### Related Formula textLength of minor axis = 2b$\text{Length of minor axis} = 2b$ textDistance between foci = 2ae$\text{Distance between foci} = 2ae$ textEccentricity: e = sqrt1 - fracb^2a^2$\text{Eccentricity: } e = \sqrt{1 - \frac{b^2}{a^2}}$ ### Core Logic We set up an algebraic equation relating b$b$, a$a$, and e$e$ from the given geometric condition, then substitute it into the eccentricity identity. ### Step 1: Set up the geometric relation Given that 2b = frac14 (2ae)$2b = \frac{1}{4} (2ae)$: b = fracae4 implies fracba = frace4$b = \frac{ae}{4} \implies \frac{b}{a} = \frac{e}{4}$ Square both sides: fracb^2a^2 = frace^216$\frac{b^2}{a^2} = \frac{e^2}{16}$ ### Step 2: Solve for eccentricity Using the eccentricity relation: e^2 = 1 - fracb^2a^2$e^2 = 1 - \frac{b^2}{a^2}$ e^2 = 1 - frace^216$e^2 = 1 - \frac{e^2}{16}$ e^2 left(1 + frac116right) = 1$e^2 \left(1 + \frac{1}{16}\right) = 1$ frac1716 e^2 = 1 implies e^2 = frac1617 implies e = frac4sqrt17$\frac{17}{16} e^2 = 1 \implies e^2 = \frac{16}{17} \implies e = \frac{4}{\sqrt{17}}$ ### Pattern Recognition Standard Ellipse relations: For standard ellipses, the ratio of axes and the eccentricity are coupled quadratic equations. Expressing b/a$b/a$ as a function of e$e$ allows direct solving of the eccentricity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Related Formula textParametric coordinates on y^2 = 4ax: quad (at^2, \, 2at)$\text{Parametric coordinates on } y^2 = 4ax: \quad (at^2, \, 2at)$ textFocal Chord relation: t_1 t_2 = -1$\text{Focal Chord relation: } t_1 t_2 = -1$ textSection Formula: (x_c, y_c) = left( fracm x_2 + n x_1m+n, \, fracm y_2 + n y_1m+n right)$\text{Section Formula: } (x_c, y_c) = \left( \frac{m x_2 + n x_1}{m+n}, \, \frac{m y_2 + n y_1}{m+n} \right)$ ### Core Logic We find the parametric parameters of coordinates P$P$ and Q$Q$, obtain their Cartesian values, and then apply the section formula with the focus S$S$ to calculate the splitting ratio. ### Step 1: Find coordinates of P and Q For parabola y^2 = 16x$y^2 = 16x$, the focal parameter is a = 4$a = 4$. Focus is S(4, 0)$S(4, 0)$. Let P$P$ be (a t_1^2, 2a t_1) = (1, -4)$(a t_1^2, 2a t_1) = (1, -4)$: 2a t_1 = -4 implies 2(4) t_1 = -4 implies t_1 = -frac12$2a t_1 = -4 \implies 2(4) t_1 = -4 \implies t_1 = -\frac{1}{2}$ Since PQ$PQ$ is a focal chord, the parametric points are coupled: t_1 t_2 = -1 implies t_2 = 2$t_1 t_2 = -1 \implies t_2 = 2$ Now, calculate the coordinates of Q$Q$: Q equiv (a t_2^2, \, 2 a t_2) = (4(4), \, 2(4)(2)) = (16, \, 16)$Q \equiv (a t_2^2, \, 2 a t_2) = (4(4), \, 2(4)(2)) = (16, \, 16)$ ### Step 2: Solve for the dividing ratio Let the focus S(4, 0)$S(4, 0)$ divide the line segment PQ$PQ$ internally in the ratio lambda : 1$\lambda : 1$. Using the y$y$-coordinate of the section formula: y_s = fraclambda y_q + 1 y_plambda + 1$y_s = \frac{\lambda y_q + 1 y_p}{\lambda + 1}$ 0 = fraclambda(16) + 1(-4)lambda + 1 implies 16lambda - 4 = 0 implies lambda = frac14$0 = \frac{\lambda(16) + 1(-4)}{\lambda + 1} \implies 16\lambda - 4 = 0 \implies \lambda = \frac{1}{4}$ Thus, the focus S$S$ divides the chord internally in the ratio 1:4$1:4$. Since gcd(1, 4) = 1$\gcd(1, 4) = 1$, we have m = 1$m = 1$ and n = 4$n = 4$: m^2 + n^2 = 1^2 + 4^2 = 1 + 16 = 17$m^2 + n^2 = 1^2 + 4^2 = 1 + 16 = 17$ ### Pattern Recognition Harmonic Mean Shortcut: In any parabola, the focus divides a focal chord internally into segments of lengths SP$SP$ and SQ$SQ$ such that the semi-latus rectum 2a$2a$ is the harmonic mean of these segments: frac1SP + frac1SQ = frac1a$\frac{1}{SP} + \frac{1}{SQ} = \frac{1}{a}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Related Formula For a standard hyperbola: Focus: (pm ae, 0)$(\pm ae, 0)$ Directrix: x = pm fracae$x = \pm \frac{a}{e}$ Eccentricity relation: (ae)^2 = a^2 + b^2$(ae)^2 = a^2 + b^2$ Length of latus rectum: l = frac2b^2a$l = \frac{2b^2}{a}$ ### Core Logic Given ae = sqrt10$ae = \sqrt{10}$ and fracae = frac9sqrt10$\frac{a}{e} = \frac{9}{\sqrt{10}}$. Multiplying these gives a^2$a^2$, which determines both parameters. ### Step 1: Find a and e a^2 = (ae) cdot left(fracaeright) = sqrt10 cdot frac9sqrt10 = 9 implies a = 3$a^2 = (ae) \cdot \left(\frac{a}{e}\right) = \sqrt{10} \cdot \frac{9}{\sqrt{10}} = 9 \implies a = 3$ Substitute a = 3$a = 3$ into ae = sqrt10$ae = \sqrt{10}$: e = fracsqrt103 implies e^2 = frac109$e = \frac{\sqrt{10}}{3} \implies e^2 = \frac{10}{9}$ ### Step 2: Find b and l Using (ae)^2 = a^2 + b^2$(ae)^2 = a^2 + b^2$: 10 = 9 + b^2 implies b^2 = 1$10 = 9 + b^2 \implies b^2 = 1$ Then the length of latus rectum l$l$ is: l = frac2b^2a = frac2(1)3 = frac23$l = \frac{2b^2}{a} = \frac{2(1)}{3} = \frac{2}{3}$ ### Step 3: Evaluate Final Expression Calculate 9(e^2 + l)$9(e^2 + l)$: 9left(frac109 + frac23right) = 10 + 6 = 16$9\left(\frac{10}{9} + \frac{2}{3}\right) = 10 + 6 = 16$ ### Pattern Recognition Multiplying focus location by directrix location immediately eliminates e$e$, giving a^2$a^2$ directly. Once a^2$a^2$ is known, b^2$b^2$ follow seamlessly via (ae)^2 = a^2+b^2$(ae)^2 = a^2+b^2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Related Formula Focal distances of any point P(acostheta, bsintheta)$P(a\cos\theta, b\sin\theta)$ on an ellipse are given by: SP = a - ex_P = a(1 - ecostheta)$SP = a - ex_P = a(1 - e\cos\theta)$ S'P = a + ex_P = a(1 + ecostheta)$S'P = a + ex_P = a(1 + e\cos\theta)$ Product of focal distances: SP cdot S'P = a^2(1 - e^2cos^2theta) = a^2 - e^2x_P^2$SP \cdot S'P = a^2(1 - e^2\cos^2\theta) = a^2 - e^2x_P^2$ ### Core Logic Compute the eccentricity e$e$, express the product SP cdot S'P$SP \cdot S'P$ in terms of cos^2theta$\cos^2\theta$, and analyze its bounds across the domain to find minimum and maximum limits.

### Step 1: Determine Ellipse Parameters Given a^2 = 18$a^2 = 18$ and b^2 = 9$b^2 = 9$. b^2 = a^2(1 - e^2) implies 9 = 18(1 - e^2) implies 1 - e^2 = frac12 implies e = frac1sqrt10$b^2 = a^2(1 - e^2) \implies 9 = 18(1 - e^2) \implies 1 - e^2 = \frac{1}{2} \implies e = \frac{1}{\sqrt{10}}$ ### Step 2: Express Focal Product The parametric coordinates are P(3sqrt2costheta, 3sintheta)$P(3\sqrt{2}\cos\theta, 3\sin\theta)$. SP cdot S'P = a^2 - (ae)^2cos^2theta$SP \cdot S'P = a^2 - (ae)^2\cos^2\theta$ Since a^2=18$a^2=18$ and (ae)^2 = a^2-b^2 = 18-9 = 9$(ae)^2 = a^2-b^2 = 18-9 = 9$: SP cdot S'P = 18 - 9cos^2theta$SP \cdot S'P = 18 - 9\cos^2\theta$ ### Step 3: Evaluate Extrema and Sum Since 0 le cos^2theta le 1$0 \le \cos^2\theta \le 1$: * Maximum value occurs when cos^2theta = 0 implies max = 18$\cos^2\theta = 0 \implies \max = 18$. * Minimum value occurs when cos^2theta = 1 implies min = 18 - 9 = 9$\cos^2\theta = 1 \implies \min = 18 - 9 = 9$. textSum = min + max = 9 + 18 = 27$\text{Sum} = \min + \max = 9 + 18 = 27$ ### Pattern Recognition The product of focal distances can also be written directly as b^2$b^2$ at the minor axis vertices (max) and a^2(1-e^2)$a^2(1-e^2)$ varying down to a^2-c^2$a^2-c^2$. Summing them up yields b^2 + a^2 = 9 + 18 = 27$b^2 + a^2 = 9 + 18 = 27$ instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### 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