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:
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:
Show 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.
-9
3
9
-3
The Correct Option is A
Solution and Explanation
The hyperbola is given as:
\[3x^2 - 4y^2 = 36.\]
The line equation is:
\[3x + 2y = 1.\]
Slope of the line (\(m\)) is:
\[m = -\frac{3}{2}.\]
To find the nearest point, the slope of the perpendicular from the hyperbola is given by:
\[m = \pm \frac{\sec \theta \cdot 3}{\sqrt{12} \cdot \tan \theta}.\]
Equating the slopes:
\[\frac{3}{\sqrt{12}} \times \frac{1}{\sin \theta} = -\frac{3}{2}.\]
Solving for \(\sin \theta\):
\[\sin \theta = -\frac{1}{\sqrt{3}}.\]
The corresponding point on the hyperbola is:
\[\left(\sqrt{12} \cdot \sec \theta, 3 \cdot \tan \theta\right).\]
Simplify:
\[\left(\sqrt{12} \cdot \frac{\sqrt{3}}{2}, -3 \cdot \frac{1}{\sqrt{2}}\right) \implies \left(\frac{6}{\sqrt{2}}, -\frac{3}{\sqrt{2}}\right).\]
The value of \(\sqrt{2}(y_0 - x_0)\) is:
\[\sqrt{2} \left(-\frac{3}{\sqrt{2}} - \frac{6}{\sqrt{2}}\right) = \sqrt{2} \cdot -\frac{9}{\sqrt{2}} = -9.\]
Conclusion: The value of \(\sqrt{2}(y_0 - x_0)\) is \(-9\).
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Concepts Used:
Hyperbola
Hyperbola is the locus of all the points in a plane such that the difference in their distances from two fixed points in the plane is constant.
Hyperbola is made up of two similar curves that resemble a parabola. Hyperbola has two fixed points which can be shown in the picture, are known as foci or focus. When we join the foci or focus using a line segment then its midpoint gives us centre. Hence, this line segment is known as the transverse axis.
