Question

If a hyperbola passes through the point P(\(\sqrt{2},\sqrt{3}\)) and has foci at (±2,0), then the tangent to this hyperbola at P is

• A. y=x\(\sqrt3-\sqrt6\)
• B. \(y = \sqrt{6}x - \sqrt{3}\)
• C. y=x\(\sqrt6+\sqrt3\)
• D. y=x\(\sqrt3+\sqrt6\)

Answer 1

The Correct Option is B

Step 1: Standard form of the hyperbola
We are told the hyperbola is centered at the origin and has foci at \((\pm 2, 0)\), so it is of the form:

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

Step 2: Use the foci to relate eccentricity and a 
Foci are at \((\pm ae, 0)\), and we are given \(ae = 2\) ⟹ \(e = \frac{2}{a}\)

Also, we know: \[ b^2 = a^2(e^2 - 1) = a^2\left(\left(\frac{2}{a}\right)^2 - 1\right) = a^2\left(\frac{4}{a^2} - 1\right) = 4 - a^2 \]

Step 3: Use the fact that point \(P(\sqrt{2}, \sqrt{3})\) lies on the hyperbola
Substitute into the hyperbola equation: \[ \frac{2}{a^2} - \frac{3}{b^2} = 1 \]
Substitute \(b^2 = 4 - a^2\) into the equation: \[ \frac{2}{a^2} - \frac{3}{4 - a^2} = 1 \]
Multiply both sides by \(a^2(4 - a^2)\):
\[ 2(4 - a^2) - 3a^2 = a^2(4 - a^2) \]
\[ 8 - 2a^2 - 3a^2 = 4a^2 - a^4 \]
\[ 8 - 5a^2 = 4a^2 - a^4 \Rightarrow a^4 - 9a^2 + 8 = 0 \]

Step 4: Solve the quadratic in terms of \(z = a^2\)
\[ z^2 - 9z + 8 = 0 \Rightarrow (z - 8)(z - 1) = 0 \Rightarrow z = 8 \text{ or } 1 \]
Try \(a^2 = 1\):
\[ b^2 = 4 - 1 = 3 \Rightarrow \text{Check if } \frac{2}{1} - \frac{3}{3} = 2 - 1 = 1 ✓ \Rightarrow \text{Valid} \]

Step 5: Equation of tangent at point \((x_0, y_0) = (\sqrt{2}, \sqrt{3})\)
Tangent to a hyperbola at \((x_0, y_0)\) is:
\[ \frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1 \]
\[ \frac{x \cdot \sqrt{2}}{1} - \frac{y \cdot \sqrt{3}}{3} = 1 \Rightarrow \sqrt{2}x - \frac{\sqrt{3}}{3}y = 1 \]
Multiply through by 3 to eliminate the denominator: \[ 3\sqrt{2}x - \sqrt{3}y = 3 \Rightarrow \boxed{y = \sqrt{6}x - \sqrt{3}} \]

Final Answer: The equation of the tangent is: \[ \boxed{y = \sqrt{6}x - \sqrt{3}} \]