). ### Step 1: Relate Centroid coordinates to P Let point P have coordinates (alpha, beta). Since P lies on 2x - 3y + 4 = 0: 2alpha - 3beta + 4 = 0 implies beta = frac2alpha + 43 Let the centroid be (h, k): h = frac6 + 6 + alpha3 = frac12 + alpha3 implies alpha = 3h - 12 k = fracsqrt12 - sqrt12 + beta3 = fracbeta3 implies beta = 3k ### Step 2: Form the Locus Equation Substitute alpha and \beta into the line equation of P: 2(3h - 12) - 3(3k) + 4 = 0 6h - 24 - 9k + 4 = 0 6h - 9k = 20 Replacing (h, k) with general coordinates (x, y) gives the locus: 6x - 9y = 20 ### Pattern Recognition Notice how the y-coordinates of intersection points A and B are symmetric (\,pmsqrt12\,), meaning their sum is zero. This simplifies the expression for k instantly to just beta/3. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Coordinate Geometry Class 11 Mathematics: Conic Sections

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