### Related Formula Area of a trapezium is given by: textArea = frac12 times (textsum of parallel sides) times (textdistance between them)$\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{distance between them})$ ### Core Logic Let the coordinates of the vertices be parameterized on the parabola y^2 = 4x$y^2 = 4x$. Since AD$AD$ and BC$BC$ are parallel to the y-axis, the coordinates take the form: A(at_1^2, 2at_1)$A(at_1^2, 2at_1)$ and D(at_1^2, -2at_1)$D(at_1^2, -2at_1)$ B(at_2^2, 2at_2)$B(at_2^2, 2at_2)$ and C(at_2^2, -2at_2)$C(at_2^2, -2at_2)$ Given a=1$a=1$, the points simplify accordingly.

### Step 1: Using Diagonal Properties The length of diagonal AC$AC$ passing through focal point (1,0)$(1,0)$ implies focal chord properties: textLength AC = aleft(t_1 + frac1t_1right)^2 = frac254$\text{Length AC} = a\left(t_1 + \frac{1}{t_1}\right)^2 = \frac{25}{4}$ t_1 + frac1t_1 = pmfrac52 implies t_1 = 2 text or frac12$t_1 + \frac{1}{t_1} = \pm\frac{5}{2} \implies t_1 = 2 \text{ or } \frac{1}{2}$ ### Step 2: Finding Coordinates and Area Substituting t_1 = 2$t_1 = 2$, we get: Aleft(frac12, 1right), Dleft(frac14, -1right), B(4, 4), C(4, -4)$A\left(\frac{1}{2}, 1\right), D\left(\frac{1}{4}, -1\right), B(4, 4), C(4, -4)$ Evaluating the area formula: textArea = frac12 times (8 + 2) times left(4 - frac14 ight) = frac754$\text{Area} = \frac{1}{2} \times (8 + 2) \times \left(4 - \frac{1}{4} ight) = \frac{75}{4}$ ### Pattern Recognition Focal chords of parabolas always satisfy t_1 t_2 = -1$t_1 t_2 = -1$. Recognizing the passage through (1,0)$(1,0)$ unlocks quick parametric simplifications. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections

### Related Formula Area of an ellipse with semi-axes a$a$ and b$b$: textArea = pi a b$\text{Area} = \pi a b$ ### Core Logic Calculate the constant eccentricity e$e$ from the initial ellipse E_1$E_1$:

e = sqrt1 - frac49 = fracsqrt53$e = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$ For any subsequent ellipse E_2$E_2$, its major axis equals the minor axis of E_1$E_1$ (2b_1 = 4 implies a_2 = 2$2b_1 = 4 \implies a_2 = 2$). Since eccentricity remains constant: frac59 = 1 - fracb_2^2a_2^2 = 1 - fracb_2^24 implies b_2^2 = frac169 implies b_2 = frac43$\frac{5}{9} = 1 - \frac{b_2^2}{a_2^2} = 1 - \frac{b_2^2}{4} \implies b_2^2 = \frac{16}{9} \implies b_2 = \frac{4}{3}$ ### Step 1: Finding the Area Sequence Terms Evaluate the area values for the initial ellipses: A_1 = pi cdot 3 cdot 2 = 6pi$A_1 = \pi \cdot 3 \cdot 2 = 6\pi$ A_2 = pi cdot 2 cdot frac43 = frac8pi3$A_2 = \pi \cdot 2 \cdot \frac{4}{3} = \frac{8\pi}{3}$ The areas form an infinite geometric progression with a common ratio r = frac49$r = \frac{4}{9}$. ### Step 2: Summing the Infinite Geometric Series sum_i=1^infty A_i = frac6pi1 - frac49 = frac6pifrac59 = frac54pi5$\sum_{i=1}^{\infty} A_i = \frac{6\pi}{1 - \frac{4}{9}} = \frac{6\pi}{\frac{5}{9}} = \frac{54\pi}{5}$ Evaluating the final scaling formula: frac5pi left( frac54pi5 right) = 54$\frac{5}{\pi} \left( \frac{54\pi}{5} \right) = 54$ ### Pattern Recognition Iterative dimensional scaling creates geometric progressions where the ratio equals the square of the linear scaling factor. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections

### Related Formula Parametric line equation relative to an offset point P(x_0, y_0)$P(x_0, y_0)$: x = x_0 + rcostheta, quad y = y_0 + rsintheta$x = x_0 + r\cos\theta, \quad y = y_0 + r\sin\theta$

### Core Logic Assume any line through P$P$ can be represented parametrically by [cite: 1277]: Q(sqrt5 + rcostheta, sqrt5 + rsintheta)$Q(\sqrt{5} + r\cos\theta, \sqrt{5} + r\sin\theta)$ [cite: 1277] Substitute coordinates into the standard ellipse formula [cite: 1278]: 25(sqrt5 + rcostheta)^2 + 36(sqrt5 + rsintheta)^2 = 900$25(\sqrt{5} + r\cos\theta)^2 + 36(sqrt{5} + r\sin\theta)^2 = 900$ [cite: 1278] Expanding and gathering powers of r$r$ yields [cite: 1280]: r^2(25cos^2theta + 36sin^2theta) + 2sqrt5r(25costheta + 36sintheta) - 595 = 0$r^2(25\cos^2\theta + 36\sin^2\theta) + 2\sqrt{5}r(25\cos\theta + 36\sin\theta) - 595 = 0$ [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$PA \cdot PB = |r_1 r_2| = \frac{595}{25\cos^2\theta + 36\sin^2\theta} = \frac{595}{25 + 11\sin^2\theta}$ [cite: 1283] ### Step 1: Maximization condition To make PA cdot PB$PA \cdot PB$ maximum, the denominator must be minimized [cite: 1284]: sin^2theta = 0 implies theta = 0$\sin^2\theta = 0 \implies \theta = 0$ [cite: 1284] This implies the chord line AB$AB$ must run parallel to the x-axis [cite: 1285]: y_A = y_B = sqrt5$y_A = y_B = \sqrt{5}$ [cite: 1285] Substitute y = sqrt5$y = \sqrt{5}$ back into the ellipse equation to calculate x-coordinates [cite: 1286]: fracx^236 + frac525 = 1 implies fracx^236 = frac45 implies x^2 = frac1445$\frac{x^2}{36} + \frac{5}{25} = 1 \implies \frac{x^2}{36} = \frac{4}{5} \implies x^2 = \frac{144}{5}$ [cite: 1287] Therefore, the coordinates are x = pm frac12sqrt5$x = \pm \frac{12}{\sqrt{5}}$. ### Step 2: Distance value summation Compute PA^2 + PB^2$PA^2 + PB^2$ using coordinates directly [cite: 1289]: PA^2 + PB^2 = left(sqrt5 - frac12sqrt5right)^2 + left(sqrt5 + frac12sqrt5right)^2$PA^2 + PB^2 = \left(\sqrt{5} - \frac{12}{\sqrt{5}}\right)^2 + \left(\sqrt{5} + \frac{12}{\sqrt{5}}\right)^2$ [cite: 1289] = 2left(5 + frac1445right) = frac3385$= 2\left(5 + \frac{144}{5}\right) = \frac{338}{5}$ [cite: 1290] Multiplying by 5 gives the target integer answer [cite: 1290]: 5(PA^2 + PB^2) = 338$5(PA^2 + PB^2) = 338$ [cite: 1290] ### Pattern Recognition Parametric distances from a point intersecting a conic configuration usually form a standard quadratic equation in r$r$. The angle parameter immediately simplifies the boundary constraint optimization. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections (Ellipse)

### Related Formula Shortest distance between symmetric profiles: The minimal spacing normal line runs completely perpendicular to the mutual line of symmetry y=x$y=x$.

### Core Logic The given curve equations reflect symmetry across line y=x$y=x$[cite: 1382, 1383]. The tangent slope at the closest matching locations must run parallel to this mirror path [cite: 1403]: fracmathrmdymathrmdx = 1$\frac{\mathrm{d}y}{\mathrm{d}x} = 1$ [cite: 1403] Differentiate curve equation y = x^2 + 2$y = x^2 + 2$ [cite: 1404]: fracmathrmdymathrmdx = 2x = 1 implies x = frac12$\frac{\mathrm{d}y}{\mathrm{d}x} = 2x = 1 \implies x = \frac{1}{2}$ [cite: 1405, 1406] Substitute back to get y-coordinate [cite: 1406]: y = left(frac12right)^2 + 2 = frac94 implies Bleft(frac12, frac94right)$y = \left(\frac{1}{2}\right)^2 + 2 = \frac{9}{4} \implies B\left(\frac{1}{2}, \frac{9}{4}\right)$ [cite: 1406, 1407] By mirror symmetry, the corresponding point on the other parabola is [cite: 1407]: Aleft(frac94, frac12right)$A\left(\frac{9}{4}, \frac{1}{2}\right)$ [cite: 1407] ### Step 1: Calculating distance and circle radius Evaluate chord distance AB$AB$ using standard metrics [cite: 1407]: AB = sqrtleft(frac94 - frac12right)^2 + left(frac12 - frac94right)^2 = sqrt2 cdot left(frac74right)^2 = frac7sqrt24$AB = \sqrt{\left(\frac{9}{4} - \frac{1}{2}\right)^2 + \left(\frac{1}{2} - \frac{9}{4}\right)^2} = \sqrt{2 \cdot \left(\frac{7}{4}\right)^2} = \frac{7\sqrt{2}}{4}$ [cite: 1407, 1408] The diameter of the smallest circle spanning between these touching curves equals distance AB$AB$ [cite: 1408]. textRadius = fracAB2 = frac7sqrt28$\text{Radius} = \frac{AB}{2} = \frac{7\sqrt{2}}{8}$ [cite: 1408] ### Pattern Recognition Mutually inverse conic curves track symmetric footprints. Their closest distance segments always align perfectly perpendicular to the main baseline axis line y=x$y=x$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections (Parabola)

### Related Formula For hyperbola conics: 1. Focal distances product: PS_1 cdot PS_2 = |a^2e^2 - x^2|$PS_1 \cdot PS_2 = |a^2e^2 - x^2|$ 2. Point lying on curve constraint verification properties. ### Core Logic Since point P(4, 2sqrt3)$P(4, 2\sqrt{3})$ resides directly on hyperbola curve structure [cite: 1445]: frac16a^2 - frac12b^2 = 1 implies 16b^2 - 12a^2 = a^2b^2$\frac{16}{a^2} - \frac{12}{b^2} = 1 \implies 16b^2 - 12a^2 = a^2b^2$ [cite: 1446, 1448] Using focal coordinate geometric spacing properties [cite: 1445, 1451]: PS_1 = ae - 4, quad PS_2 = ae + 4 implies PS_1 cdot PS_2 = a^2e^2 - 16 = 32$PS_1 = ae - 4, \quad PS_2 = ae + 4 \implies PS_1 \cdot PS_2 = a^2e^2 - 16 = 32$ [cite: 1445, 1451] a^2e^2 = 48 implies a^2 + b^2 = 48$a^2e^2 = 48 \implies a^2 + b^2 = 48$ [cite: 1452, 1453] ### Step 1: Solving axis components values Substitute b^2 = 48 - a^2$b^2 = 48 - a^2$ back into original parameter product template [cite: 1448]: 16(48 - a^2) - 12a^2 = a^2(48 - a^2)$16(48 - a^2) - 12a^2 = a^2(48 - a^2)$ 768 - 16a^2 - 12a^2 = 48a^2 - a^4 implies a^4 - 76a^2 + 768 = 0$768 - 16a^2 - 12a^2 = 48a^2 - a^4 \implies a^4 - 76a^2 + 768 = 0$ (a^2 - 64)(a^2 - 12) = 0$(a^2 - 64)(a^2 - 12) = 0$ Testing parameters [cite: 1454]: From relation b^2 - a^2 = 4$b^2 - a^2 = 4$ [cite: 1454], we resolve the dimensions [cite: 1457, 1458]: a^2 = 8, quad b^2 = 12$a^2 = 8, \quad b^2 = 12$ [cite: 1457, 1458] ### Step 2: Total Calculation Length formulas for targeted metrics [cite: 1459]: p = 2b implies p^2 = 4b^2 = 4(12) = 48$p = 2b \implies p^2 = 4b^2 = 4(12) = 48$ q = frac2b^2a implies q^2 = frac4b^4a^2 = frac4(144)8 = 72$q = \frac{2b^2}{a} \implies q^2 = \frac{4b^4}{a^2} = \frac{4(144)}{8} = 72$ textFinal Metric Total = p^2 + q^2 = 48 + 72 = 120$\text{Final Metric Total} = p^2 + q^2 = 48 + 72 = 120$ [cite: 1459, 1460] ### Pattern Recognition Focal calculations relative to specific points simplify elegantly under eccentricity conversions. Solving quadratic frames sequentially ensures structural accuracy. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections (Hyperbola)