In the line alpha x + 4y = sqrt7$\alpha x + 4y = \sqrt{7}$, where alpha in R$\alpha \in R$, touches the ellipse 3x^2 + 4y^2 = 1$3x^{2} + 4y^{2} = 1$ at the point P$P$ in the first quadrant, then one of the focal distances of P$P$ is:

### Related Formula textCondition of tangency for ellipse fracx^2a^2 + fracy^2b^2 = 1 text is c^2 = a^2m^2 + b^2$$\text{Condition of tangency for ellipse } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \text{ is } c^2 = a^2m^2 + b^2$$ textFocal distance SP = a pm ex$$\text{Focal distance } SP = a \pm ex$$ textEccentricity e = sqrt1 - fracb^2a^2$$\text{Eccentricity } e = \sqrt{1 - \frac{b^2}{a^2}}$$ ### Core Logic

Identify the slope and intercepts of the line to find alpha$\alpha$. Use the point of contact formula to locate P(x_1, y_1)$P(x_1, y_1)$ and apply focal distance definitions. ### Step 1: Determine alpha Rewrite the ellipse: fracx^21/3 + fracy^21/4 = 1 implies a^2 = frac13, b^2 = frac14$\frac{x^2}{1/3} + \frac{y^2}{1/4} = 1 \implies a^2 = \frac{1}{3}, b^2 = \frac{1}{4}$. The line is y = -fracalpha4x + fracsqrt74$y = -\frac{\alpha}{4}x + \frac{\sqrt{7}}{4}$. Using c^2 = a^2m^2 + b^2$c^2 = a^2m^2 + b^2$: left(fracsqrt74right)^2 = frac13 left(-fracalpha4right)^2 + frac14$$\left(\frac{\sqrt{7}}{4}\right)^2 = \frac{1}{3} \left(-\frac{\alpha}{4}\right)^2 + \frac{1}{4}$$ frac716 = fracalpha^248 + frac416 implies frac316 = fracalpha^248 implies alpha^2 = 9 implies alpha = pm 3$$\frac{7}{16} = \frac{\alpha^2}{48} + \frac{4}{16} \implies \frac{3}{16} = \frac{\alpha^2}{48} \implies \alpha^2 = 9 \implies \alpha = \pm 3$$ Since P$P$ is in the first quadrant, coordinates x, y$x, y$ are positive, so we use the tangent 3x + 4y - sqrt7 = 0$3x + 4y - \sqrt{7} = 0$. ### Step 2: Find Point of Contact P The tangent at P(x_1, y_1)$P(x_1, y_1)$ is 3xx_1 + 4yy_1 = 1$3xx_1 + 4yy_1 = 1$. Comparing this with 3x + 4y = sqrt7$3x + 4y = \sqrt{7}$ (divided by sqrt7$\sqrt{7}$ to match constant 1$1$): frac3xsqrt7 + frac4ysqrt7 = 1$\frac{3x}{\sqrt{7}} + \frac{4y}{\sqrt{7}} = 1$. Comparing coefficients: 3x_1 = frac3sqrt7 implies x_1 = frac1sqrt7$$3x_1 = \frac{3}{\sqrt{7}} \implies x_1 = \frac{1}{\sqrt{7}}$$ 4y_1 = frac4sqrt7 implies y_1 = frac1sqrt7$$4y_1 = \frac{4}{\sqrt{7}} \implies y_1 = \frac{1}{\sqrt{7}}$$ So P = left(frac1sqrt7, frac1sqrt7right)$P = \left(\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right)$. ### Step 3: Calculate Focal Distance Find eccentricity: e = sqrt1 - frac1/41/3 = sqrt1 - frac34 = frac12$$e = \sqrt{1 - \frac{1/4}{1/3}} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$$ The focal distances are a - ex$a - ex$ and a + ex$a + ex$. Since a^2 = 1/3 implies a = 1/sqrt3$a^2 = 1/3 \implies a = 1/\sqrt{3}$. SP = a - ex_1 = frac1sqrt3 - frac12left(frac1sqrt7right) = frac1sqrt3 - frac12sqrt7$$SP = a - ex_1 = \frac{1}{\sqrt{3}} - \frac{1}{2}\left(\frac{1}{\sqrt{7}}\right) = \frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{7}}$$ S'P = a + ex_1 = frac1sqrt3 + frac12left(frac1sqrt7right) = frac1sqrt3 + frac12sqrt7$$S'P = a + ex_1 = \frac{1}{\sqrt{3}} + \frac{1}{2}\left(\frac{1}{\sqrt{7}}\right) = \frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$$ Matching with the options, the focal distance is frac1sqrt3 + frac12sqrt7$\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$. ### Pattern Recognition For tangency lx+my+n=0$lx+my+n=0$ to x^2/a^2 + y^2/b^2 = 1$x^2/a^2 + y^2/b^2 = 1$, use a^2 l^2 + b^2 m^2 = n^2$a^2 l^2 + b^2 m^2 = n^2$. Points of contact can be quickly evaluated by comparing T=0$T=0$ to the normalized tangent equation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections