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 y = mx - 2am - am^3 quad text(Equation of normal to y^2 = 4ax)$$y = mx - 2am - am^3 \quad \text{(Equation of normal to } y^2 = 4ax)$$ ### Core Logic First, extract the centre C$C$ of the circle x^2 + y^2 - 4x - 16y + 64 = 0$x^2 + y^2 - 4x - 16y + 64 = 0$: C equiv (2, 8)$C \equiv (2, 8)$ For the parabola y^2 = 4x$y^2 = 4x$, comparing with y^2 = 4ax$y^2 = 4ax$, we get a = 1$a = 1$. The shortest distance between a point and a curve is measured along the normal to the curve passing through that point. We need the normal to the parabola that passes through the circle's centre (2, 8)$(2, 8)$. ### Step 1: Applying Normal Condition Equation of normal in slope form (a=1$a=1$): y = mx - 2m - m^3$$y = mx - 2m - m^3$$ Since it passes through (2, 8)$(2, 8)$: 8 = m(2) - 2m - m^3$$8 = m(2) - 2m - m^3$$ 8 = -m^3 Rightarrow m^3 = -8 Rightarrow m = -2$$8 = -m^3 \Rightarrow m^3 = -8 \Rightarrow m = -2$$ ### Step 2: Finding Coordinate Point on Parabola The foot of the normal on the parabola is given by the coordinate P(am^2, -2am)$P(am^2, -2am)$. Substitute a=1, m=-2$a=1, m=-2$: P equiv (1(-2)^2, -2(1)(-2))$$P \equiv (1(-2)^2, -2(1)(-2))$$ P equiv (4, 4)$P \equiv (4, 4)$ ### Step 3: Calculating Shortest Distance The distance d$d$ is between the point P(4, 4)$P(4, 4)$ and the centre C(2, 8)$C(2, 8)$: d = sqrt(4-2)^2 + (4-8)^2$$d = \sqrt{(4-2)^2 + (4-8)^2}$$ d = sqrt2^2 + (-4)^2 = sqrt4 + 16 = sqrt20$$d = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20}$$ Thus, d^2 = 20$d^2 = 20$. ### Pattern Recognition Shortest distance between a curve and a fixed point (like a circle's center) ALWAYS lies exactly along their common normal. Write the parameterized normal equation and force it through the coordinate to extract the slope root. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Circles Class 11 Maths: Parabola

### Related Formula T = S_1$T = S_1$ Equation of chord with a given midpoint (x_1, y_1)$(x_1, y_1)$ for an ellipse fracx^2a^2 + fracy^2b^2 = 1$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is: fracx x_1a^2 + fracy y_1b^2 = fracx_1^2a^2 + fracy_1^2b^2$$\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2}$$ ### Core Logic Given the ellipse fracx^225 + fracy^216 = 1$\frac{x^2}{25} + \frac{y^2}{16} = 1$ and midpoint (x_1, y_1) = (1, frac25)$(x_1, y_1) = (1, \frac{2}{5})$. Substitute these into the T = S_1$T = S_1$ formula: fracx(1)25 + fracy(2/5)16 = frac1^225 + frac(2/5)^216$$\frac{x(1)}{25} + \frac{y(2/5)}{16} = \frac{1^2}{25} + \frac{(2/5)^2}{16}$$ fracx25 + fracy40 = frac125 + frac4/2516$$\frac{x}{25} + \frac{y}{40} = \frac{1}{25} + \frac{4/25}{16}$$ frac8x + 5y200 = frac125 + frac1100$$\frac{8x + 5y}{200} = \frac{1}{25} + \frac{1}{100}$$ frac8x + 5y200 = frac5100 = frac10200$$\frac{8x + 5y}{200} = \frac{5}{100} = \frac{10}{200}$$ This simplifies to the chord equation: 8x + 5y = 10 Rightarrow y = frac10 - 8x5$$8x + 5y = 10 \Rightarrow y = \frac{10 - 8x}{5}$$ ### Step 1: Finding Points of Intersection Substitute the chord equation back into the ellipse equation to find intersection coordinates: fracx^225 + frac(frac10 - 8x5)^216 = 1$$\frac{x^2}{25} + \frac{(\frac{10 - 8x}{5})^2}{16} = 1$$ fracx^225 + frac(10 - 8x)^2400 = 1$$\frac{x^2}{25} + \frac{(10 - 8x)^2}{400} = 1$$ Multiply the entire equation by 400: 16x^2 + (10 - 8x)^2 = 400$$16x^2 + (10 - 8x)^2 = 400$$ 16x^2 + 100 - 160x + 64x^2 = 400$$16x^2 + 100 - 160x + 64x^2 = 400$$ 80x^2 - 160x - 300 = 0$$80x^2 - 160x - 300 = 0$$ Divide by 20: 4x^2 - 8x - 15 = 0$$4x^2 - 8x - 15 = 0$$ ### Step 2: Solving the Quadratic Solve for roots x_1$x_1$ and x_2$x_2$: x = frac-(-8) pm sqrt(-8)^2 - 4(4)(-15)2(4)$$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(4)(-15)}}{2(4)}$$ x = frac8 pm sqrt64 + 2408 = frac8 pm sqrt3048$$x = \frac{8 \pm \sqrt{64 + 240}}{8} = \frac{8 \pm \sqrt{304}}{8}$$ Difference of x-coordinates: |x_1 - x_2| = fracsqrt3044$$|x_1 - x_2| = \frac{\sqrt{304}}{4}$$ ### Step 3: Calculating Chord Length Because the points lie on the line y = frac10 - 8x5$y = \frac{10 - 8x}{5}$, the difference in y-coordinates relates to the slope m = -frac85$m = -\frac{8}{5}$: |y_1 - y_2| = |-frac85| |x_1 - x_2|$$|y_1 - y_2| = |-\frac{8}{5}| |x_1 - x_2|$$ The distance D$D$ between the points is: D = sqrt(x_1 - x_2)^2 + (y_1 - y_2)^2 = |x_1 - x_2| sqrt1 + m^2$$D = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} = |x_1 - x_2| \sqrt{1 + m^2}$$ D = fracsqrt3044 sqrt1 + left(frac-85right)^2$$D = \frac{\sqrt{304}}{4} \sqrt{1 + \left(\frac{-8}{5}\right)^2}$$ D = fracsqrt3044 sqrt1 + frac6425 = fracsqrt3044 fracsqrt895$$D = \frac{\sqrt{304}}{4} \sqrt{1 + \frac{64}{25}} = \frac{\sqrt{304}}{4} \frac{\sqrt{89}}{5}$$ D = fracsqrt16 times 194 fracsqrt895 = frac4sqrt194 fracsqrt895 = fracsqrt16915$$D = \frac{\sqrt{16 \times 19}}{4} \frac{\sqrt{89}}{5} = \frac{4\sqrt{19}}{4} \frac{\sqrt{89}}{5} = \frac{\sqrt{1691}}{5}$$ ### Pattern Recognition For intersecting lines and conics, you rarely need to explicitly find y_1$y_1$ and y_2$y_2$. Use the slope property: Distance = |x_1 - x_2| sqrt1+m^2$= |x_1 - x_2| \sqrt{1+m^2}$ to bypass entirely the substitution of messy quadratic roots back into the linear equation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Ellipse Class 11 Maths: Straight Lines

### 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 textEccentricity of Hyperbola: e^2 = 1 + fracb^2a^2$$\text{Eccentricity of Hyperbola: } e^2 = 1 + \frac{b^2}{a^2}$$ textFoci coordinates: (pm ae, 0)$$\text{Foci coordinates: } (\pm ae, 0)$$ textArea of Triangle: frac12 times textbase times textheight$$\text{Area of Triangle: } \frac{1}{2} \times \text{base} \times \text{height}$$ ### Core Logic For the hyperbola fracx^29 - fracy^24 = 1$\frac{x^2}{9} - \frac{y^2}{4} = 1$, we have a^2 = 9$a^2 = 9$ and b^2 = 4$b^2 = 4$. e^2 = 1 + frac49 = frac139 Rightarrow e = fracsqrt133$$e^2 = 1 + \frac{4}{9} = \frac{13}{9} \Rightarrow e = \frac{\sqrt{13}}{3}$$ The distance between the two foci S_1, S_2$S_1, S_2$ is 2ae$2ae$: 2ae = 2(3)left(fracsqrt133right) = 2sqrt13$$2ae = 2(3)\left(\frac{\sqrt{13}}{3}\right) = 2\sqrt{13}$$ ### Step 1: Using the Triangle Area

Let point P on the hyperbola be (alpha, beta)$(\alpha, \beta)$ in the first quadrant (so alpha, beta gt 0$\alpha, \beta \gt 0$). The base of the triangle is the segment between foci, length = 2sqrt13$= 2\sqrt{13}$. The height of the triangle is the y-coordinate of P, which is beta$\beta$. Area of Delta PS_1S_2 = frac12 times textbase times textheight$\Delta PS_1S_2 = \frac{1}{2} \times \text{base} \times \text{height}$ 2sqrt13 = frac12 times (2sqrt13) times beta$$2\sqrt{13} = \frac{1}{2} \times (2\sqrt{13}) \times \beta$$ 2sqrt13 = sqrt13 cdot beta Rightarrow beta = 2$$2\sqrt{13} = \sqrt{13} \cdot \beta \Rightarrow \beta = 2$$ ### Step 2: Finding P's coordinates and Distance Since P lies on the hyperbola: fracalpha^29 - fracbeta^24 = 1$$\frac{\alpha^2}{9} - \frac{\beta^2}{4} = 1$$ Substitute beta = 2$\beta = 2$: fracalpha^29 - frac44 = 1$$\frac{\alpha^2}{9} - \frac{4}{4} = 1$$ fracalpha^29 - 1 = 1 Rightarrow fracalpha^29 = 2 Rightarrow alpha^2 = 18$$\frac{\alpha^2}{9} - 1 = 1 \Rightarrow \frac{\alpha^2}{9} = 2 \Rightarrow \alpha^2 = 18$$ We need the square of the distance of P from the origin, which is alpha^2 + beta^2$\alpha^2 + \beta^2$: textDistance^2 = alpha^2 + beta^2 = 18 + 2^2 = 18 + 4 = 22$$\text{Distance}^2 = \alpha^2 + \beta^2 = 18 + 2^2 = 18 + 4 = 22$$ ### Pattern Recognition Area formed by a point on a conic and its foci uses the interfocal distance 2ae$2ae$ as a flat base on the x-axis, directly exposing the point's y-coordinate. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections