Let mathbfe_1$\mathbf{e}_1$ and mathbfe_2$\mathbf{e}_2$ be the eccentricities of the ellipse fracmathrmx^2mathrmb^2 + fracmathrmy^225 = 1$\frac{\mathrm{x}^2}{\mathrm{b}^2} + \frac{\mathrm{y}^2}{25} = 1$ and the hyperbola fracmathrmx^216 - fracmathrmy^2mathrmb^2 = 1$\frac{\mathrm{x}^2}{16} - \frac{\mathrm{y}^2}{\mathrm{b}^2} = 1$, respectively. If mathrmb < 5$\mathrm{b} < 5$ and mathrme_1mathrme_2 = 1$\mathrm{e}_1\mathrm{e}_2 = 1$, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is:

### Related Formula The condition for a line y = mx + p$y = mx + p$ to be tangent to an ellipse fracx^2a^2 + fracy^2b^2 = 1$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is: p^2 = a^2m^2 + b^2$$p^2 = a^2m^2 + b^2$$ The coordinate of the point of contact is given by left(-fraca^2mp, fracb^2pright)$\left(-\frac{a^2m}{p}, \frac{b^2}{p}\right)$. ### Core Logic Given the ellipse parameter values a^2 = 16$a^2 = 16$ and b^2 = 9$b^2 = 9$, and tangent line slope m = 1$m = 1$: p^2 = 16(1)^2 + 9 = 25 implies p = pm 5$$p^2 = 16(1)^2 + 9 = 25 \implies p = \pm 5$$ Thus, the two values of p$p$ are 5$5$ and -5$-5$. The points of contact A$A$ and B$B$ are: - For p = 5$p = 5$: A = left(-frac16(1)5, frac95right) = left(-frac165, frac95right)$A = \left(-\frac{16(1)}{5}, \frac{9}{5}\right) = \left(-\frac{16}{5}, \frac{9}{5}\right)$ - For p = -5$p = -5$: B = left(-frac16(1)-5, frac9-5right) = left(frac165, -frac95right)$B = \left(-\frac{16(1)}{-5}, \frac{9}{-5}\right) = \left(\frac{16}{5}, -\frac{9}{5}\right)$ ### Step 1: Intersecting line with Ellipse The line y = x$y = x$ intersects the ellipse fracx^216 + fracy^29 = 1$\frac{x^2}{16} + \frac{y^2}{9} = 1$: fracx^216 + fracx^29 = 1 implies frac25x^2144 = 1 implies x^2 = frac14425 implies x = pm frac125$$\frac{x^2}{16} + \frac{x^2}{9} = 1 \implies \frac{25x^2}{144} = 1 \implies x^2 = \frac{144}{25} \implies x = \pm \frac{12}{5}$$ Since y = x$y = x$, the intersection points C$C$ and D$D$ are: C = left(-frac125, -frac125right) quad textand quad D = left(frac125, frac125right)$$C = \left(-\frac{12}{5}, -\frac{12}{5}\right) \quad \text{and} \quad D = \left(\frac{12}{5}, \frac{12}{5}\right)$$ ### Step 2: Calculating Quadrilateral Area The area of quadrilateral ABCD$ABCD$ with vertices mapped symmetrically can be computed using the standard coordinate determinant matrix layout formula: textArea = frac12 beginvmatrix x_A & y_A & 1 \\ x_B & y_B & 1 \\ x_C & y_C & 1 endvmatrix + dots = 24$$\text{Area} = \frac{1}{2} \begin{vmatrix} x_A & y_A & 1 \\ x_B & y_B & 1 \\ x_C & y_C & 1 \end{vmatrix} + \dots = 24$$ ### Pattern Recognition Notice that the tangent lines are parallel and symmetric (p = pm 5$p = \pm 5$), and the intersecting line passes through the origin. This symmetry creates a geometric parallelogram, simplifying your area calculation by doubling the area of triangle ABD$ABD$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Related Formula Distance between foci = 2be$= 2be$ (for vertical major axis). Length of Latus Rectum (textL.R.) = frac2a^2b$(\text{L.R.}) = \frac{2a^2}{b}$. ### Core Logic Foci are F_1(2,5)$F_1(2,5)$ and F_2(2,-3)$F_2(2,-3)$. The x$x$-coordinates are identical, indicating a vertical ellipse. Distance between foci: 2be = 5 - (-3) = 8 implies be = 4$$2be = 5 - (-3) = 8 \implies be = 4$$ Given eccentricity e = frac45$e = \frac{4}{5}$: bleft(frac45right) = 4 implies b = 5$$b\left(\frac{4}{5}\right) = 4 \implies b = 5$$

### Step 1: Calculate Minor Axis length Use eccentricity relation: a^2 = b^2(1 - e^2) implies a^2 = 25left(1 - frac1625right) = 25 times frac925 = 9 implies a = 3$$a^2 = b^2(1 - e^2) \implies a^2 = 25\left(1 - \frac{16}{25}\right) = 25 \times \frac{9}{25} = 9 \implies a = 3$$ ### Step 2: Evaluate Latus Rectum textL.R. = frac2a^2b = frac2 times 95 = frac185$$\text{L.R.} = \frac{2a^2}{b} = \frac{2 \times 9}{5} = \frac{18}{5}$$ ### Pattern Recognition Always verify axis orientation (horizontal vs vertical) from coordinates before blindly substituting into standard formulas. Symmetrical components match axis lengths parameters directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### 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