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

### Core Logic For the ellipse, foci are at (pm 5, 0) implies ae = 5$(\pm 5, 0) \implies ae = 5$. Latus rectum = frac2b^2a = sqrt50 = 5sqrt2 implies b^2 = frac5sqrt2a2$\frac{2b^2}{a} = \sqrt{50} = 5\sqrt{2} \implies b^2 = \frac{5\sqrt{2}a}{2}$. ### Step 1: Solve for a and b Using b^2 = a^2(1 - e^2)$b^2 = a^2(1 - e^2)$: a^2 - (ae)^2 = b^2 implies a^2 - 25 = frac5sqrt2a2$$a^2 - (ae)^2 = b^2 \implies a^2 - 25 = \frac{5\sqrt{2}a}{2}$$ 2a^2 - 5sqrt2a - 50 = 0$$2a^2 - 5\sqrt{2}a - 50 = 0$$ 2a^2 - 10sqrt2a + 5sqrt2a - 50 = 0$$2a^2 - 10\sqrt{2}a + 5\sqrt{2}a - 50 = 0$$ 2a(a - 5sqrt2) + 5sqrt2(a - 5sqrt2) = 0$$2a(a - 5\sqrt{2}) + 5\sqrt{2}(a - 5\sqrt{2}) = 0$$ a = 5sqrt2$a = 5\sqrt{2}$ (since a > 0$a > 0$). Now, b^2 = frac5sqrt2(5sqrt2)2 = 25 implies b = 5$b^2 = \frac{5\sqrt{2}(5\sqrt{2})}{2} = 25 \implies b = 5$. ### Step 2: Hyperbola Eccentricity The hyperbola is fracx^2b^2 - fracy^2a^2b^2 = 1$\frac{x^2}{b^2} - \frac{y^2}{a^2b^2} = 1$. Here, semi-major axis A = b$A = b$ and semi-minor axis B = ab$B = ab$. Using eccentricity formula for hyperbola e_H^2 = 1 + fracB^2A^2$e_H^2 = 1 + \frac{B^2}{A^2}$: e_H^2 = 1 + fraca^2 b^2b^2 = 1 + a^2$$e_H^2 = 1 + \frac{a^2 b^2}{b^2} = 1 + a^2$$ Since a = 5sqrt2$a = 5\sqrt{2}$, a^2 = 50$a^2 = 50$. e_H^2 = 1 + 50 = 51$$e_H^2 = 1 + 50 = 51$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections