### Related Formula textArea of a rectangle bounded by x = pm x_1, y = pm y_1 text is (2x_1)(2y_1) = 4x_1y_1$\text{Area of a rectangle bounded by } x = \pm x_1, y = \pm y_1 \text{ is } (2x_1)(2y_1) = 4x_1y_1$ ### Core Logic Since the points of intersection of the circle x^2+y^2=4b$x^2+y^2=4b$ and the ellipse fracx^216+fracy^2b^2=1$\frac{x^2}{16}+\frac{y^2}{b^2}=1$ satisfy the curve y^2 = 3x^2$y^2 = 3x^2$, we can substitute y^2 = 3x^2$y^2 = 3x^2$ into the circle equation to find these coordinates entirely in terms of b$b$. Substitute y^2 = 3x^2$y^2 = 3x^2$ into x^2 + y^2 = 4b$x^2 + y^2 = 4b$: x^2 + 3x^2 = 4b$x^2 + 3x^2 = 4b$ 4x^2 = 4b Rightarrow x^2 = b$4x^2 = 4b \Rightarrow x^2 = b$ Since y^2 = 3x^2$y^2 = 3x^2$, substituting x^2=b$x^2=b$ gives: y^2 = 3b$y^2 = 3b$ Now, these points (x^2=b, y^2=3b)$(x^2=b, y^2=3b)$ must also lie perfectly on the ellipse. Substitute them into the ellipse equation: fracb16 + frac3bb^2 = 1$\frac{b}{16} + \frac{3b}{b^2} = 1$ fracb16 + frac3b = 1$\frac{b}{16} + \frac{3}{b} = 1$ ### Step 1: Solve for Parameter b Multiply the equation by 16b$16b$: b^2 + 48 = 16b$b^2 + 48 = 16b$ b^2 - 16b + 48 = 0$b^2 - 16b + 48 = 0$ (b-4)(b-12) = 0$(b-4)(b-12) = 0$ This gives two candidate values: b=4$b=4$ and b=12$b=12$. Check conditions: The problem specifically declares "two distinct conics". If b=4$b=4$, the circle becomes x^2+y^2=16$x^2+y^2=16$ and the ellipse becomes fracx^216+fracy^216=1$\frac{x^2}{16}+\frac{y^2}{16}=1$ (which is identically x^2+y^2=16$x^2+y^2=16$). The conics would coincide and not be distinct. Thus, reject b=4$b=4$. We must use b=12$b=12$. ### Step 2: Calculate Coordinates and Area With b=12$b=12$, extract the point coordinates: x^2 = 12 Rightarrow x = pm 2sqrt3$x^2 = 12 \Rightarrow x = \pm 2\sqrt{3}$ y^2 = 36 Rightarrow y = pm 6$y^2 = 36 \Rightarrow y = \pm 6$ The four intersection points form a symmetrical rectangle in the cartesian plane with dimensions: Length = 2|x| = 2(2sqrt3) = 4sqrt3$= 2|x| = 2(2\sqrt{3}) = 4\sqrt{3}$ Width = 2|y| = 2(6) = 12$= 2|y| = 2(6) = 12$ Area of the rectangle = (4sqrt3) times 12 = 48sqrt3$= (4\sqrt{3}) \times 12 = 48\sqrt{3}$. ### Step 3: Evaluate Final Expression The question asks for 3sqrt3$3\sqrt{3}$ times the area of the rectangle: 3sqrt3 times (48sqrt3) = 3 times 48 times 3 = 144 times 3 = 432$3\sqrt{3} \times (48\sqrt{3}) = 3 \times 48 \times 3 = 144 \times 3 = 432$ ### Pattern Recognition When intersection points of shapes are constrained to lie on a third curve (like y^2=kx^2$y^2=kx^2$), substitute the 3rd curve into the simplest conic first to lock x^2$x^2$ and y^2$y^2$ as scalars, then test those scalars strictly on the remaining complex conic. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Related Formula textSection Formula: (x, y) = left(fracmx_2 + nx_1m+n, fracmy_2 + ny_1m+nright)$\text{Section Formula: } (x, y) = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$ textDirectrix of Ellipse: x = fracae$\text{Directrix of Ellipse: } x = \frac{a}{e}$ textLength of Latus Rectum = frac2b^2a$\text{Length of Latus Rectum } = \frac{2b^2}{a}$ ### Core Logic First, find points A and B on 5x + 7y = 50$5x + 7y = 50$. For A on x-axis: y = 0 Rightarrow 5x = 50 Rightarrow alpha = 10 Rightarrow A(10, 0)$y = 0 \Rightarrow 5x = 50 \Rightarrow \alpha = 10 \Rightarrow A(10, 0)$ For B on y-axis: x = 0 Rightarrow 7y = 50 Rightarrow beta = frac507 Rightarrow Bleft(0, frac507right)$x = 0 \Rightarrow 7y = 50 \Rightarrow \beta = \frac{50}{7} \Rightarrow B\left(0, \frac{50}{7}\right)$ ### Step 1: Finding Coordinates of P P divides AB in ratio 7:3. Using section formula: x_P = frac7(0) + 3(10)7 + 3 = frac3010 = 3$x_P = \frac{7(0) + 3(10)}{7 + 3} = \frac{30}{10} = 3$ y_P = frac7left(frac507right) + 3(0)7 + 3 = frac5010 = 5$y_P = \frac{7\left(\frac{50}{7}\right) + 3(0)}{7 + 3} = \frac{50}{10} = 5$ So, P = (3, 5)$P = (3, 5)$. ### Step 2: Finding a and b for the Ellipse Directrix is 3x - 25 = 0 Rightarrow x = frac253$3x - 25 = 0 \Rightarrow x = \frac{25}{3}$. Thus, fracae = frac253$\frac{a}{e} = \frac{25}{3}$.

Focus S has coordinates (ae, 0)$(ae, 0)$. A perpendicular from S to the x-axis is simply the vertical line x = ae$x = ae$. Since this line passes through P(3, 5)$P(3, 5)$, the x-coordinate of S must equal the x-coordinate of P. ae = 3$ae = 3$ Now, solving for a$a$: a = left(fracaeright) cdot e = frac253 cdot e$a = \left(\frac{a}{e}\right) \cdot e = \frac{25}{3} \cdot e$ Multiply the two equations: ae cdot fracae = 3 cdot frac253 Rightarrow a^2 = 25 Rightarrow a = 5$ae \cdot \frac{a}{e} = 3 \cdot \frac{25}{3} \Rightarrow a^2 = 25 \Rightarrow a = 5$. Also, ae = 3 Rightarrow 5e = 3 Rightarrow e = frac35$ae = 3 \Rightarrow 5e = 3 \Rightarrow e = \frac{3}{5}$. Calculate b^2$b^2$: b^2 = a^2(1 - e^2) = 25left(1 - frac925right) = 25 cdot frac1625 = 16 Rightarrow b = 4$b^2 = a^2(1 - e^2) = 25\left(1 - \frac{9}{25}\right) = 25 \cdot \frac{16}{25} = 16 \Rightarrow b = 4$ ### Step 3: Calculating Latus Rectum textLength of Latus Rectum = frac2b^2a = frac2(16)5 = frac325$\text{Length of Latus Rectum} = \frac{2b^2}{a} = \frac{2(16)}{5} = \frac{32}{5}$ ### Pattern Recognition Intersection of perpendicular from focus passing through a point purely locks the x-coordinate of the focus to the x-coordinate of that point. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Straight Lines Class 11 Maths: Conic Sections

### Related Formula |r_1 - r_2| < C_1 C_2 < r_1 + r_2$|r_1 - r_2| < C_1 C_2 < r_1 + r_2$ where C_1 C_2$C_1 C_2$ is the distance between the centers and r_1, r_2$r_1, r_2$ are the radii. ### Core Logic Circle 1: (x + 1)^2 + (y + 2)^2 = r^2$(x + 1)^2 + (y + 2)^2 = r^2$ Center C_1(-1, -2)$C_1(-1, -2)$, Radius r_1 = r$r_1 = r$ Circle 2: x^2 + y^2 - 4x - 4y + 4 = 0$x^2 + y^2 - 4x - 4y + 4 = 0$ Convert to standard form: (x - 2)^2 - 4 + (y - 2)^2 - 4 + 4 = 0 Rightarrow (x - 2)^2 + (y - 2)^2 = 4$(x - 2)^2 - 4 + (y - 2)^2 - 4 + 4 = 0 \Rightarrow (x - 2)^2 + (y - 2)^2 = 4$ Center C_2(2, 2)$C_2(2, 2)$, Radius r_2 = 2$r_2 = 2$ ### Step 1: Distance between centers Distance C_1C_2 = sqrt(2 - (-1))^2 + (2 - (-2))^2 = sqrt3^2 + 4^2 = sqrt9 + 16 = 5$C_1C_2 = \sqrt{(2 - (-1))^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$. ### Step 2: Applying inequality condition For exactly two intersection points: |r_1 - r_2| < C_1 C_2 < r_1 + r_2$|r_1 - r_2| < C_1 C_2 < r_1 + r_2$ |r - 2| < 5 < r + 2$|r - 2| < 5 < r + 2$ Part A: r + 2 > 5 Rightarrow r > 3 quad dots (1)$r + 2 > 5 \Rightarrow r > 3 \quad \dots (1)$ Part B: |r - 2| < 5$|r - 2| < 5$ -5 < r - 2 < 5$-5 < r - 2 < 5$ -3 < r < 7 quad dots (2)$-3 < r < 7 \quad \dots (2)$ ### Step 3: Intersection of inequalities Taking the intersection of (1) and (2): r in (3, infty) cap (-3, 7) = (3, 7)$r \in (3, \infty) \cap (-3, 7) = (3, 7)$ 3 < r < 7$3 < r < 7$ ### Pattern Recognition Intersecting circles always strictly obey the triangle inequality formed by their radii and the distance between their centers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections

### Related Formula b^2 = a^2(1 - e^2)$b^2 = a^2(1 - e^2)$ Length of minor axis = 2b$= 2b$ Distance between foci = 2ae$= 2ae$ ### Core Logic Given condition: Length of minor axis = Half of the distance between foci. 2b = frac12(2ae)$2b = \frac{1}{2}(2ae)$ 2b = ae$2b = ae$ fracba = frace2$\frac{b}{a} = \frac{e}{2}$ ### Step 1: Squaring and finding eccentricity Squaring both sides: fracb^2a^2 = frace^24$\frac{b^2}{a^2} = \frac{e^2}{4}$ We know fracb^2a^2 = 1 - e^2$\frac{b^2}{a^2} = 1 - e^2$. 1 - e^2 = frace^24$1 - e^2 = \frac{e^2}{4}$ 1 = e^2 + frace^24 = frac5e^24$1 = e^2 + \frac{e^2}{4} = \frac{5e^2}{4}$ e^2 = frac45$e^2 = \frac{4}{5}$ e = frac2sqrt5$e = \frac{2}{\sqrt{5}}$ ### Pattern Recognition Proportional relationships between a, b,$a, b,$ and ae$ae$ in an ellipse can be immediately plugged into the fundamental identity b^2 = a^2(1-e^2)$b^2 = a^2(1-e^2)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections