Question:

<p>The ellipse is \(x^2 + 2y^2 = 6\) or \(\frac{x^2}{6} + \frac{y^2}{3} = 1\)</p><p>We need to find the point on the ellipse closest to the line \(x + y = 7\).</p><p>The point closest to the line will lie on a normal to the ellipse that is parallel to the normal of the line \(x + y = 7\).</p><p>The normal to the line is in direction \((1, 1)\).</p><p>For a point \((x, y)\) on the ellipse, the gradient is \(\nabla(x^2 + 2y^2) = (2x, 4y)\)</p><p>For this to be parallel to \((1, 1)\): \(\frac{2x}{1} = \frac{4y}{1} \Rightarrow x = 2y\)</p><p>Substituting into the ellipse equation: \((2y)^2 + 2y^2 = 6 \Rightarrow 4y^2 + 2y^2 = 6 \Rightarrow 6y^2 = 6 \Rightarrow y = 1\)</p><p>So \(x = 2\), giving the point \((2, 1)\)</p><p>∴ Answer is (b).</p>