### Related Formula For any point P$P$ on a parabola, its distance to the focus S$S$ equals its perpendicular distance to the directrix line M$M$: PS = PM$PS = PM$ ### Core Logic The axis line is y = x$y = x$. The vertex lies along this line at a distance of sqrt2$\sqrt{2}$ from the origin. Since it's in the first quadrant, its coordinates are (1,1)$(1,1)$. The focus also lies along y=x$y=x$ at a distance of 2sqrt2$2\sqrt{2}$ from the origin, which gives coordinates (2,2)$(2,2)$. ### Step 1: Finding the Equation of the Directrix The distance from the vertex to the focus is a = sqrt(2-1)^2 + (2-1)^2 = sqrt2$a = \sqrt{(2-1)^2 + (2-1)^2} = \sqrt{2}$. The directrix is perpendicular to the axis line y = x$y = x$ (slope = 1$1$), so the slope of the directrix is -1$-1$. The directrix is located at a distance a = sqrt2$a = \sqrt{2}$ behind the vertex, which brings it exactly to the origin (0,0)$(0,0)$. Therefore, the equation of the directrix line is: y - 0 = -1(x - 0) implies x + y = 0$y - 0 = -1(x - 0) \implies x + y = 0$

### Step 2: Utilizing the Focus-Directrix Property Let the point P(1,k)$P(1,k)$ lie on the parabola. Applying PS = PM$PS = PM$: sqrt(1 - 2)^2 + (k - 2)^2 = frac|1 + k|sqrt1^2 + 1^2$\sqrt{(1 - 2)^2 + (k - 2)^2} = \frac{|1 + k|}{\sqrt{1^2 + 1^2}}$ Squaring both sides: 1 + (k - 2)^2 = frac(1 + k)^22$1 + (k - 2)^2 = \frac{(1 + k)^2}{2}$ 2big(1 + k^2 - 4k + 4big) = 1 + k^2 + 2k$2\big(1 + k^2 - 4k + 4\big) = 1 + k^2 + 2k$ 2k^2 - 8k + 10 = k^2 + 2k + 1$2k^2 - 8k + 10 = k^2 + 2k + 1$ k^2 - 10k + 9 = 0$k^2 - 10k + 9 = 0$ Factoring the quadratic equations: (k - 1)(k - 9) = 0 implies k = 1 text or k = 9$(k - 1)(k - 9) = 0 \implies k = 1 \text{ or } k = 9$ ### Pattern Recognition When a vertex and focus both sit perfectly on a symmetric line like y=x$y=x$, notice that the foot of the directrix often lands on a clean coordinate intersection (like the origin here), heavily simplifying geometric distance steps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Related Formula For a point P(x_1, y_1)$P(x_1, y_1)$ on a hyperbola branch, the focal distances are ex_1 + a$ex_1 + a$ and ex_1 - a$ex_1 - a$. Their sum is 2ex_1$2ex_1$, and their product is e^2x_1^2 - a^2$e^2x_1^2 - a^2$. ### Core Logic Given the point P(4,3)$P(4,3)$, the x-coordinate is x_1 = 4$x_1 = 4$. The sum of focal distances is: 2ex_1 = 8sqrtfrac53 implies 2e(4) = 8sqrtfrac53 implies e = sqrtfrac53$2ex_1 = 8\sqrt{\frac{5}{3}} \implies 2e(4) = 8\sqrt{\frac{5}{3}} \implies e = \sqrt{\frac{5}{3}}$ Using the eccentricity relation b^2 = a^2(e^2 - 1)$b^2 = a^2(e^2 - 1)$: b^2 = a^2left(frac53 - 1right) = frac23a^2$b^2 = a^2\left(\frac{5}{3} - 1\right) = \frac{2}{3}a^2$ ### Step 1: Finding the Ellipse Parameters Since P(4,3)$P(4,3)$ lies on the hyperbola fracx^2a^2 - fracy^2b^2 = 1$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: frac16a^2 - frac9frac23a^2 = 1 implies frac16a^2 - frac272a^2 = 1$\frac{16}{a^2} - \frac{9}{\frac{2}{3}a^2} = 1 \implies \frac{16}{a^2} - \frac{27}{2a^2} = 1$ frac32 - 272a^2 = 1 implies frac52a^2 = 1 implies a^2 = frac52$\frac{32 - 27}{2a^2} = 1 \implies \frac{5}{2a^2} = 1 \implies a^2 = \frac{5}{2}$ Now calculate b^2$b^2$: b^2 = frac23left(frac52right) = frac53$b^2 = \frac{2}{3}\left(\frac{5}{2}\right) = \frac{5}{3}$ ### Step 2: Calculating l^2 and m The length of the latus rectum l$l$ is given by l = frac2b^2a$l = \frac{2b^2}{a}$: l^2 = frac4b^4a^2 = frac4left(frac259right)frac52 = frac1009 times frac25 = frac409 implies 9l^2 = 40$l^2 = \frac{4b^4}{a^2} = \frac{4\left(\frac{25}{9}\right)}{\frac{5}{2}} = \frac{100}{9} \times \frac{2}{5} = \frac{40}{9} \implies 9l^2 = 40$ The product of focal distances m$m$ is: m = e^2x_1^2 - a^2 = left(frac53right)(16) - frac52 = frac803 - frac52 = frac160 - 156 = frac1456$m = e^2x_1^2 - a^2 = \left(\frac{5}{3}\right)(16) - \frac{5}{2} = \frac{80}{3} - \frac{5}{2} = \frac{160 - 15}{6} = \frac{145}{6}$ 6m = 145$6m = 145$ ### Step 3: Final Computation Evaluating the targeted expression: 9l^2 + 6m = 40 + 145 = 185$9l^2 + 6m = 40 + 145 = 185$ ### Pattern Recognition Using focal property formulas directly (2ex_1$2ex_1$ for sum and e^2x_1^2 - a^2$e^2x_1^2 - a^2$ for product) avoids the lengthy process of finding focus coordinate values and executing distance formulas explicitly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Related Formula The condition for a line y = mx + p$y = mx + p$ to be tangent to an ellipse fracx^2a^2 + fracy^2b^2 = 1$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is: p^2 = a^2m^2 + b^2$p^2 = a^2m^2 + b^2$ The coordinate of the point of contact is given by left(-fraca^2mp, fracb^2pright)$\left(-\frac{a^2m}{p}, \frac{b^2}{p}\right)$. ### Core Logic Given the ellipse parameter values a^2 = 16$a^2 = 16$ and b^2 = 9$b^2 = 9$, and tangent line slope m = 1$m = 1$: p^2 = 16(1)^2 + 9 = 25 implies p = pm 5$p^2 = 16(1)^2 + 9 = 25 \implies p = \pm 5$ Thus, the two values of p$p$ are 5$5$ and -5$-5$. The points of contact A$A$ and B$B$ are: - For p = 5$p = 5$: A = left(-frac16(1)5, frac95right) = left(-frac165, frac95right)$A = \left(-\frac{16(1)}{5}, \frac{9}{5}\right) = \left(-\frac{16}{5}, \frac{9}{5}\right)$ - For p = -5$p = -5$: B = left(-frac16(1)-5, frac9-5right) = left(frac165, -frac95right)$B = \left(-\frac{16(1)}{-5}, \frac{9}{-5}\right) = \left(\frac{16}{5}, -\frac{9}{5}\right)$ ### Step 1: Intersecting line with Ellipse The line y = x$y = x$ intersects the ellipse fracx^216 + fracy^29 = 1$\frac{x^2}{16} + \frac{y^2}{9} = 1$: fracx^216 + fracx^29 = 1 implies frac25x^2144 = 1 implies x^2 = frac14425 implies x = pm frac125$\frac{x^2}{16} + \frac{x^2}{9} = 1 \implies \frac{25x^2}{144} = 1 \implies x^2 = \frac{144}{25} \implies x = \pm \frac{12}{5}$ Since y = x$y = x$, the intersection points C$C$ and D$D$ are: C = left(-frac125, -frac125right) quad textand quad D = left(frac125, frac125right)$C = \left(-\frac{12}{5}, -\frac{12}{5}\right) \quad \text{and} \quad D = \left(\frac{12}{5}, \frac{12}{5}\right)$ ### Step 2: Calculating Quadrilateral Area The area of quadrilateral ABCD$ABCD$ with vertices mapped symmetrically can be computed using the standard coordinate determinant matrix layout formula: textArea = frac12 beginvmatrix x_A & y_A & 1 \\ x_B & y_B & 1 \\ x_C & y_C & 1 endvmatrix + dots = 24$\text{Area} = \frac{1}{2} \begin{vmatrix} x_A & y_A & 1 \\ x_B & y_B & 1 \\ x_C & y_C & 1 \end{vmatrix} + \dots = 24$ ### Pattern Recognition Notice that the tangent lines are parallel and symmetric (p = pm 5$p = \pm 5$), and the intersecting line passes through the origin. This symmetry creates a geometric parallelogram, simplifying your area calculation by doubling the area of triangle ABD$ABD$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Core Logic The circle C$C$ has its center at the origin and passes through the foci F_1(-ae, 0)$F_1(-ae, 0)$ and F_2(ae, 0)$F_2(ae, 0)$. This means F_1F_2$F_1F_2$ is the diameter of the circle. Any point P$P$ lying on this circle satisfies the property that the angle subtended by the diameter is a right angle: angle F_1PF_2 = 90^circ$\angle F_1PF_2 = 90^\circ$

### Step 1: Using the Area and Ellipse Definition Since triangle PF_1F_2$\triangle PF_1F_2$ is a right-angled triangle at P$P$: textArea = frac12 cdot PF_1 cdot PF_2 = 30 implies PF_1 cdot PF_2 = 60$\text{Area} = \frac{1}{2} \cdot PF_1 \cdot PF_2 = 30 \implies PF_1 \cdot PF_2 = 60$ By the definition of an ellipse, the sum of the focal distances to any point on the curve is equal to the length of the major axis (2a = 17$2a = 17$): PF_1 + PF_2 = 17$PF_1 + PF_2 = 17$ ### Step 2: Calculating Distance between Foci Applying Pythagoras' theorem in right-angled triangle PF_1F_2$\triangle PF_1F_2$: F_1F_2^2 = PF_1^2 + PF_2^2 = (PF_1 + PF_2)^2 - 2(PF_1 cdot PF_2)$F_1F_2^2 = PF_1^2 + PF_2^2 = (PF_1 + PF_2)^2 - 2(PF_1 \cdot PF_2)$ Substitute the known values from our equations block: F_1F_2^2 = (17)^2 - 2(60) = 289 - 120 = 169$F_1F_2^2 = (17)^2 - 2(60) = 289 - 120 = 169$ F_1F_2 = sqrt169 = 13$F_1F_2 = \sqrt{169} = 13$ Therefore, the distance between the foci is 13$13$. ### Pattern Recognition Whenever a circle is circumscribed around the foci of an ellipse, remember Thales' theorem: any intersection point with the ellipse forms a right triangle with the focal diameter, linking focal properties directly to Pythagoras. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Related Formula Distance between foci = 2be$= 2be$ (for vertical major axis). Length of Latus Rectum (textL.R.) = frac2a^2b$(\text{L.R.}) = \frac{2a^2}{b}$. ### Core Logic Foci are F_1(2,5)$F_1(2,5)$ and F_2(2,-3)$F_2(2,-3)$. The x$x$-coordinates are identical, indicating a vertical ellipse. Distance between foci: 2be = 5 - (-3) = 8 implies be = 4$2be = 5 - (-3) = 8 \implies be = 4$ Given eccentricity e = frac45$e = \frac{4}{5}$: bleft(frac45right) = 4 implies b = 5$b\left(\frac{4}{5}\right) = 4 \implies b = 5$

### Step 1: Calculate Minor Axis length Use eccentricity relation: a^2 = b^2(1 - e^2) implies a^2 = 25left(1 - frac1625right) = 25 times frac925 = 9 implies a = 3$a^2 = b^2(1 - e^2) \implies a^2 = 25\left(1 - \frac{16}{25}\right) = 25 \times \frac{9}{25} = 9 \implies a = 3$ ### Step 2: Evaluate Latus Rectum textL.R. = frac2a^2b = frac2 times 95 = frac185$\text{L.R.} = \frac{2a^2}{b} = \frac{2 \times 9}{5} = \frac{18}{5}$ ### Pattern Recognition Always verify axis orientation (horizontal vs vertical) from coordinates before blindly substituting into standard formulas. Symmetrical components match axis lengths parameters directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections