If A and B are the points of intersection of the circle x^2+y^2-8x=0$x^{2}+y^{2}-8x=0$ and the hyperbola fracx^29-fracy^24=1$\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$ and a point P moves on the line 2x-3y+4=0$2x-3y+4=0$, then the centroid of Delta PAB$\Delta PAB$ lies on the line:

### Related Formula Product of focal distances for a point on a hyperbola satisfies: PF_1 cdot PF_2 = e^2alpha^2 - a^2$$PF_1 \cdot PF_2 = e^2\alpha^2 - a^2$$ ### Core Logic Given focus ae = 5$ae = 5$ and directrix fracae = frac95$\frac{a}{e} = \frac{9}{5}$. Multiplying gives a^2 = 9 implies a = 3$a^2 = 9 \implies a = 3$. Then 3e = 5 implies e = frac53$3e = 5 \implies e = \frac{5}{3}$. Using hyperbola identity: b^2 = a^2(e^2 - 1) = 9left(frac259 - 1right) = 16 implies b = 4$b^2 = a^2(e^2 - 1) = 9\left(\frac{25}{9} - 1\right) = 16 \implies b = 4$. The equation of the hyperbola is: fracx^29 - fracy^216 = 1$$\frac{x^2}{9} - \frac{y^2}{16} = 1$$ ### Step 1: Point Substitution Since point (alpha, 2sqrt5)$(\alpha, 2\sqrt{5})$ lies on the hyperbola: fracalpha^29 - frac2016 = 1 implies fracalpha^29 = 1 + frac54 = frac94 implies alpha^2 = frac814$$\frac{\alpha^2}{9} - \frac{20}{16} = 1 \implies \frac{\alpha^2}{9} = 1 + \frac{5}{4} = \frac{9}{4} \implies \alpha^2 = \frac{81}{4}$$ ### Step 2: Focal Product Calculation Evaluating p$p$: p = e^2alpha^2 - a^2 = left(frac259right)left(frac814right) - 9 = frac2254 - 9 = frac1894$$p = e^2\alpha^2 - a^2 = \left(\frac{25}{9}\right)\left(\frac{81}{4}\right) - 9 = \frac{225}{4} - 9 = \frac{189}{4}$$ 4p = 189$4p = 189$ ### Pattern Recognition Combining the metric coordinates ae$ae$ and fracae$\frac{a}{e}$ via simple multiplication locks in the basic structural axis parameter a^2$a^2$ immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Related Formula The equation of a chord of an ellipse whose midpoint (x_1, y_1)$(x_1, y_1)$ is given is: T = S_1$T = S_1$ fracxx_1a^2 + fracyy_1b^2 = fracx_1^2a^2 + fracy_1^2b^2$$\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2}$$ ### Core Logic For the given ellipse fracx^225 + fracy^216 = 1$\frac{x^2}{25} + \frac{y^2}{16} = 1$ and midpoint (x_1, y_1) = (3, 1)$(x_1, y_1) = (3, 1)$ , we substitute these values into the T = S_1$T = S_1$ expression. ### Step 1: Substitution and Expansion Substituting the coordinates into the formula: frac3x25 + frac1y16 - 1 = frac3^225 + frac1^216 - 1$$\frac{3x}{25} + \frac{1y}{16} - 1 = \frac{3^2}{25} + \frac{1^2}{16} - 1$$ frac3x25 + fracy16 = frac925 + frac116$$\frac{3x}{25} + \frac{y}{16} = \frac{9}{25} + \frac{1}{16}$$ ### Step 2: Simplification Multiply both sides by the least common multiple of 25 and 16, which is 400: 16(3x) + 25(y) = 16(9) + 25(1)$$16(3x) + 25(y) = 16(9) + 25(1)$$ 48x + 25y = 144 + 25$$48x + 25y = 144 + 25$$ 48x + 25y = 169$48x + 25y = 169$ ### Pattern Recognition Whenever a midpoint is given for a chord of any second-degree conic curve, the relation T = S_1$T = S_1$ simplifies the process instantaneously without finding the individual intersection points. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Related Formula The locus definition of a parabola states that the squared distance from any point P(x,y)$P(x,y)$ to the focus S(x_0, y_0)$S(x_0, y_0)$ equals the squared perpendicular distance to the directrix line Ax + By + C = 0$Ax + By + C = 0$: (x - x_0)^2 + (y - y_0)^2 = frac(Ax + By + C)^2A^2 + By^2$$(x - x_0)^2 + (y - y_0)^2 = \frac{(Ax + By + C)^2}{A^2 + By^2}$$ ### Step 1: Determine the Focus coordinates The axis line of the parabola is perpendicular to the directrix x + 2y = 0$x + 2y = 0$ and passes through the vertex V(1.5, 3)$V(1.5, 3)$ [cite: 3354, 3355]. Slope of directrix = -0.5 Rightarrow$-0.5 \Rightarrow$ Slope of axis = 2. Equation of axis : y - 3 = 2left(x - frac32right) Rightarrow y - 2x = 0$$y - 3 = 2\left(x - \frac{3}{2}\right) \Rightarrow y - 2x = 0$$ [cite: 4059, 4060] The intersection of the axis (y - 2x = 0$y - 2x = 0$) and directrix (x + 2y = 0$x + 2y = 0$) gives the foot of the directrix, which is (0, 0)$(0, 0)$ [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)$$\left(\frac{3}{2}, 3\right) = \left(\frac{x_f + 0}{2}, \frac{y_f + 0}{2}\right) \Rightarrow \text{Focus } S = (3, 6)$$ ### Step 2: Derive the Parabola Locus Equation Equate the distance equations from point P(x,y)$P(x,y)$ : (x - 3)^2 + (y - 6)^2 = frac(x + 2y)^21^2 + 2^2$$(x - 3)^2 + (y - 6)^2 = \frac{(x + 2y)^2}{1^2 + 2^2}$$ 5left(x^2 - 6x + 9 + y^2 - 12y + 36right) = x^2 + 4xy + 4y^2$$5\left(x^2 - 6x + 9 + y^2 - 12y + 36\right) = x^2 + 4xy + 4y^2$$ 5x^2 - 30x + 45 + 5y^2 - 60y + 180 = 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$$4x^2 + y^2 - 4xy - 30x - 60y + 225 = 0$$ ### Step 3: Coefficient Extraction Compare with the equation template alpha x^2 + beta y^2 - gamma xy - 30x - 60y + 225 = 0$\alpha x^2 + \beta y^2 - \gamma xy - 30x - 60y + 225 = 0$ : alpha = 4, quad beta = 1, quad gamma = 4$$\alpha = 4, \quad \beta = 1, \quad \gamma = 4$$ alpha + beta + gamma = 4 + 1 + 4 = 9$$\alpha + \beta + \gamma = 4 + 1 + 4 = 9$$ ### 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)$(0,0)$ as the foot quickly reveals the focus coordinates via doubling. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Related Formula 1. For standard hyperbola fracx^2a^2 - fracy^2b^2 = 1$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: Latus Rectum = frac2b^2a$= \frac{2b^2}{a}$, transverse axis length = 2a$= 2a$, eccentricity relation b^2 = a^2(e^2 - 1)$b^2 = a^2(e^2 - 1)$. 2. For conjugate hyperbola -fracx^2A^2 + fracy^2B^2 = 1$-\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$: Latus Rectum = frac2A^2B$= \frac{2A^2}{B}$, transverse axis length = 2B$= 2B$, eccentricity relation A^2 = B^2(e^2 - 1)$A^2 = B^2(e^2 - 1)$. ### Step 1: Solve Parameters for Hyperbola H_1$H_1$ Given latus rectum and eccentricity parameters [cite: 3416, 3417]: frac2b^2a = 15sqrt2$$\frac{2b^2}{a} = 15\sqrt{2}$$ e_1^2 = 1 + fracb^2a^2 = frac52 Rightarrow fracb^2a^2 = frac32 Rightarrow b^2 = frac32a^2$$e_1^2 = 1 + \frac{b^2}{a^2} = \frac{5}{2} \Rightarrow \frac{b^2}{a^2} = \frac{3}{2} \Rightarrow b^2 = \frac{3}{2}a^2$$ Substitute b^2$b^2$ into latus rectum equation: frac2left(frac32a^2right)a = 3a = 15sqrt2 Rightarrow a = 5sqrt2$$\frac{2\left(\frac{3}{2}a^2\right)}{a} = 3a = 15\sqrt{2} \Rightarrow a = 5\sqrt{2}$$ b^2 = frac32(50) = 75 Rightarrow b = 5sqrt3$$b^2 = \frac{3}{2}(50) = 75 \Rightarrow b = 5\sqrt{3}$$ Transverse axis length of H_1 = 2a = 10sqrt2$H_1 = 2a = 10\sqrt{2}$. ### Step 2: Solve Parameters for Hyperbola H_2$H_2$ The product of their transverse axes lengths equals 100sqrt10$100\sqrt{10}$ [cite: 3418, 4094]: 2a cdot 2B = 100sqrt10 Rightarrow 10sqrt2 cdot 2B = 100sqrt10 Rightarrow 2B = 10sqrt5 Rightarrow B = 5sqrt5$$2a \cdot 2B = 100\sqrt{10} \Rightarrow 10\sqrt{2} \cdot 2B = 100\sqrt{10} \Rightarrow 2B = 10\sqrt{5} \Rightarrow B = 5\sqrt{5}$$ [cite: 4094, 4095, 4096] Given latus rectum for conjugate hyperbola H_2$H_2$ [cite: 3416, 4093]: frac2A^2B = 12sqrt5 Rightarrow frac2A^25sqrt5 = 12sqrt5 Rightarrow 2A^2 = 60 times 5 = 300 Rightarrow A^2 = 150$$\frac{2A^2}{B} = 12\sqrt{5} \Rightarrow \frac{2A^2}{5\sqrt{5}} = 12\sqrt{5} \Rightarrow 2A^2 = 60 \times 5 = 300 \Rightarrow A^2 = 150$$ [cite: 4093, 4097] ### Step 3: Calculate 25e_2^2$25e_2^2$ Find e_2^2$e_2^2$ using the conjugate eccentricity relation : e_2^2 = 1 + fracA^2B^2 = 1 + frac150(5sqrt5)^2 = 1 + frac150125 = 1 + frac65 = frac115$$e_2^2 = 1 + \frac{A^2}{B^2} = 1 + \frac{150}{(5\sqrt{5})^2} = 1 + \frac{150}{125} = 1 + \frac{6}{5} = \frac{11}{5}$$ [cite: 4104, 4106, 4107] Compute 25e_2^2$25e_2^2$ [cite: 3418, 4107]: 25e_2^2 = 25 times frac115 = 55$$25e_2^2 = 25 \times \frac{11}{5} = 55$$ [cite: 4105, 4107] ### Pattern Recognition Pay extra attention to conjugate-type equations (-fracx^2A^2 + fracy^2B^2 = 1$-\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$). For these vertical hyperbolas, the transverse axis corresponds to the variable with the positive sign (2B$2B$), and the components inside the latus rectum swap positions proportionally. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Related Formula For a point P(x_1, y_1)$P(x_1, y_1)$ on a standard horizontal ellipse, the focal distances are given by (a + ex_1)$(a + ex_1)$ and (a - ex_1)$(a - ex_1)$. Their algebraic product equals: textProduct = a^2 - e^2 x_1^2$$\text{Product} = a^2 - e^2 x_1^2$$ ### Core Logic Substitute the given coordinate x_1 = sqrt3$x_1 = \sqrt{3}$ into our focal distance product value: a^2 - e^2(sqrt3)^2 = frac74 implies a^2 - 3e^2 = frac74$$a^2 - e^2(\sqrt{3})^2 = \frac{7}{4} \implies a^2 - 3e^2 = \frac{7}{4}$$ 4a^2 = 7 + 12e^2 quad dots (1)$$4a^2 = 7 + 12e^2 \quad \dots (1)$$ ### Step 1: Apply Point Ingestion into Conic Equation Since the point left(sqrt3, frac12right)$\left(\sqrt{3}, \frac{1}{2}\right)$ lies directly on the ellipse perimeter: frac3a^2 + frac(1/2)^2b^2 = 1 implies frac3a^2 + frac14b^2 = 1$$\frac{3}{a^2} + \frac{(1/2)^2}{b^2} = 1 \implies \frac{3}{a^2} + \frac{1}{4b^2} = 1$$ Using the standard eccentricity identity b^2 = a^2(1-e^2)$b^2 = a^2(1-e^2)$: frac3a^2 + frac14a^2(1-e^2) = 1 implies 12(1-e^2) + 1 = 4a^2(1-e^2)$$\frac{3}{a^2} + \frac{1}{4a^2(1-e^2)} = 1 \implies 12(1-e^2) + 1 = 4a^2(1-e^2)$$ 13 - 12e^2 = 4a^2(1-e^2) quad dots (2)$$13 - 12e^2 = 4a^2(1-e^2) \quad \dots (2)$$ ### Step 2: Solve the Bi-quadratic Equation for Eccentricity Substitute 4a^2$4a^2$ from equation (1) directly into equation (2): 13 - 12e^2 = (7 + 12e^2)(1-e^2)$$13 - 12e^2 = (7 + 12e^2)(1-e^2)$$ 13 - 12e^2 = 7 - 7e^2 + 12e^2 - 12e^4$$13 - 12e^2 = 7 - 7e^2 + 12e^2 - 12e^4$$ 12e^4 - 17e^2 + 6 = 0$$12e^4 - 17e^2 + 6 = 0$$ Factorize the quadratic form in terms of e^2$e^2$: (4e^2 - 3)(3e^2 - 2) = 0 implies e^2 = frac34 text or e^2 = frac23$$(4e^2 - 3)(3e^2 - 2) = 0 \implies e^2 = \frac{3}{4} \text{ or } e^2 = \frac{2}{3}$$ This yields two distinct valid eccentricity parameters: e_1 = fracsqrt32, quad e_2 = sqrtfrac23$$e_1 = \frac{\sqrt{3}}{2}, \quad e_2 = \sqrt{\frac{2}{3}}$$ ### Step 3: Evaluate Target Absolute Difference Find the difference between the two eccentricity parameters: textDifference = fracsqrt32 - fracsqrt2sqrt3 = frac3 - 2sqrt22sqrt3$$\text{Difference} = \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{\sqrt{3}} = \frac{3 - 2\sqrt{2}}{2\sqrt{3}}$$ ### Pattern Recognition Converting b^2$b^2$ to a^2(1-e^2)$a^2(1-e^2)$ 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