### Related Formula The general equation of a circle is: x^2 + y^2 + 2gx + 2fy + c = 0$x^2 + y^2 + 2gx + 2fy + c = 0$ Radius of the circle: r = sqrtg^2 + f^2 - c$r = \sqrt{g^2 + f^2 - c}$ If a set of points lies on this circle, their coordinates must satisfy the equation. ### Core Logic Since (0,0)$(0,0)$ lies on the circle: 0^2 + 0^2 + 2g(0) + 2f(0) + c = 0 implies c = 0$0^2 + 0^2 + 2g(0) + 2f(0) + c = 0 \implies c = 0$ Thus, the equation simplifies to: x^2 + y^2 + 2gx + 2fy = 0$x^2 + y^2 + 2gx + 2fy = 0$ ### Step 1: Finding g$g$, f$f$ and r^2$r^2$ Substitute (4,6)$(4,6)$: 16 + 36 + 8g + 12f = 0 implies 2g + 3f = -13 quad text--- (1)$16 + 36 + 8g + 12f = 0 \implies 2g + 3f = -13 \quad \text{--- (1)}$ Substitute (-1,5)$(-1,5)$: 1 + 25 - 2g + 10f = 0 implies -g + 5f = -13 implies g = 5f + 13 quad text--- (2)$1 + 25 - 2g + 10f = 0 \implies -g + 5f = -13 \implies g = 5f + 13 \quad \text{--- (2)}$ Substituting g$g$ from (2) into (1): 2(5f + 13) + 3f = -13$2(5f + 13) + 3f = -13$ 13f + 26 = -13 implies f = -3$13f + 26 = -13 \implies f = -3$ g = 5(-3) + 13 = -2$g = 5(-3) + 13 = -2$ The circle equation is: x^2 + y^2 - 4x - 6y = 0$x^2 + y^2 - 4x - 6y = 0$ Calculating radius squared r^2$r^2$: r^2 = g^2 + f^2 - c = (-2)^2 + (-3)^2 - 0 = 13$r^2 = g^2 + f^2 - c = (-2)^2 + (-3)^2 - 0 = 13$

### Step 2: Solving for k$k$ The point (k, 3k)$(k, 3k)$ lies on this circle: k^2 + (3k)^2 - 4k - 6(3k) = 0$k^2 + (3k)^2 - 4k - 6(3k) = 0$ 10k^2 - 22k = 0 implies k(10k - 22) = 0$10k^2 - 22k = 0 \implies k(10k - 22) = 0$ Since the points must be distinct and k=0$k=0$ gives (0,0)$(0,0)$ which is already a given point, we must have: 10k = 22 implies k = frac115$10k = 22 \implies k = \frac{11}{5}$ Now, calculate 10k + r^2$10k + r^2$: 10k + r^2 = 10left(frac115right) + 13 = 22 + 13 = 35$10k + r^2 = 10\left(\frac{11}{5}\right) + 13 = 22 + 13 = 35$ ### Pattern Recognition Notice that the slope of the line joining origin (0,0)$(0,0)$ to the general point is y = 3x$y = 3x$. For three given coordinates, if origin is one of them, the circle equation lacks the constant c$c$. It is always faster to first solve for parameters g, f$g, f$ and then check geometry. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections Class 10 Mathematics: Coordinate Geometry

### Related Formula For an ellipse E$E$: - Foci: (pm ae, 0)$(\pm ae, 0)$ - Eccentricity: b^2 = a^2(1 - e^2)$b^2 = a^2(1 - e^2)$ - The circle of minimum area enclosing a centered ellipse has diameter equal to the major axis of the ellipse (R = a$R = a$). - Area of triangle: textArea = frac12 cdot textbase cdot textheight$\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}$ ### Core Logic Let's first find coordinates a$a$ and b$b$: - ae = 2$ae = 2$ - e = frac12 implies aleft(frac12right) = 2 implies a = 4$e = \frac{1}{2} \implies a\left(\frac{1}{2}\right) = 2 \implies a = 4$ - b^2 = a^2(1 - e^2) = 16left(1 - frac14right) = 12 implies b = 2sqrt3$b^2 = a^2(1 - e^2) = 16\left(1 - \frac{1}{4}\right) = 12 \implies b = 2\sqrt{3}$ ### Step 1: Setting Circle and Triangle geometry The enclosing circle C$C$ has radius R = a = 4$R = a = 4$, centered at (0,0)$(0,0)$. Thus, its equation is: x^2 + y^2 = 16 implies P = (4costheta, 4sintheta)$x^2 + y^2 = 16 \implies P = (4\cos\theta, 4\sin\theta)$ The intersection of the ellipse with the negative y$y$-axis is (0, -b) = (0, -2sqrt3)$(0, -b) = (0, -2\sqrt{3})$. Since side QR$QR$ (length = 8$= 8$) is parallel to the major axis (x$x$-axis) and contains (0, -2sqrt3)$(0, -2\sqrt{3})$, the equation of the line containing QR$QR$ is: y = -2sqrt3$y = -2\sqrt{3}$

### Step 2: Maximizing Area of Delta PQR$\Delta PQR$ The perpendicular height of vertex P(4costheta, 4sintheta)$P(4\cos\theta, 4\sin\theta)$ from the base line y = -2sqrt3$y = -2\sqrt{3}$ is: H = 4sintheta - (-2sqrt3) = 4sintheta + 2sqrt3$H = 4\sin\theta - (-2\sqrt{3}) = 4\sin\theta + 2\sqrt{3}$ To maximize the area, we maximize height H$H$ by choosing sintheta = 1$\sin\theta = 1$: H_max = 4 + 2sqrt3$H_{max} = 4 + 2\sqrt{3}$ textMaximum Area = frac12 cdot textbase QR cdot H_max$\text{Maximum Area} = \frac{1}{2} \cdot \text{base } QR \cdot H_{max}$ textMaximum Area = frac12 cdot 8 cdot (4 + 2sqrt3) = 4(4 + 2sqrt3) = 8(2 + sqrt3)$\text{Maximum Area} = \frac{1}{2} \cdot 8 \cdot (4 + 2\sqrt{3}) = 4(4 + 2\sqrt{3}) = 8(2 + \sqrt{3})$ ### Pattern Recognition The enclosing circle with minimum area is called the auxiliary circle. Its radius is equal to the semi-major axis a$a$. Max height of a triangle with a base fixed at line y=-k$y=-k$ and vertex on the circle is R + k$R + k$. This directly gives textArea = frac12 cdot textbase cdot (a+b)$\text{Area} = \frac{1}{2} \cdot \text{base} \cdot (a+b)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Related Formula For a circle x^2 + (y-k)^2 = R^2$x^2 + (y-k)^2 = R^2$ and any smooth curve, the shortest distance lies along the normal to the curve passing through the center of the circle C(h,k)$C(h,k)$: textShortest Distance = textDistance(P, C) - R$\text{Shortest Distance} = \text{Distance}(P, C) - R$ where P$P$ is the point of normal intersection on the curve. ### Core Logic Let's first identify the circle parameters: x^2 + y^2 + 12y + 35 = 0 implies x^2 + (y+6)^2 = 36 - 35 = 1$x^2 + y^2 + 12y + 35 = 0 \implies x^2 + (y+6)^2 = 36 - 35 = 1$ Center C = (0, -6)$C = (0, -6)$ and radius R = 1$R = 1$. The first curve is the parabola y^2 = 8x$y^2 = 8x$, where a = 2$a = 2$. Normal equation of y^2 = 4ax$y^2 = 4ax$ in slope form: y = mx - 2am - am^3$y = mx - 2am - am^3$ Substituting a=2$a=2$: y = mx - 4m - 2m^3$y = mx - 4m - 2m^3$ ### Step 1: Find normal passing through circle center Normal passes through C(0, -6)$C(0, -6)$: -6 = m(0) - 4m - 2m^3$-6 = m(0) - 4m - 2m^3$ 2m^3 + 4m - 6 = 0 implies m^3 + 2m - 3 = 0$2m^3 + 4m - 6 = 0 \implies m^3 + 2m - 3 = 0$ By inspection, m=1$m=1$ is a real solution: (m-1)(m^2 + m + 3) = 0$(m-1)(m^2 + m + 3) = 0$ Since m^2 + m + 3 = 0$m^2 + m + 3 = 0$ has complex roots, the unique real normal slope is m=1$m=1$.

### Step 2: Point calculation and shortest distance For m=1$m=1$ and a=2$a=2$, normal intersection point P(am^2, -2am)$P(am^2, -2am)$ is: P = (2(1)^2, -2(2)(1)) = (2, -4)$P = (2(1)^2, -2(2)(1)) = (2, -4)$ Distance from P(2,-4)$P(2,-4)$ to center C(0,-6)$C(0,-6)$: PC = sqrt(2-0)^2 + (-4 - (-6))^2 = sqrt4 + 4 = 2sqrt2$PC = \sqrt{(2-0)^2 + (-4 - (-6))^2} = \sqrt{4 + 4} = 2\sqrt{2}$ Shortest distance: textSD = PC - R = 2sqrt2 - 1$\text{SD} = PC - R = 2\sqrt{2} - 1$ ### Pattern Recognition The shortest distance between a parabola and a circle is always along the common normal of the parabola passing through the circle's center. Finding the normal in slope form and solving for m$m$ avoids complex calculus. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### 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