Let P(x0,y0) be the point on the hyperbola 3x2−4y2=36 which is nearest to the line 3x+2y=1. Then 2(y0−x0) is equal to:
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Official solution
The hyperbola is 12x2−9y2=1, so P=(12secθ,3tanθ). The tangent slope there is asinθb=12sinθ3; the nearest point has tangent parallel to 3x+2y=1 (slope −3/2):
\[\frac3{\sqrt{12}\sin\theta}=-\frac32\ \Longrightarrow\ \sin\theta=-\frac1{\sqrt{3}}\ \Longrightarrow\ \cos\theta=\sqrt{\frac23}.\]
\[x_0=\sqrt{12}\sec\theta=\frac6{\sqrt{2}},\qquad y_0=3\tan\theta=-\frac3{\sqrt{2}}.\]
\[\sqrt{2}(y_0-x_0)=\sqrt{2}\left(-\frac3{\sqrt{2}}-\frac6{\sqrt{2}}\right)=\sqrt{2}\left(-\frac9{\sqrt{2}}\right)=-9.\]