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

Question

Accepted Answer

C. 1. Write the ellipse in **standard form**: $3x^2+4y^2=1$ means $\frac{x^2}{1/3}+\frac{y^2}{1/4}=1$, so $a^2=\frac13$, $b^2=\frac14$.

2. For the line $\alpha x+4y-\sqrt7=0$ to be tangent to this ellipse, use the **tangency condition** $c^2=a^2m^2+b^2$ after writing the line in slope form. Applying this condition here gives:
$$
\frac{7}{16}=\frac13\cdot\frac{\alpha^2}{16}+\frac14 \;\Rightarrow\; \alpha=\pm3
$$

3. Taking $\alpha=3$, the tangent line is $3x+4y-\sqrt7=0$. Working out the **point of contact** for a tangent line touching this ellipse gives:
$$
P=\left(\frac{1}{\sqrt7},\frac{1}{\sqrt7}\right)
$$

4. The **eccentricity** of the ellipse is:
$$
e=\sqrt{1-\frac{1/4}{1/3}}=\sqrt{1-\frac34}=\frac12
$$

5. The **focal distance** of a point from a focus equals $e$ times the distance from that point to the corresponding **directrix** — a defining property of conics. If $M$ and $M'$ are the feet of perpendiculars from $P$ to the two directrices at $x=\pm\frac{a}{e}$, then:
$$
PS=e\left(\frac{a}{e}-\frac{1}{\sqrt7}\right),\qquad PS'=e\left(\frac{a}{e}+\frac{1}{\sqrt7}\right)
$$

6. Substituting $a=\frac{1}{\sqrt3}$ and $e=\frac12$:
$$
PS=\frac12\left(\frac{2}{\sqrt3}-\frac{1}{\sqrt7}\right)=\frac{1}{\sqrt3}-\frac{1}{2\sqrt7}
$$
$$
PS'=\frac12\left(\frac{2}{\sqrt3}+\frac{1}{\sqrt7}\right)=\frac{1}{\sqrt3}+\frac{1}{2\sqrt7}
$$

Hence, one focal distance of $P$ is $\frac{1}{\sqrt3}+\frac{1}{2\sqrt7}$ — **Option C**.

If the line αx+4y=7\alpha x+4 y=\sqrt{7}αx+4y=7​, where α∈R\alpha \in \mathbb{R}α∈R, touches the ellipse 3x2+4y2=13 x^2+4 y^2=13x2+4y2=1 at the point P in the first quadrant, then one of the focal distances of PPP is :

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