Question

Let \(P(x_0, y_0)\) be the point on the hyperbola \(3x^2 - 4y^2 = 36\), which is nearest to the line \(3x + 2y = 1\). Then \(\sqrt{2}(y_0 - x_0)\) is equal to:

• A. -9
• B. 3
• C. 9
• D. -3

Hint:

To find the nearest point on a hyperbola to a line, solve the equations by ensuring the slopes match, or apply Lagrange multipliers for optimization.

Answer 1

The Correct Option is A

To find the point \(P(x_0, y_0)\) on the hyperbola \(3x^2 - 4y^2 = 36\) nearest to the line \(3x + 2y = 1\), we first need to understand the problem geometrically.

 

1. \(3x^2 - 4y^2 = 36\) represents the hyperbola. We can rewrite it in the standard form as \(\frac{x^2}{12} - \frac{y^2}{9} = 1\). This is the equation of the hyperbola centered at the origin, with transverse and conjugate axes along the x-axis and y-axis, respectively.
2. The line \(3x + 2y = 1\) can be rewritten in slope-intercept form as \(y = -\frac{3}{2}x + \frac{1}{2}\).
3. The distance \(d\) from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\) is given by: \(d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\). For the given line, \(A = 3\), \(B = 2\), and \(C = -1\).
4. We need to minimize the distance \(d\), which is equivalent to minimizing \(|3x_0 + 2y_0 - 1|\).
5. To find the point of tangency, use Lagrange multipliers by setting up the equations: \[ \begin{align*} \nabla f &= \lambda \nabla g \\ (6x, -8y) &= \lambda (3, 2) \end{align*} \] This gives two equations: \[ \begin{align*} 6x &= 3\lambda \\ -8y &= 2\lambda \end{align*} \] Solve to find \(\lambda\). From these, \(x = \frac{\lambda}{2}\) and \(y = -\frac{\lambda}{4}\).
6. Substitute these expressions of \(x\) and \(y\) into the hyperbola equation: \[ 3\left(\frac{\lambda}{2}\right)^2 - 4\left(-\frac{\lambda}{4}\right)^2 = 36 \] Solving for \(\lambda\): \[ \frac{3\lambda^2}{4} - \frac{\lambda^2}{4}= 36 \implies \frac{2\lambda^2}{4} = 36 \implies \lambda^2 = 72 \implies \lambda = 6\sqrt{2} \]
7. Subsequently, \(x_0 = \frac{6\sqrt{2}}{2} = 3\sqrt{2}\) and \(y_0 = -\frac{6\sqrt{2}}{4} = -\frac{3\sqrt{2}}{2}\).
8. Find \(\sqrt{2}(y_0 - x_0)\): \[ \sqrt{2}\left(-\frac{3\sqrt{2}}{2} - 3\sqrt{2} \right) = \sqrt{2}\left(-\frac{9\sqrt{2}}{2} \right) = -9 \]
9. Therefore, the value is \(-9\).

Thus, \(\sqrt{2}(y_0 - x_0) = -9\), making the correct answer \(-9\).