Question

If the line $ax + 4y = \sqrt{7}$, where $a \in \mathbb{R}$, 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 :

• A. $\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{5}}$
• B. $\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$
• C. $\frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{5}}$
• D. $\frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{11}}$

Hint:

For any ellipse, the sum of focal distances of any point is equal to the length of the major axis ($2a$).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The given ellipse is $\frac{x^2}{1/3} + \frac{y^2}{1/4} = 1$. Focal distance of a point $(x_1, y_1)$ on an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (with $a>b$) is $|a \pm ex_1|$.

Step 2: Key Formula or Approach:
Compare the given tangent $ax + 4y = \sqrt{7}$ with the standard tangent equation $3xx_1 + 4yy_1 = 1$.

Step 3: Detailed Explanation:
Standard form: $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, where $A^2 = 1/3, B^2 = 1/4$.
Tangent at $(x_1, y_1)$ is $3xx_1 + 4yy_1 = 1$.
Given tangent: $\frac{a}{\sqrt{7}}x + \frac{4}{\sqrt{7}}y = 1$.
Comparing coefficients: $4y_1 = \frac{4}{\sqrt{7}} \implies y_1 = \frac{1}{\sqrt{7}}$.
Since $(x_1, y_1)$ lies on $3x^2 + 4y^2 = 1$:
\[ 3x_1^2 + 4(1/7) = 1 \implies 3x_1^2 = 1 - 4/7 = 3/7 \implies x_1 = \frac{1}{\sqrt{7}} \]
(Taking positive value as $P$ is in the first quadrant).
Now, calculate eccentricity $e$:
\[ e^2 = 1 - \frac{B^2}{A^2} = 1 - \frac{1/4}{1/3} = 1 - 3/4 = 1/4 \implies e = 1/2. \]
Focal distances are $A \pm ex_1$:
\[ \text{Focal Distance} = \frac{1}{\sqrt{3}} \pm \frac{1}{2} \cdot \frac{1}{\sqrt{7}} = \frac{1}{\sqrt{3}} \pm \frac{1}{2\sqrt{7}}. \]

Step 4: Final Answer:
One focal distance is $\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$.