Ellipse Tangency Point P Focal Distance Calculation
The problem asks for one of the focal distances of the point P where the line $\alpha x + 4y = \sqrt{7}$ touches the ellipse $3x^2 + 4y^2 = 1$, with P in the first quadrant.
Ellipse Parameters
First, let's identify the parameters of the ellipse $3x^2 + 4y^2 = 1$. We rewrite it in standard form:
$ \frac{x^2}{1/3} + \frac{y^2}{1/4} = 1 $Comparing this to $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we get:
- $a^2 = \frac{1}{3} \implies a = \frac{1}{\sqrt{3}}$
- $b^2 = \frac{1}{4} \implies b = \frac{1}{2}$
Since $a^2 > b^2$, the major axis is along the x-axis.
The eccentricity $e$ is calculated as:
$ e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{1/4}{1/3} = 1 - \frac{3}{4} = \frac{1}{4} $ $ e = \sqrt{\frac{1}{4}} = \frac{1}{2} $Focal Distances of Point P
For any point P$(x_1, y_1)$ on an ellipse with the major axis along the x-axis, the focal distances are given by $a \pm ex_1$.
Let's examine the correct answer provided: $\frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{5}}$. This has the form $a - ex_1$.
- The term $a$ matches our calculated $a = \frac{1}{\sqrt{3}}$.
- Therefore, $ex_1$ must correspond to $\frac{1}{2\sqrt{5}}$.
Determining Coordinates of Point P
Using $e = \frac{1}{2}$ and $ex_1 = \frac{1}{2\sqrt{5}}$, we find the x-coordinate of P:
$ x_1 = \frac{ex_1}{e} = \frac{1/(2\sqrt{5})}{1/2} = \frac{1}{\sqrt{5}} $Since point P is in the first quadrant, $x_1 > 0$ and $y_1 > 0$. We use the ellipse equation to find $y_1$:
$ 3x_1^2 + 4y_1^2 = 1 $ $ 3\left(\frac{1}{\sqrt{5}}\right)^2 + 4y_1^2 = 1 $ $ 3\left(\frac{1}{5}\right) + 4y_1^2 = 1 $ $ \frac{3}{5} + 4y_1^2 = 1 $ $ 4y_1^2 = 1 - \frac{3}{5} = \frac{2}{5} $ $ y_1^2 = \frac{2}{20} = \frac{1}{10} $Since P is in the first quadrant, $y_1 = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}}$.
Thus, the point P is $\left(\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{10}}\right)$. This point lies on the ellipse and is in the first quadrant.
Focal Distances of P
The two focal distances of P$(\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{10}})$ are:
- $d_1 = a + ex_1 = \frac{1}{\sqrt{3}} + \frac{1}{2} \times \frac{1}{\sqrt{5}} = \frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{5}}$
- $d_2 = a - ex_1 = \frac{1}{\sqrt{3}} - \frac{1}{2} \times \frac{1}{\sqrt{5}} = \frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{5}}$
The question asks for one of the focal distances. The value $\frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{5}}$ matches one of these calculated focal distances.
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