Animated Solution for Mathematics - Conic Sections: If the line $\alpha x+4y=\sqrt{7}$, where $\alpha \in \mathbb{R}$, touches the ellipse $3x^2+4y^2=1$ at the point $P$ in the first quadrant, then one of the focal distances of $P$ is :

Imagine you are standing before the elegant curve of an ellipse, defined by the equation $3x^2+4y^2=1$. To truly understand this problem, we must first bring this ellipse into its standard, recognizable form.

We divide the entire equation by $1$, giving us:

Here, we immediately see that $a^2=1/3$ and $b^2=1/4$. Because $a^2>b^2$, we know we are dealing with a horizontal ellipse, stretched along the $x$-axis. This is our foundation.

Now, consider a line $\alpha x+4y=\sqrt{7}$ that kisses the ellipse at a single point $P(x_0,y_0)$ in the first quadrant. This is not just any line; it is a tangent.

In the language of coordinate geometry, the tangent at any point $(x_0,y_0)$ on our ellipse $3x^2+4y^2=1$ is given by the elegant formula:

The line provided in the problem is $\alpha x+4y=\sqrt{7}$. To make these two equations speak the same language, we must normalize the given line by dividing the entire equation by $\sqrt{7}$:

Now, the comparison is trivial. By matching the coefficients of $y$, we find that $4y_0=4/\sqrt{7}$, which simplifies beautifully to $y_0=1/\sqrt{7}$.

With $y_0$ in hand, finding $x_0$ is just a matter of substitution. Since $P(x_0,y_0)$ lies on the ellipse, it must satisfy $3x_0^2+4y_0^2=1$.

Plugging in our value for $y_0$, we get $3x_0^2+4(1/7)=1$, which leads us to $3x_0^2=3/7$, or $x_0^2=1/7$. Since $P$ is in the first quadrant, $x_0$ must be positive, so $x_0=1/\sqrt{7}$.

Finally, we arrive at the focal distance. The distance of any point $(x_0,y_0)$ from the foci is given by the formula $r=a\pm e{x}_0$. We already know $a=1/\sqrt{3}$.

We calculate the eccentricity $e$ using $b^2=a^2(1-e^2)$:

Substituting these values, the focal distance becomes:

Looking at our options, the positive case matches perfectly. You have navigated the geometry, mastered the algebra, and uncovered the hidden focal distance. The final result is $\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{7}}$.