Question

The line $x-y+4=0$ touches the ellipse $x^{2}+3y^{2}=12$ at ________.

• A. $(1, 3)$
• B. $(3, 1)$
• C. $(0, 2)$
• D. $(0, -2)$
• E. $(-3, 1)$

Hint:

Verify by checking if the point satisfies both the line and the ellipse equations.

Answer 1

Answer: (E)

Step 1: Concept
A point $(x_1, y_1)$ on the ellipse touches the tangent line.

Step 2: Meaning
Equation of ellipse: $\frac{x^2}{12} + \frac{y^2}{4} = 1$. Tangent at $(x_1, y_1)$ is $\frac{xx_1}{12} + \frac{yy_1}{4} = 1$.

Step 3: Analysis
Comparing $\frac{x_1}{12}x + \frac{y_1}{4}y = 1$ with the given line $x - y = -4$ (which is $-\frac{1}{4}x + \frac{1}{4}y = 1$).
$\frac{x_1}{12} = -\frac{1}{4} \implies x_1 = -3$.
$\frac{y_1}{4} = \frac{1}{4} \implies y_1 = 1$.

Step 4: Conclusion
The point of tangency is $(-3, 1)$. Final Answer: (E)