### Related Formula For a standard hyperbola: - Focus positions are (pm ae, 0)$(\pm ae, 0)$. - Length of semi-latus rectum is fracb^2a$\frac{b^2}{a}$. - Eccentricity identity linkage: b^2 = a^2(e^2 - 1) implies a^2e^2 = a^2 + b^2$b^2 = a^2(e^2 - 1) \implies a^2e^2 = a^2 + b^2$. ### Core Logic Given focus F_1 equiv (-ae, 0) equiv P(-3, 0)$F_1 \equiv (-ae, 0) \equiv P(-3, 0)$, so ae = 3$ae = 3$. The other focus is F_2 equiv (ae, 0) equiv (3, 0)$F_2 \equiv (ae, 0) \equiv (3, 0)$. The latus rectum passes vertically through F_2$F_2$, with endpoints L_1left(ae, fracb^2aright)$L_1\left(ae, \frac{b^2}{a}\right)$ and L_2left(ae, -fracb^2aright)$L_2\left(ae, -\frac{b^2}{a}\right)$. This segment subtends a right angle at P(-ae, 0)$P(-ae, 0)$. By symmetry, the top half angle at P$P$ must be exactly 45^circ$45^\circ$. ### Step 1: Set Up Slope Relationship

Using the geometric slope relationship: tan 45^circ = fractextheighttextbase = fracb^2/a2ae$\tan 45^\circ = \frac{\text{height}}{\text{base}} = \frac{b^2/a}{2ae}$ 1 = fracb^22a^2e implies 2a^2e = b^2 implies b^2 = 6a quad (textsince ae = 3)$1 = \frac{b^2}{2a^2e} \implies 2a^2e = b^2 \implies b^2 = 6a \quad (\text{since } ae = 3)$ ### Step 2: Solve the Quadratic Excentricity Equation Substitute ae = 3$ae = 3$ and b^2 = 6a$b^2 = 6a$ into the eccentricity identity a^2e^2 = a^2 + b^2$a^2e^2 = a^2 + b^2$: 9 = a^2 + 6a implies a^2 + 6a - 9 = 0$9 = a^2 + 6a \implies a^2 + 6a - 9 = 0$ Solving for a$a$ using the quadratic formula (taking the positive root since a > 0$a > 0$): a = frac-6 pm sqrt36 - 4(1)(-9)2 = frac-6 + sqrt722 = -3 + 3sqrt2 = 3(sqrt2 - 1)$a = \frac{-6 \pm \sqrt{36 - 4(1)(-9)}}{2} = \frac{-6 + \sqrt{72}}{2} = -3 + 3\sqrt{2} = 3(\sqrt{2} - 1)$ ### Step 3: Evaluate product and sum coefficients Now compute a^2b^2$a^2b^2$: a^2b^2 = a^2(6a) = 6a^3$a^2b^2 = a^2(6a) = 6a^3$ 6a^3 = 6left[3(sqrt2 - 1)right]^3 = 6 times 27 times (sqrt2 - 1)^3$6a^3 = 6\left[3(\sqrt{2} - 1)\right]^3 = 6 \times 27 \times (\sqrt{2} - 1)^3$ 6a^3 = 162 times (2sqrt2 - 6 + 3sqrt2 - 1) = 162 times (5sqrt2 - 7)$6a^3 = 162 \times (2\sqrt{2} - 6 + 3\sqrt{2} - 1) = 162 \times (5\sqrt{2} - 7)$ 6a^3 = 810sqrt2 - 1134$6a^3 = 810\sqrt{2} - 1134$ Matching with alphasqrt2 - beta$\alpha\sqrt{2} - \beta$ gives: alpha = 810 quad textand quad beta = 1134$\alpha = 810 \quad \text{and} \quad \beta = 1134$ Calculate the final required sum: alpha + beta = 810 + 1134 = 1944$\alpha + \beta = 810 + 1134 = 1944$ ### Pattern Recognition Recognizing that the right angle subtended at the opposite focus implies a perfect tan(45^circ)$\tan(45^\circ)$ right triangle instantly yields the key linear constraint b^2 = 2a(ae)$b^2 = 2a(ae)$, avoiding the need for lengthy distance-formula tracking. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Related Formula textFocal Distance Product on Vertical Ellipse = b^2 - e^2 k^2$\text{Focal Distance Product on Vertical Ellipse} = b^2 - e^2 k^2$ ### Core Logic Extract the coordinate center of the target circle, substitute it directly to locate the missing parameter p$p$, and resolve eccentricity metrics. ### Step 1: Extract Circle Metric Values For circle x^2 + y^2 - 2x - 4y - 11 = 0$x^{2} + y^{2} - 2x - 4y - 11 = 0$: textCentre C(1, 2), quad textRadius r = sqrt1 + 4 + 11 = 4$\text{Centre } C(1, 2), \quad \text{Radius } r = \sqrt{1 + 4 + 11} = 4$ ### Step 2: Standardize Ellipse Formulation Ellipse passes through point C(1,2)$C(1,2)$: 3(1)^2 + p(2)^2 = 4 implies 3 + 4p = 4 implies p = frac14$3(1)^2 + p(2)^2 = 4 \implies 3 + 4p = 4 \implies p = \frac{1}{4}$ Standard model form: fracx^24/3 + fracy^216 = 1$\frac{x^2}{4/3} + \frac{y^2}{16} = 1$ (b > a$b > a$, vertical configuration axis). e = sqrt1 - frac4/316 = sqrt1 - frac112 = sqrtfrac1112$e = \sqrt{1 - \frac{4/3}{16}} = \sqrt{1 - \frac{1}{12}} = \sqrt{\frac{11}{12}}$ ### Step 3: Evaluate Product Chain Focal distance elements at ordinate coordinate height k=2$k=2$ are bounded by b pm ek$b \pm ek$: f_1 f_2 = b^2 - e^2 k^2 = 16 - left(frac1112right) times 4 = 16 - frac113 = frac373$f_1 f_2 = b^2 - e^2 k^2 = 16 - \left(\frac{11}{12}\right) \times 4 = 16 - \frac{11}{3} = \frac{37}{3}$ Target evaluation expression response string: 6f_1 f_2 - r = 6 left(frac373right) - 4 = 74 - 4 = 70$6f_1 f_2 - r = 6 \left(\frac{37}{3}\right) - 4 = 74 - 4 = 70$ ### Pattern Recognition Pay attention to whether b > a$b > a$ or a > b$a > b$ when analyzing ellipse forms. Focal distance definitions swap directions immediately across major horizontal/vertical configurations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections Class 11 Mathematics: Circles

### Related Formula textTangent line at point (x_1, y_1) implies yy_1 = 2a(x+x_1)$\text{Tangent line at point } (x_1, y_1) \implies yy_1 = 2a(x+x_1)$ ### Core Logic Establish the tangent vector expression at the parabola intersection mark. Since this path line functions as a shared contact tangent boundaries sheet for the circular arc, impose radius equations. ### Step 1: Derive Shared Parabola Tangent Line Tangent line profile for y^2 = 9x$y^2 = 9x$ at coordinate indicator (4,6)$(4,6)$: 6y = 9 cdot left( fracx+42 right) implies 3x - 4y + 12 = 0$6y = 9 \cdot \left( \frac{x+4}{2} \right) \implies 3x - 4y + 12 = 0$ ### Step 2: Build Geometric Metric Connections Circle touches axis at (a,0)$(a,0)$, mapping coordinates center directly to C(a,r)$C(a,r)$. Perpendicular boundary constraint steps require: frac3a - 4r + 125 = pm r implies 3a + 12 = 4r pm 5r$\frac{3a - 4r + 12}{5} = \pm r \implies 3a + 12 = 4r \pm 5r$ ### Step 3: Solve for Radius Matrix Bounds Enforce circle equation intersection constraint profile (x-a)^2 + (y-r)^2 = r^2$(x-a)^2 + (y-r)^2 = r^2$ at point (4,6)$(4,6)$: a^2 - 8a - 12r + 52 = 0$a^2 - 8a - 12r + 52 = 0$ Evaluating the target systems from structural logic tracks rejects positive value parameters, providing: a = -14, quad r = 30$a = -14, \quad r = 30$ {{SOL_IMG_75}} ### Pattern Recognition Shared tangent elements connect independent conic fields. Locating circular center boundaries using axial coordinate tracking simplifies secondary equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections Class 11 Mathematics: Circles

### 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 Properties of focal chord parameter metrics in parabolas y^2 = 4ax$y^2 = 4ax$: t_1 cdot t_2 = -1$t_1 \cdot t_2 = -1$ Distance to the directrix property: SP = a(1 + t^2), quad SQ = aleft(1 + frac1t^2right)$SP = a(1 + t^2), \quad SQ = a\left(1 + \frac{1}{t^2}\right)$ ### Core Logic Given parabola y^2 = 12x implies a = 3$y^2 = 12x \implies a = 3$. Focus S = (3, 0)$S = (3, 0)$. Set up focal segments product equation: SP cdot SQ = 3(1+t^2) cdot 3left(1+frac1t^2right) = frac1474$SP \cdot SQ = 3(1+t^2) \cdot 3\left(1+\frac{1}{t^2}\right) = \frac{147}{4}$ 9 cdot frac(1+t^2)^2t^2 = frac1474 implies frac(1+t^2)^2t^2 = frac4912$9 \cdot \frac{(1+t^2)^2}{t^2} = \frac{147}{4} \implies \frac{(1+t^2)^2}{t^2} = \frac{49}{12}$ Solving for t^2$t^2$: 12t^4 - 25t^2 + 12 = 0 implies t^2 = frac34 quad textor quad frac43$12t^4 - 25t^2 + 12 = 0 \implies t^2 = \frac{3}{4} \quad \text{or} \quad \frac{4}{3}$ ### Step 1: Compute Endpoint Coordinate Bounds Choosing t = -fracsqrt32$t = -\frac{\sqrt{3}}{2}$ allows defining both chord coordinates symmetrically: P(3t^2, 6t) implies Pleft(frac94, -3sqrt3right)$P(3t^2, 6t) \implies P\left(\frac{9}{4}, -3\sqrt{3}\right)$ Qleft(frac3t^2, -frac6tright) implies Q(4, 4sqrt3)$Q\left(\frac{3}{t^2}, -\frac{6}{t}\right) \implies Q(4, 4\sqrt{3})$ ### Step 2: Derive Circle Equation Write the diameter circle form equation: (x - 4)left(x - frac94right) + (y - 4sqrt3)(y + 3sqrt3) = 0$(x - 4)\left(x - \frac{9}{4}\right) + (y - 4\sqrt{3})(y + 3\sqrt{3}) = 0$ x^2 + y^2 - frac254x - sqrt3y - 27 = 0$x^2 + y^2 - \frac{25}{4}x - \sqrt{3}y - 27 = 0$ Multiply by 64 to clear the fractions and match the given equation template structure: 64x^2 + 64y^2 - 400x - 64sqrt3y - 1728 = 0$64x^2 + 64y^2 - 400x - 64\sqrt{3}y - 1728 = 0$ Comparing directly with 64x^2 + 64y^2 - alpha x - 64sqrt3y = beta$64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta$ yields: alpha = 400, quad beta = 1728$\alpha = 400, \quad \beta = 1728$ beta - alpha = 1728 - 400 = 1328$\beta - \alpha = 1728 - 400 = 1328$ ### Pattern Recognition The distance from focal chord endpoints to the focus equals their perpendicular distance to the directrix. This property connects parameter metrics to geometric lengths cleanly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections Class 11 Mathematics: Circles