### Related Formula Parametric equation of a straight line: x = x_1 + rcostheta, quad y = y_1 + rsintheta$x = x_1 + r\cos\theta, \quad y = y_1 + r\sin\theta$ ### Core Logic Step 1: Find point P(x_1,y_1)$P(x_1,y_1)$ on circle C$C$. Equation of tangent at P$P$ on x^2 + y^2 - 2y - 1 = 0$x^2 + y^2 - 2y - 1 = 0$ is xx_1 + y(y_1 - 1) - (y_1 + 1) = 0$xx_1 + y(y_1 - 1) - (y_1 + 1) = 0$. Comparing with line x + y = 3 implies fracx_11 = fracy_1 - 11 = fracy_1 + 13$x + y = 3 \implies \frac{x_1}{1} = \frac{y_1 - 1}{1} = \frac{y_1 + 1}{3}$. Solving gives x_1 = 1, y_1 = 2$x_1 = 1, y_1 = 2$. Thus, P = (1, 2)$P = (1, 2)$. ### Step 1: Use Line Parametrics for Q and R Line x + y = 3$x + y = 3$ makes an angle theta = 135^circ$\theta = 135^{\circ}$ with the positive x-axis. Using parametric distances from P(1,2)$P(1,2)$ with r = PQ = frac2sqrt23$r = PQ = \frac{2\sqrt{2}}{3}$: x = 1 pm rcos(135^circ) = 1 mp fracrsqrt2$x = 1 \pm r\cos(135^{\circ}) = 1 \mp \frac{r}{\sqrt{2}}$ y = 2 pm rsin(135^circ) = 2 pm fracrsqrt2$y = 2 \pm r\sin(135^{\circ}) = 2 \pm \frac{r}{\sqrt{2}}$ Substitute r = frac2sqrt23$r = \frac{2\sqrt{2}}{3}$: For Q$Q$: x_2 = 1 + frac23 = frac53, y_2 = 2 - frac23 = frac43$x_2 = 1 + \frac{2}{3} = \frac{5}{3}, y_2 = 2 - \frac{2}{3} = \frac{4}{3}$. For R$R$: x_3 = 1 - frac23 = frac13, y_3 = 2 + frac23 = frac83$x_3 = 1 - \frac{2}{3} = \frac{1}{3}, y_3 = 2 + \frac{2}{3} = \frac{8}{3}$. ### Step 2: Evaluate Final Expression Calculate the products: x_1y_1 = 1 times 2 = 2$x_1y_1 = 1 \times 2 = 2$ x_2y_2 = frac53 times frac43 = frac209$x_2y_2 = \frac{5}{3} \times \frac{4}{3} = \frac{20}{9}$ x_3y_3 = frac13 times frac83 = frac89$x_3y_3 = \frac{1}{3} \times \frac{8}{3} = \frac{8}{9}$ 9(x_1y_1 + x_2y_2 + x_3y_3) = 9left(2 + frac209 + frac89right) = 18 + 20 + 8 = 46$9(x_1y_1 + x_2y_2 + x_3y_3) = 9\left(2 + \frac{20}{9} + \frac{8}{9}\right) = 18 + 20 + 8 = 46$ ### Pattern Recognition Parametric distance equations are perfect for lines containing midpoints. This approach bypasses calculating the individual ellipse equations a^2, b^2$a^2, b^2$ completely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Circles Class 11 Mathematics: Conic Sections

### Related Formula Length of latus rectum of an ellipse and its eccentricity relation are: textLR = frac2b^2a$\text{LR} = \frac{2b^2}{a}$ e^2 = 1 - fracb^2a^2$e^2 = 1 - \frac{b^2}{a^2}$ ### Core Logic Given length of textLR = 10 implies frac2b^2a = 10 implies b^2 = 5a quad dots text(i)$\text{LR} = 10 \implies \frac{2b^2}{a} = 10 \implies b^2 = 5a \quad \dots \text{(i)}$ Now, let's find the minimum value of f(t) = t^2 + t + frac1112$f(t) = t^2 + t + \frac{11}{12}$. Differentiating: f'(t) = 2t + 1 = 0 implies t = -frac12$f'(t) = 2t + 1 = 0 \implies t = -\frac{1}{2}$. textMinimum value e = fleft(-frac12 ight) = left(-frac12 ight)^2 + left(-frac12 ight) + frac1112 = frac14 - frac12 + frac1112 = frac3 - 6 + 1112 = frac812 = frac23$\text{Minimum value } e = f\left(-\frac{1}{2} ight) = \left(-\frac{1}{2} ight)^2 + \left(-\frac{1}{2} ight) + \frac{11}{12} = \frac{1}{4} - \frac{1}{2} + \frac{11}{12} = \frac{3 - 6 + 11}{12} = \frac{8}{12} = \frac{2}{3}$ ### Step 1: Solve for a and b Using eccentricity formula: e^2 = frac49 = 1 - fracb^2a^2 implies fracb^2a^2 = frac59 implies b^2 = frac5a^29 quad dots text(ii)$e^2 = \frac{4}{9} = 1 - \frac{b^2}{a^2} \implies \frac{b^2}{a^2} = \frac{5}{9} \implies b^2 = \frac{5a^2}{9} \quad \dots \text{(ii)}$ Equating (i) and (ii): 5a = frac5a^29 implies a = 9$5a = \frac{5a^2}{9} \implies a = 9$ Then from (i): b^2 = 5(9) = 45 implies b = 3sqrt5$b^2 = 5(9) = 45 \implies b = 3\sqrt{5}$ Hence, a^2 = 81$a^2 = 81$. ### Step 2: Calculate a^2 + b^2 a^2 + b^2 = 81 + 45 = 126$a^2 + b^2 = 81 + 45 = 126$ ### Pattern Recognition A quadratic function at^2+bt+c$at^2+bt+c$ reaches its extreme value at t = -fracb2a$t = -\frac{b}{2a}$. Using this layout avoids full calculus derivation steps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections Class 11 Mathematics: Quadratic Equations

### 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