Question

For which of the following curves, the line x + √3 y = 2√3 is the tangent at the point (3√3 / 2, 1/2) ?

• A. x² + y² = 7
• B. y² = (1 / 6√3) x
• C. 2x² - 18y² = 9
• D. x² + 9y² = 9

Hint:

The tangent to any second-degree curve $ax^2+2hxy+by^2+2gx+2fy+c=0$ at $(x_1, y_1)$ can be written by replacing $x^2 \to xx_1$, $y^2 \to yy_1$, and $x \to \frac{x+x_1}{2}$.

Answer 1

The Correct Option is D

Step 1: Check if point $P( \frac{3\sqrt{3}}{2}, \frac{1}{2})$ lies on the curve in (D): $(\frac{3\sqrt{3}}{2})^2 + 9(\frac{1}{2})^2 = \frac{27}{4} + \frac{9}{4} = \frac{36}{4} = 9$. Yes.
Step 2: Find tangent to $x^2 + 9y^2 = 9$ at $(x_1, y_1)$ using $xx_1 + 9yy_1 = 9$.
Step 3: $x(\frac{3\sqrt{3}}{2}) + 9y(\frac{1}{2}) = 9$.
Step 4: Multiply by $2$: $3\sqrt{3}x + 9y = 18$.
Step 5: Divide by $3\sqrt{3}$: $x + \frac{3}{\sqrt{3}}y = \frac{6}{\sqrt{3}} \Rightarrow x + \sqrt{3}y = 2\sqrt{3}$. This matches the given line.