### 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

### Related Formula Equation of a chord of an ellipse with given midpoint (x_1, y_1)$(x_1, y_1)$ is given by 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 Given ellipse: fracx^29 + fracy^24 = 1$\frac{x^2}{9} + \frac{y^2}{4} = 1$ and midpoint (x_1, y_1) = (sqrt2, frac43)$(x_1, y_1) = (\sqrt{2}, \frac{4}{3})$. Applying T = S_1$T = S_1$: fracxsqrt29 + fracy(4/3)4 = frac(sqrt2)^29 + frac(4/3)^24$\frac{x\sqrt{2}}{9} + \frac{y(4/3)}{4} = \frac{(\sqrt{2})^2}{9} + \frac{(4/3)^2}{4}$ fracsqrt2x9 + fracy3 = frac29 + frac1636$\frac{\sqrt{2}x}{9} + \frac{y}{3} = \frac{2}{9} + \frac{16}{36}$ fracsqrt2x9 + fracy3 = frac29 + frac49 = frac69$\frac{\sqrt{2}x}{9} + \frac{y}{3} = \frac{2}{9} + \frac{4}{9} = \frac{6}{9}$ Multiplying through by 9: sqrt2x + 3y = 6 implies 3y = 6 - sqrt2x$\sqrt{2}x + 3y = 6 \implies 3y = 6 - \sqrt{2}x$ ### Step 1: Find Intersection Points with Ellipse Substitute 3y = 6 - sqrt2x$3y = 6 - \sqrt{2}x$ into the multiplied form of ellipse 4x^2 + 9y^2 = 36$4x^2 + 9y^2 = 36$: 4x^2 + (3y)^2 = 36$4x^2 + (3y)^2 = 36$ 4x^2 + (6 - sqrt2x)^2 = 36$4x^2 + (6 - \sqrt{2}x)^2 = 36$ 4x^2 + 36 + 2x^2 - 12sqrt2x = 36$4x^2 + 36 + 2x^2 - 12\sqrt{2}x = 36$ 6x^2 - 12sqrt2x = 0$6x^2 - 12\sqrt{2}x = 0$ 6x(x - 2sqrt2) = 0$6x(x - 2\sqrt{2}) = 0$ Thus, x = 0$x = 0$ or x = 2sqrt2$x = 2\sqrt{2}$. ### Step 2: Find y-coordinates and Chord Length If x_1 = 0 implies 3y_1 = 6 implies y_1 = 2$x_1 = 0 \implies 3y_1 = 6 \implies y_1 = 2$ If x_2 = 2sqrt2 implies 3y_2 = 6 - sqrt2(2sqrt2) = 6 - 4 = 2 implies y_2 = frac23$x_2 = 2\sqrt{2} \implies 3y_2 = 6 - \sqrt{2}(2\sqrt{2}) = 6 - 4 = 2 \implies y_2 = \frac{2}{3}$ The end points of the chord are A(0, 2)$A(0, 2)$ and B(2sqrt2, frac23)$B(2\sqrt{2}, \frac{2}{3})$. textLength of chord AB = sqrt(2sqrt2 - 0)^2 + left(frac23 - 2right)^2$\text{Length of chord } AB = \sqrt{(2\sqrt{2} - 0)^2 + \left(\frac{2}{3} - 2\right)^2}$ AB = sqrt8 + left(-frac43right)^2 = sqrt8 + frac169 = sqrtfrac889 = fracsqrt4 times 223 = frac2sqrt223$AB = \sqrt{8 + \left(-\frac{4}{3}\right)^2} = \sqrt{8 + \frac{16}{9}} = \sqrt{\frac{88}{9}} = \frac{\sqrt{4 \times 22}}{3} = \frac{2\sqrt{22}}{3}$ Comparing with frac2sqrtalpha3$\frac{2\sqrt{\alpha}}{3}$, we get alpha = 22$\alpha = 22$. ### Pattern Recognition When the intersection equation results in a simple factoring like 6x^2 - 12sqrt2x = 0$6x^2 - 12\sqrt{2}x = 0$, 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)

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