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

### Core Logic The circle C$C$ has its center at the origin and passes through the foci F_1(-ae, 0)$F_1(-ae, 0)$ and F_2(ae, 0)$F_2(ae, 0)$. This means F_1F_2$F_1F_2$ is the diameter of the circle. Any point P$P$ lying on this circle satisfies the property that the angle subtended by the diameter is a right angle: angle F_1PF_2 = 90^circ$$\angle F_1PF_2 = 90^\circ$$

### Step 1: Using the Area and Ellipse Definition Since triangle PF_1F_2$\triangle PF_1F_2$ is a right-angled triangle at P$P$: textArea = frac12 cdot PF_1 cdot PF_2 = 30 implies PF_1 cdot PF_2 = 60$$\text{Area} = \frac{1}{2} \cdot PF_1 \cdot PF_2 = 30 \implies PF_1 \cdot PF_2 = 60$$ By the definition of an ellipse, the sum of the focal distances to any point on the curve is equal to the length of the major axis (2a = 17$2a = 17$): PF_1 + PF_2 = 17$$PF_1 + PF_2 = 17$$ ### Step 2: Calculating Distance between Foci Applying Pythagoras' theorem in right-angled triangle PF_1F_2$\triangle PF_1F_2$: F_1F_2^2 = PF_1^2 + PF_2^2 = (PF_1 + PF_2)^2 - 2(PF_1 cdot PF_2)$$F_1F_2^2 = PF_1^2 + PF_2^2 = (PF_1 + PF_2)^2 - 2(PF_1 \cdot PF_2)$$ Substitute the known values from our equations block: F_1F_2^2 = (17)^2 - 2(60) = 289 - 120 = 169$$F_1F_2^2 = (17)^2 - 2(60) = 289 - 120 = 169$$ F_1F_2 = sqrt169 = 13$$F_1F_2 = \sqrt{169} = 13$$ Therefore, the distance between the foci is 13$13$. ### Pattern Recognition Whenever a circle is circumscribed around the foci of an ellipse, remember Thales' theorem: any intersection point with the ellipse forms a right triangle with the focal diameter, linking focal properties directly to Pythagoras. ### 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

### Related Formula Eccentricity for ellipse (a$a) and hyperbola are calculated as: e_1^2 = 1 - fraca^2b^2$$e_1^2 = 1 - \frac{a^2}{b^2}$$ e_2^2 = 1 + fracb^2a^2$$e_2^2 = 1 + \frac{b^2}{a^2}$$ ### Core Logic Since b < 5$b < 5$, for the ellipse fracx^2b^2 + fracy^225 = 1$\frac{x^2}{b^2} + \frac{y^2}{25} = 1$, the major axis is along the y$y$-axis: e_1^2 = 1 - fracb^225$$e_1^2 = 1 - \frac{b^2}{25}$$ For the hyperbola fracx^216 - fracy^2b^2 = 1$\frac{x^2}{16} - \frac{y^2}{b^2} = 1$: e_2^2 = 1 + fracb^216$$e_2^2 = 1 + \frac{b^2}{16}$$ Given e_1^2 e_2^2 = 1$e_1^2 e_2^2 = 1$: left(1 - fracb^225right)left(1 + fracb^216right) = 1$$\left(1 - \frac{b^2}{25}\right)\left(1 + \frac{b^2}{16}\right) = 1$$ 1 + fracb^216 - fracb^225 - fracb^4400 = 1 implies frac9b^2400 = fracb^4400 implies b^2 = 9$$1 + \frac{b^2}{16} - \frac{b^2}{25} - \frac{b^4}{400} = 1 \implies \frac{9b^2}{400} = \frac{b^4}{400} \implies b^2 = 9$$ ### Step 1: Calculate Foci Locations Substituting b^2 = 9$b^2 = 9$: Ellipse foci: ae_1 = 5 cdot sqrt1 - frac925 = 5 cdot frac45 = 4$ae_1 = 5 \cdot \sqrt{1 - \frac{9}{25}} = 5 \cdot \frac{4}{5} = 4$. Foci lie along y$y$-axis: (0, pm 4)$(0, \pm 4)$. Hyperbola foci: ae_2 = 4 cdot sqrt1 + frac916 = 4 cdot frac54 = 5$ae_2 = 4 \cdot \sqrt{1 + \frac{9}{16}} = 4 \cdot \frac{5}{4} = 5$. Foci lie along x$x$-axis: (pm 5, 0)$(\pm 5, 0)$. ### Step 2: Construct the New Ellipse The new ellipse passes through (0, pm 4)$(0, \pm 4)$ and (pm 5, 0)$(\pm 5, 0)$. Thus, its semi-major axis is A = 5$A = 5$ along the x$x$-axis and semi-minor axis is B = 4$B = 4$ along the y$y$-axis: E = sqrt1 - fracB^2A^2 = sqrt1 - frac1625 = frac35$$E = \sqrt{1 - \frac{B^2}{A^2}} = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$$ ### Pattern Recognition When a conic passes through points directly on the axes like (pm alpha, 0)$(\pm alpha, 0)$ and (0, pm beta)$(0, \pm \beta)$, those points immediately represent the semi-axes values A$A$ and B$B$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections