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 - Eccentricity of hyperbola fracx^2a^2 - fracy^2b^2 = 1$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: e_h = sqrt1 + fracb^2a^2$e_h = \sqrt{1 + \frac{b^2}{a^2}}$ - Eccentricity of ellipse fracx^2B^2 + fracy^2A^2 = 1$\frac{x^2}{B^2} + \frac{y^2}{A^2} = 1$ (where A > B$A > B$): e_e = sqrt1 - fracB^2A^2$e_e = \sqrt{1 - \frac{B^2}{A^2}}$ ### Core Logic Let's rewrite both equations in their standard forms: 1. **Hyperbola:** x^2 - y^2csc^2theta = 5 implies fracx^25 - fracy^25sin^2theta = 1$x^2 - y^2\csc^2\theta = 5 \implies \frac{x^2}{5} - \frac{y^2}{5\sin^2\theta} = 1$ Here, a^2 = 5$a^2 = 5$ and b^2 = 5sin^2theta$b^2 = 5\sin^2\theta$. e_h = sqrt1 + frac5sin^2theta5 = sqrt1 + sin^2theta$$e_h = \sqrt{1 + \frac{5\sin^2\theta}{5}} = \sqrt{1 + \sin^2\theta}$$ 2. **Ellipse:** x^2csc^2theta + y^2 = 5 implies fracx^25sin^2theta + fracy^25 = 1$x^2\csc^2\theta + y^2 = 5 \implies \frac{x^2}{5\sin^2\theta} + \frac{y^2}{5} = 1$ Since 0 < theta < pi/2$0 < \theta < \pi/2$, we know 0 < sintheta < 1 implies 5sin^2theta < 5$0 < \sin\theta < 1 \implies 5\sin^2\theta < 5$. Thus, the major axis is along the y-axis, meaning A^2 = 5$A^2 = 5$ and B^2 = 5sin^2theta$B^2 = 5\sin^2\theta$. e_e = sqrt1 - frac5sin^2theta5 = sqrt1 - sin^2theta = costheta$$e_e = \sqrt{1 - \frac{5\sin^2\theta}{5}} = \sqrt{1 - \sin^2\theta} = \cos\theta$$ ### Step 1: Setting up the Equation and Solving Given e_h = sqrt7e_e$e_h = \sqrt{7}e_e$, squaring both sides yields: e_h^2 = 7e_e^2$e_h^2 = 7e_e^2$ 1 + sin^2theta = 7(1 - sin^2theta)$$1 + \sin^2\theta = 7(1 - \sin^2\theta)$$ 1 + sin^2theta = 7 - 7sin^2theta$$1 + \sin^2\theta = 7 - 7\sin^2\theta$$ 8sin^2theta = 6 implies sin^2theta = frac34$$8\sin^2\theta = 6 \implies \sin^2\theta = \frac{3}{4}$$ Since theta in (0, pi/2)$\theta \in (0, \pi/2)$: sintheta = fracsqrt32 implies theta = fracpi3$$\sin\theta = \frac{\sqrt{3}}{2} \implies \theta = \frac{\pi}{3}$$ ### Pattern Recognition Sees: Conic parameters tied dynamically to trigonometry. Trap: In the ellipse equation, do not automatically assume the x-axis holds the major axis. Since sin^2theta < 1$\sin^2\theta < 1$, the denominator under y^2$y^2$ is larger, making it a vertical ellipse. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections Class 11 Mathematics: Trigonometric Functions

### Related Formula - Eccentricity of an ellipse (a>b$a>b$): e_E^2 = 1 - fracb^2a^2$e_E^2 = 1 - \frac{b^2}{a^2}$ - Eccentricity of a hyperbola: e_H^2 = 1 + fracb^2a^2$e_H^2 = 1 + \frac{b^2}{a^2}$ ### Core Logic Given the eccentricity of the ellipse is e_E = frac1sqrt2$e_E = \frac{1}{\sqrt{2}}$: e_E^2 = 1 - fracb^2a^2 implies left(frac1sqrt2right)^2 = 1 - fracb^2a^2$$e_E^2 = 1 - \frac{b^2}{a^2} \implies \left(\frac{1}{\sqrt{2}}\right)^2 = 1 - \frac{b^2}{a^2}$$ frac12 = 1 - fracb^2a^2 implies fracb^2a^2 = frac12$$\frac{1}{2} = 1 - \frac{b^2}{a^2} \implies \frac{b^2}{a^2} = \frac{1}{2}$$ ### Step 1: Finding Hyperbola Eccentricity The required equation asks for the square of the eccentricity of the conjugate parameter hyperbola fracx^2a^2 - fracy^2b^2 = 1$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: e_H^2 = 1 + fracb^2a^2$$e_H^2 = 1 + \frac{b^2}{a^2}$$ Substituting the derived ratio fracb^2a^2 = frac12$\frac{b^2}{a^2} = \frac{1}{2}$: e_H^2 = 1 + frac12 = frac32$$e_H^2 = 1 + \frac{1}{2} = \frac{3}{2}$$ Thus, the square of the eccentricity is frac32$\frac{3}{2}$. ### Pattern Recognition Sees: Linked parameters between ellipse and hyperbola. Shortcut: Notice that information about the latus rectum length (sqrt14$\sqrt{14}$) is a deliberate distractor! Since the definition of hyperbola eccentricity squared (e_H^2 = 1 + b^2/a^2$e_H^2 = 1 + b^2/a^2$) relies solely on the dimensionless ratio fracb^2a^2$\frac{b^2}{a^2}$, you don't need to compute absolute scales for a$a$ or b$b$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

### Related Formula For two circles with centers C_1, C_2$C_1, C_2$ and radii r_1, r_2$r_1, r_2$ to intersect at two distinct real points, the distance between their centers must satisfy: |r_1 - r_2| < C_1C_2 < r_1 + r_2$$|r_1 - r_2| < C_1C_2 < r_1 + r_2$$ ### Core Logic Extracting properties from the circle equations: - Circle C$C$: x^2 + y^2 = 4 implies$x^2 + y^2 = 4 \implies$ Center C_1 = (0,0)$C_1 = (0,0)$, Radius r_1 = 2$r_1 = 2$ - Circle C'$C'$: x^2 + y^2 - 4lambda x + 9 = 0 implies$x^2 + y^2 - 4\lambda x + 9 = 0 \implies$ Center C_2 = (2lambda, 0)$C_2 = (2\lambda, 0)$, Radius r_2 = sqrt(-2lambda)^2 - 9 = sqrt4lambda^2 - 9$r_2 = \sqrt{(-2\lambda)^2 - 9} = \sqrt{4\lambda^2 - 9}$ For real intersection conditions, the radius must be defined: 4lambda^2 - 9 > 0 implies lambda^2 > frac94 quad implies (1)$$4\lambda^2 - 9 > 0 \implies \lambda^2 > \frac{9}{4} \quad \implies (1)$$ Distance between centers: C_1C_2 = sqrt(2lambda - 0)^2 + 0^2 = |2lambda|$C_1C_2 = \sqrt{(2\lambda - 0)^2 + 0^2} = |2\lambda|$. ### Step 1: Formulate the Triangle Inequality Solutions Applying the intersection condition: |2 - sqrt4lambda^2 - 9| < |2lambda| < 2 + sqrt4lambda^2 - 9$$|2 - \sqrt{4\lambda^2 - 9}| < |2\lambda| < 2 + \sqrt{4\lambda^2 - 9}$$ The right side inequality |2lambda| < 2 + sqrt4lambda^2 - 9$|2\lambda| < 2 + \sqrt{4\lambda^2 - 9}$ is always valid for real radii. Solving the left side inequality by squaring: left(2 - sqrt4lambda^2 - 9right)^2 < (2lambda)^2$$\left(2 - \sqrt{4\lambda^2 - 9}\right)^2 < (2\lambda)^2$$ 4 + (4lambda^2 - 9) - 4sqrt4lambda^2 - 9 < 4lambda^2$$4 + (4\lambda^2 - 9) - 4\sqrt{4\lambda^2 - 9} < 4\lambda^2$$ -5 - 4sqrt4lambda^2 - 9 < 0 implies 4sqrt4lambda^2 - 9 > -5$$-5 - 4\sqrt{4\lambda^2 - 9} < 0 \implies 4\sqrt{4\lambda^2 - 9} > -5$$ Since a square root is always non-negative, square both sides to get the strict bound: 16(4lambda^2 - 9) > 25 implies 64lambda^2 - 144 > 25$$16(4\lambda^2 - 9) > 25 \implies 64\lambda^2 - 144 > 25$$ 64lambda^2 > 169 implies lambda^2 > frac16964 quad implies (2)$$64\lambda^2 > 169 \implies \lambda^2 > \frac{169}{64} \quad \implies (2)$$ Combining bounds (1) and (2), condition (2) is more restrictive, meaning: lambda in left(-infty, -frac138right) cup left(frac138, inftyright)$$\lambda \in \left(-\infty, -\frac{13}{8}\right) \cup \left(\frac{13}{8}, \infty\right)$$ This can be written as mathbbR - left[-frac138, frac138right]$\mathbb{R} - \left[-\frac{13}{8}, \frac{13}{8}\right]$. ### Step 2: Coordinate Analysis of the Target Point Comparing our interval with mathbbR - [a, b]$\mathbb{R} - [a, b]$, we identify: a = -frac138, quad b = frac138$$a = -\frac{13}{8}, \quad b = \frac{13}{8}$$ Now, substitute these bounds to locate the coordinates of our target point (8a+12, \, 16b-20)$(8a+12, \, 16b-20)$: - x$x$-coordinate: 8left(-frac138right) + 12 = -13 + 12 = -1$8\left(-\frac{13}{8}\right) + 12 = -13 + 12 = -1$ - y$y$-coordinate: 16left(frac138right) - 20 = 26 - 20 = 6$16\left(\frac{13}{8}\right) - 20 = 26 - 20 = 6$ Hence, the point is (-1, 6)$(-1, 6)$. ### Step 3: Test Point against the Options Substitute (-1, 6)$(-1, 6)$ into the given curves to find a match: Testing Option (4): 6x^2 + y^2 = 42$6x^2 + y^2 = 42$ 6(-1)^2 + (6)^2 = 6(1) + 36 = 42$$6(-1)^2 + (6)^2 = 6(1) + 36 = 42$$ Since LHS = RHS, the point satisfies the curve in Option (4). ### Pattern Recognition Sees: Dynamic variable interval limits forming loci requirements. Shortcut: When checking inequalities like |2 - sqrtz| < 2lambda$|2 - \sqrt{z}| < 2\lambda$, algebraic squaring helps simplify the terms quickly by eliminating the parameter 4lambda^2$4\lambda^2$ from both sides. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections (Circles) Class 11 Mathematics: Linear Inequalities

### Related Formula - Distance from a point (x_0, y_0)$(x_0, y_0)$ to a straight line Ax + By + C = 0$Ax + By + C = 0$ is: d = frac|Ax_0 + By_0 + C|sqrtA^2 + B^2$$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$ - Tangency Condition: Perpendicular distance from the center of a circle to a line must equal its radius (d = r$d = r$). ### Core Logic First, find the point of intersection P$P$ between the circle x^2 + y^2 = 3$x^2 + y^2 = 3$ and the parabola x^2 = 2y$x^2 = 2y$: 2y + y^2 = 3 implies y^2 + 2y - 3 = 0$$2y + y^2 = 3 \implies y^2 + 2y - 3 = 0$$ (y + 3)(y - 1) = 0 implies y = 1 quad text(since P text lies in the first quadrant)$$(y + 3)(y - 1) = 0 \implies y = 1 \quad \text{(since } P \text{ lies in the first quadrant)}$$ Substituting y = 1$y = 1$ back: x^2 = 2(1) implies x = sqrt2$x^2 = 2(1) \implies x = \sqrt{2}$. Thus, point P = (sqrt2, 1)$P = (\sqrt{2}, 1)$. ### Step 1: Solve for Line Parameter alpha Since P(sqrt2, 1)$P(\sqrt{2}, 1)$ lies on line L: sqrt2x + y = alpha$L: \sqrt{2}x + y = \alpha$: sqrt2(sqrt2) + 1 = alpha implies 2 + 1 = alpha implies alpha = 3$$\sqrt{2}(\sqrt{2}) + 1 = \alpha \implies 2 + 1 = \alpha \implies \alpha = 3$$ So the equation of line L$L$ is sqrt2x + y - 3 = 0$\sqrt{2}x + y - 3 = 0$. ### Step 2: Find Centers Q1 and Q2 The centers lie on the y-axis, so let their coordinates be (0, k)$(0, k)$. The perpendicular distance to line L$L$ is equal to the radius 2sqrt3$2\sqrt{3}$: frac|sqrt2(0) + k - 3|sqrt(sqrt2)^2 + 1^2 = 2sqrt3$$\frac{|\sqrt{2}(0) + k - 3|}{\sqrt{(\sqrt{2})^2 + 1^2}} = 2\sqrt{3}$$ frac|k - 3|sqrt3 = 2sqrt3 implies |k - 3| = 6$$\frac{|k - 3|}{\sqrt{3}} = 2\sqrt{3} \implies |k - 3| = 6$$ Unfolding the absolute parameter: - k - 3 = 6 implies k = 9 implies Q_1 = (0, 9)$k - 3 = 6 \implies k = 9 \implies Q_1 = (0, 9)$ - k - 3 = -6 implies k = -3 implies Q_2 = (0, -3)$k - 3 = -6 \implies k = -3 \implies Q_2 = (0, -3)$ ### Step 3: Evaluate Triangle Area Squared Using the matrix determinant method for the area of triangle PQ_1Q_2$\triangle PQ_1Q_2$: textArea = frac12 beginvmatrix sqrt2 & 1 & 1 \\ 0 & 9 & 1 \\ 0 & -3 & 1 endvmatrix = frac12 left| sqrt2 cdot (9 - (-3)) right| = 6sqrt2$$\text{Area} = \frac{1}{2} \begin{vmatrix} \sqrt{2} & 1 & 1 \\ 0 & 9 & 1 \\ 0 & -3 & 1 \end{vmatrix} = \frac{1}{2} \left| \sqrt{2} \cdot (9 - (-3)) \right| = 6\sqrt{2}$$ Squaring the final area value: textArea^2 = (6sqrt2)^2 = 72$$\text{Area}^2 = (6\sqrt{2})^2 = 72$$ ### Pattern Recognition Sees: Intersecting conics interacting via tangent logic matrix calculations. Shortcut: Since the base Q_1Q_2$Q_1Q_2$ sits entirely on the y-axis, its length is simply 9 - (-3) = 12$9 - (-3) = 12$. The height corresponds to the x-coordinate of point P$P$, which is sqrt2$\sqrt{2}$. Area is frac12 times 12 times sqrt2 = 6sqrt2$\frac{1}{2} \times 12 \times \sqrt{2} = 6\sqrt{2}$, saving determinant steps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections (Circles and Parabolas) Class 10 Coordinate Geometry: Area of Triangles

### Related Formula Equation of chord with given midpoint (x_1, y_1)$(x_1, y_1)$ is T = S_1$T = S_1$. ### Core Logic For the parabola x^2 = 8y$x^2 = 8y$, the equation of the chord with midpoint left(1, frac54right)$\left(1, \frac{5}{4}\right)$ is: x(1) - 4left(y + frac54right) = 1^2 - 8left(frac54right)$$x(1) - 4\left(y + \frac{5}{4}\right) = 1^2 - 8\left(\frac{5}{4}\right)$$ x - 4y - 5 = 1 - 10 = -9$$x - 4y - 5 = 1 - 10 = -9$$ x - 4y + 4 = 0 quad dots (i)$$x - 4y + 4 = 0 \quad \dots (i)$$ ### Step 1: Point Intersection with Paraboloid Curve Since P(alpha, beta)$P(\alpha, \beta)$ lies on this chord and also on y^2 = 4x$y^2 = 4x$: 1) alpha - 4beta + 4 = 0 implies alpha = 4beta - 4$\alpha - 4\beta + 4 = 0 \implies \alpha = 4\beta - 4$ 2) beta^2 = 4alpha$\beta^2 = 4\alpha$ Substituting alpha$\alpha$ into the equation: beta^2 = 4(4beta - 4) implies beta^2 - 16beta + 16 = 0$$\beta^2 = 4(4\beta - 4) \implies \beta^2 - 16\beta + 16 = 0$$ ### Step 2: Calculating the Target Value We need to find the value of (alpha - 28)(beta - 8)$(\alpha - 28)(\beta - 8)$. Substitute alpha = 4beta - 4$\alpha = 4\beta - 4$ into this targeted expression: textValue = (4beta - 4 - 28)(beta - 8) = (4beta - 32)(beta - 8) = 4(beta - 8)(beta - 8) = 4(beta^2 - 16beta + 64)$$\text{Value} = (4\beta - 4 - 28)(\beta - 8) = (4\beta - 32)(\beta - 8) = 4(\beta - 8)(\beta - 8) = 4(\beta^2 - 16\beta + 64)$$ From the quadratic step, we know beta^2 - 16beta = -16$\beta^2 - 16\beta = -16$. Substituting this: textValue = 4(-16 + 64) = 4(48) = 192$$\text{Value} = 4(-16 + 64) = 4(48) = 192$$ ### Pattern Recognition Do not solve for ugly root combinations explicitly if the target expression can be algebraically mapped back to the defining quadratic equations. This prevents unnecessary fractional math steps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections