Question

If the least distance between a point on $x^2+2y^2=6$ and $x+y-7=0$ is $\frac{k}{\sqrt{2}}$ unit, then $k=$

Answer 1

Equation of tangent to the ellipse parallel to $x+y-7=0$ will be
$y=-x\pm \sqrt{6(-1{)}^2+3}\Rightarrow x+y\mp 3=0$
distance between the tangent and the given line is
$d=\frac{|\mp 3+7|}{\sqrt{1^2+1^2}}=\frac{4}{\sqrt{2}},\frac{10}{\sqrt{2}}$
Hence minimum distance will be
$=\frac{4}{\sqrt{2}}$ unit