Let r$r$ be the radius of the circle, which touches x$x$ -axis at point (a, 0)$(a, 0)$ , a < 0$a < 0$ and the parabola y^2 = 9x$y^2 = 9x$ at the point (4, 6)$(4, 6)$ . Then r$r$ is equal to

### Related Formula Shortest distance between symmetric profiles: The minimal spacing normal line runs completely perpendicular to the mutual line of symmetry y=x$y=x$.

### Core Logic The given curve equations reflect symmetry across line y=x$y=x$[cite: 1382, 1383]. The tangent slope at the closest matching locations must run parallel to this mirror path [cite: 1403]: fracmathrmdymathrmdx = 1$$\frac{\mathrm{d}y}{\mathrm{d}x} = 1$$ [cite: 1403] Differentiate curve equation y = x^2 + 2$y = x^2 + 2$ [cite: 1404]: fracmathrmdymathrmdx = 2x = 1 implies x = frac12$$\frac{\mathrm{d}y}{\mathrm{d}x} = 2x = 1 \implies x = \frac{1}{2}$$ [cite: 1405, 1406] Substitute back to get y-coordinate [cite: 1406]: y = left(frac12right)^2 + 2 = frac94 implies Bleft(frac12, frac94right)$$y = \left(\frac{1}{2}\right)^2 + 2 = \frac{9}{4} \implies B\left(\frac{1}{2}, \frac{9}{4}\right)$$ [cite: 1406, 1407] By mirror symmetry, the corresponding point on the other parabola is [cite: 1407]: Aleft(frac94, frac12right)$$A\left(\frac{9}{4}, \frac{1}{2}\right)$$ [cite: 1407] ### Step 1: Calculating distance and circle radius Evaluate chord distance AB$AB$ using standard metrics [cite: 1407]: AB = sqrtleft(frac94 - frac12right)^2 + left(frac12 - frac94right)^2 = sqrt2 cdot left(frac74right)^2 = frac7sqrt24$$AB = \sqrt{\left(\frac{9}{4} - \frac{1}{2}\right)^2 + \left(\frac{1}{2} - \frac{9}{4}\right)^2} = \sqrt{2 \cdot \left(\frac{7}{4}\right)^2} = \frac{7\sqrt{2}}{4}$$ [cite: 1407, 1408] The diameter of the smallest circle spanning between these touching curves equals distance AB$AB$ [cite: 1408]. textRadius = fracAB2 = frac7sqrt28$$\text{Radius} = \frac{AB}{2} = \frac{7\sqrt{2}}{8}$$ [cite: 1408] ### Pattern Recognition Mutually inverse conic curves track symmetric footprints. Their closest distance segments always align perfectly perpendicular to the main baseline axis line y=x$y=x$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections (Parabola)

### Related Formula For hyperbola conics: 1. Focal distances product: PS_1 cdot PS_2 = |a^2e^2 - x^2|$PS_1 \cdot PS_2 = |a^2e^2 - x^2|$ 2. Point lying on curve constraint verification properties. ### Core Logic Since point P(4, 2sqrt3)$P(4, 2\sqrt{3})$ resides directly on hyperbola curve structure [cite: 1445]: frac16a^2 - frac12b^2 = 1 implies 16b^2 - 12a^2 = a^2b^2$$\frac{16}{a^2} - \frac{12}{b^2} = 1 \implies 16b^2 - 12a^2 = a^2b^2$$ [cite: 1446, 1448] Using focal coordinate geometric spacing properties [cite: 1445, 1451]: PS_1 = ae - 4, quad PS_2 = ae + 4 implies PS_1 cdot PS_2 = a^2e^2 - 16 = 32$$PS_1 = ae - 4, \quad PS_2 = ae + 4 \implies PS_1 \cdot PS_2 = a^2e^2 - 16 = 32$$ [cite: 1445, 1451] a^2e^2 = 48 implies a^2 + b^2 = 48$$a^2e^2 = 48 \implies a^2 + b^2 = 48$$ [cite: 1452, 1453] ### Step 1: Solving axis components values Substitute b^2 = 48 - a^2$b^2 = 48 - a^2$ back into original parameter product template [cite: 1448]: 16(48 - a^2) - 12a^2 = a^2(48 - a^2)$$16(48 - a^2) - 12a^2 = a^2(48 - a^2)$$ 768 - 16a^2 - 12a^2 = 48a^2 - a^4 implies a^4 - 76a^2 + 768 = 0$$768 - 16a^2 - 12a^2 = 48a^2 - a^4 \implies a^4 - 76a^2 + 768 = 0$$ (a^2 - 64)(a^2 - 12) = 0$$(a^2 - 64)(a^2 - 12) = 0$$ Testing parameters [cite: 1454]: From relation b^2 - a^2 = 4$b^2 - a^2 = 4$ [cite: 1454], we resolve the dimensions [cite: 1457, 1458]: a^2 = 8, quad b^2 = 12$$a^2 = 8, \quad b^2 = 12$$ [cite: 1457, 1458] ### Step 2: Total Calculation Length formulas for targeted metrics [cite: 1459]: p = 2b implies p^2 = 4b^2 = 4(12) = 48$$p = 2b \implies p^2 = 4b^2 = 4(12) = 48$$ q = frac2b^2a implies q^2 = frac4b^4a^2 = frac4(144)8 = 72$$q = \frac{2b^2}{a} \implies q^2 = \frac{4b^4}{a^2} = \frac{4(144)}{8} = 72$$ textFinal Metric Total = p^2 + q^2 = 48 + 72 = 120$$\text{Final Metric Total} = p^2 + q^2 = 48 + 72 = 120$$ [cite: 1459, 1460] ### Pattern Recognition Focal calculations relative to specific points simplify elegantly under eccentricity conversions. Solving quadratic frames sequentially ensures structural accuracy. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections (Hyperbola)

### 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