Let y^2 = 12x$y^2 = 12x$ the parabola and S$S$ be its focus. Let PQ$PQ$ be a focal chord of the parabola such that (SP) (SQ) = frac1474$(SQ) = \frac{147}{4}$. Let C$C$ be the circle described taking PQ$PQ$ as a diameter. If the equation of a circle C$C$ is 64x^2 + 64y^2 - alpha x - 64sqrt3y = beta$64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta$, then \beta - \alpha is equal to

### Related Formula Equation of a chord of a conic section with a given midpoint (x_1, y_1)$(x_1, y_1)$ is: T = S_1$T = S_1$ ### Core Logic Given midpoint Mleft(frac52, frac12 ight)$M\left(\frac{5}{2}, \frac{1}{2} ight)$ and ellipse fracx^29 + fracy^24 = 1$\frac{x^2}{9} + \frac{y^2}{4} = 1$.

Write T$T$ and S_1$S_1$ terms: T: fracxleft(frac52right)9 + fracyleft(frac12right)4$$T: \frac{x\left(\frac{5}{2}\right)}{9} + \frac{y\left(\frac{1}{2}\right)}{4}$$ S_1: fracleft(frac52right)^29 + fracleft(frac12right)^24$$S_1: \frac{\left(\frac{5}{2}\right)^2}{9} + \frac{\left(\frac{1}{2}\right)^2}{4}$$ Equating both sides: frac5x18 + fracy8 = frac2536 + frac116$$\frac{5x}{18} + \frac{y}{8} = \frac{25}{36} + \frac{1}{16}$$ ### Step 1: Simplify to Standard Form Multiply the entire equation by 144 to eliminate fractions: 144left(frac5x18right) + 144left(fracy8right) = 144left(frac2536right) + 144left(frac116right)$$144\left(\frac{5x}{18}\right) + 144\left(\frac{y}{8}\right) = 144\left(\frac{25}{36}\right) + 144\left(\frac{1}{16}\right)$$ 40x + 18y = 4(25) + 9(1)$$40x + 18y = 4(25) + 9(1)$$ 40x + 18y = 109$40x + 18y = 109$ Comparing this directly with alpha x + beta y = 109$\alpha x + \beta y = 109$ provides: alpha = 40, quad beta = 18$$\alpha = 40, \quad \beta = 18$$ alpha + beta = 40 + 18 = 58$$\alpha + \beta = 40 + 18 = 58$$ ### Pattern Recognition Whenever you see 'chord whose midpoint is given', write T = S_1$T = S_1$ automatically. Match coefficients directly at the final step after equating constant integers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### 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)