Let C$C$ be the circle of minimum area enclosing the ellipse E: fracx^2a^2 + fracy^2b^2 = 1$E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with eccentricity frac12$\frac{1}{2}$ and foci (pm 2, 0)$(pm 2, 0)$. Let PQR$PQR$ be a variable triangle, whose vertex P$P$ is on the circle C$C$ and the side QR$QR$ of length 8$8$ is parallel to the major axis of E$E$ and contains the point of intersection of E$E$ with the negative y$y$-axis. Then the maximum area of the triangle PQR$PQR$ is:

### Related Formula For a horizontal hyperbola centered at (h,k)$(h,k)$: frac(x-h)^2a^2 - frac(y-k)^2b^2 = 1$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$ - Foci: (h pm ae, k)$(h \pm ae, k)$ - Eccentricity relation: b^2 = a^2(e^2 - 1) = a^2 e^2 - a^2$b^2 = a^2(e^2 - 1) = a^2 e^2 - a^2$ ### Core Logic Foci are S_1 = (4,2)$S_1 = (4,2)$ and S_2 = (8,2)$S_2 = (8,2)$. - Center C(h,k)$C(h,k)$ is the midpoint: h = frac4 + 82 = 6, quad k = 2 implies C = (6, 2)$$h = \frac{4 + 8}{2} = 6, \quad k = 2 \implies C = (6, 2)$$ - Distance between foci: 2ae = 8 - 4 = 4 implies ae = 2$$2ae = 8 - 4 = 4 \implies ae = 2$$ Thus, b^2 = 4 - a^2$b^2 = 4 - a^2$. ### Step 1: Expanding standard equation The equation is: frac(x-6)^2a^2 - frac(y-2)^24-a^2 = 1$$\frac{(x-6)^2}{a^2} - \frac{(y-2)^2}{4-a^2} = 1$$ (4-a^2)(x-6)^2 - a^2(y-2)^2 = a^2(4-a^2)$$(4-a^2)(x-6)^2 - a^2(y-2)^2 = a^2(4-a^2)$$ Comparing with 3x^2 - y^2 - alpha x + beta y + gamma = 0$3x^2 - y^2 - \alpha x + \beta y + \gamma = 0$, the ratio of coefficients of x^2$x^2$ and y^2$y^2$ is frac3-1 = -3$\frac{3}{-1} = -3$: frac4 - a^2-a^2 = -3 implies 4 - a^2 = 3a^2 implies 4a^2 = 4 implies a^2 = 1$$\frac{4 - a^2}{-a^2} = -3 \implies 4 - a^2 = 3a^2 \implies 4a^2 = 4 \implies a^2 = 1$$ Thus, b^2 = 4 - 1 = 3$b^2 = 4 - 1 = 3$. ### Step 2: Finding values of coefficients alpha, beta, gamma$\alpha, \beta, \gamma$ Substituting a^2 = 1$a^2 = 1$ back into standard form equation: 3(x-6)^2 - (y-2)^2 = 3$$3(x-6)^2 - (y-2)^2 = 3$$ 3(x^2 - 12x + 36) - (y^2 - 4y + 4) = 3$$3(x^2 - 12x + 36) - (y^2 - 4y + 4) = 3$$ 3x^2 - 36x + 108 - y^2 + 4y - 4 = 3$$3x^2 - 36x + 108 - y^2 + 4y - 4 = 3$$ 3x^2 - y^2 - 36x + 4y + 101 = 0$$3x^2 - y^2 - 36x + 4y + 101 = 0$$ Comparing coefficients: - alpha = 36$\alpha = 36$ - beta = 4$\beta = 4$ - gamma = 101$\gamma = 101$ alpha + beta + gamma = 36 + 4 + 101 = 141$$\alpha + \beta + \gamma = 36 + 4 + 101 = 141$$

### Pattern Recognition Symmetric focal coordinates (y=2$y=2$) indicate the hyperbola is horizontal. Identifying coordinates of the center (6,2)$(6,2)$ quickly and using coefficient ratio comparison restricts parameters immediately without requiring complex algebraic systems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Conic Sections

### Related Formula By the definition of a parabola, the distance from any point (x, y)$(x, y)$ on the curve to the focus (x_f, y_f)$(x_f, y_f)$ equals its perpendicular distance to the directrix line Ax + By + C = 0$Ax + By + C = 0$: (x - x_f)^2 + (y - y_f)^2 = frac(Ax + By + C)^2A^2 + B^2$$(x - x_f)^2 + (y - y_f)^2 = \frac{(Ax + By + C)^2}{A^2 + B^2}$$ ### Core Logic Substituting the focus (-2, 1)$(-2, 1)$ and directrix 2x + y + 2 = 0$2x + y + 2 = 0$ into the definition equation: (x + 2)^2 + (y - 1)^2 = frac(2x + y + 2)^22^2 + 1^2$$(x + 2)^2 + (y - 1)^2 = \frac{(2x + y + 2)^2}{2^2 + 1^2}$$ 5left[(x + 2)^2 + (y - 1)^2right] = (2x + y + 2)^2$$5\left[(x + 2)^2 + (y - 1)^2\right] = (2x + y + 2)^2$$ ### Step 1: Substitute the Given Abscissa

We need the points whose abscissa (x-coordinate) is x = -2$x = -2$. Substitute x = -2$x = -2$ into the general equation: 5left[(-2 + 2)^2 + (y - 1)^2right] = (2(-2) + y + 2)^2$$5\left[(-2 + 2)^2 + (y - 1)^2\right] = (2(-2) + y + 2)^2$$ 5left[0 + (y - 1)^2right] = (-4 + y + 2)^2$$5\left[0 + (y - 1)^2\right] = (-4 + y + 2)^2$$ 5(y - 1)^2 = (y - 2)^2$$5(y - 1)^2 = (y - 2)^2$$ 5(y^2 - 2y + 1) = y^2 - 4y + 4$$5(y^2 - 2y + 1) = y^2 - 4y + 4$$ 5y^2 - 10y + 5 = y^2 - 4y + 4$$5y^2 - 10y + 5 = y^2 - 4y + 4$$ 4y^2 - 6y + 1 = 0$$4y^2 - 6y + 1 = 0$$ ### Step 2: Find the Sum of Ordinates The ordinates y_1$y_1$ and y_2$y_2$ are the roots of the quadratic equation 4y^2 - 6y + 1 = 0$4y^2 - 6y + 1 = 0$. The sum of the ordinates is: y_1 + y_2 = -frac-64 = frac64 = frac32$$y_1 + y_2 = -\frac{-6}{4} = \frac{6}{4} = \frac{3}{2}$$ ### Pattern Recognition Notice how evaluating the intersection layout directly simplifies when the substitution value matches the coordinate of the focus, converting the entire quadratic horizontal layout component to 0 immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### 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