If S$S$ and S'$S'$ are the foci of the ellipse fracx^218 + fracy^29 = 1$\frac{x^2}{18} + \frac{y^2}{9} = 1$ and P$P$ be a point on the ellipse, then min(SP cdot S'P) + max(SP cdot S'P)$\min(SP \cdot S'P) + \max(SP \cdot S'P)$ is equal to:

### Related Formula For a standard parabola y^2 = 4ax$y^2 = 4ax$: Focus: S(a, 0)$S(a, 0)$ Parametric coordinates: (at^2, 2at)$(at^2, 2at)$ Equation of a circle on diametric endpoints (x_1, y_1)$(x_1, y_1)$ and (x_2, y_2)$(x_2, y_2)$: (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$ ### Core Logic Find the point P$P$ using the slope of the focal chord, write the equation of the circle with diameter PS$PS$, and find its y$y$-intercept.

### Step 1: Determine P Coordinates For y^2 = 4x$y^2 = 4x$, parameter a=1 implies S(1,0)$a=1 \implies S(1,0)$ and P(t^2, 2t)$P(t^2, 2t)$. Slope of focal chord PS$PS$: tan 60^circ = frac2t - 0t^2 - 1 = sqrt3 implies 2t = sqrt3t^2 - sqrt3$$\tan 60^\circ = \frac{2t - 0}{t^2 - 1} = \sqrt{3} \implies 2t = \sqrt{3}t^2 - \sqrt{3}$$ sqrt3t^2 - 2t - sqrt3 = 0 implies (sqrt3t + 1)(t - sqrt3) = 0$$\sqrt{3}t^2 - 2t - \sqrt{3} = 0 \implies (\sqrt{3}t + 1)(t - \sqrt{3}) = 0$$ Since P$P$ is in the first quadrant, t > 0 implies t = sqrt3$t > 0 \implies t = \sqrt{3}$. Thus, P((sqrt3)^2, 2sqrt3) = P(3, 2sqrt3)$P((\sqrt{3})^2, 2\sqrt{3}) = P(3, 2\sqrt{3})$. ### Step 2: Construct the Diametric Circle Equation Endpoints are S(1,0)$S(1,0)$ and P(3, 2sqrt3)$P(3, 2\sqrt{3})$: (x - 1)(x - 3) + (y - 0)(y - 2sqrt3) = 0$$(x - 1)(x - 3) + (y - 0)(y - 2\sqrt{3}) = 0$$ ### Step 3: Solve for y-intercept The circle touches/intersects the y$y$-axis at x = 0$x = 0$: (0 - 1)(0 - 3) + y(y - 2sqrt3) = 0 implies 3 + y^2 - 2sqrt3y = 0$$(0 - 1)(0 - 3) + y(y - 2\sqrt{3}) = 0 \implies 3 + y^2 - 2\sqrt{3}y = 0$$ This is a perfect square expression (y - sqrt3)^2 = 0 implies y = sqrt3$(y - \sqrt{3})^2 = 0 \implies y = \sqrt{3}$. Thus, the intercept value is a = sqrt3$a = \sqrt{3}$. ### Step 4: Compute Final Target Value 5a^2 = 5(sqrt3)^2 = 15$$5a^2 = 5(\sqrt{3})^2 = 15$$ ### Pattern Recognition A circle whose diameter is a focal radius always touches the tangent at the vertex (y$y$-axis for a standard parabola). The coordinate of the contact point is simply given by a t = 1 cdot sqrt3 = sqrt3$a t = 1 \cdot \sqrt{3} = \sqrt{3}$, bypasses the full equation construction entirely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### 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