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:

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:

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:

Since point P is in the first quadrant, $x_1 > 0$ and $y_1 > 0$. We use the ellipse equation to find $y_1$:

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.