Question

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

• A. $1$
• B. $4$
• C. $3$
• D. $2$

Answer 1

The correct option is B $4$

REF.Image

$\frac{x^2}{6}+\frac{y^2}{3}=1$ ___ (i)

$x+y=7$ ___ (ii)

Let any point $P(\sqrt{6}cos\theta ,\sqrt{3}sin\theta )$

Equation of tangent at 'P'

$\frac{x\times \sqrt{6}cos\theta }{6}+\frac{y\times \sqrt{3}sin\theta }{3}=1$ ___ (iii)

$E{q}^n$ (iii) & (ii) are parallel

$\therefore -\frac{\frac{cos\theta }{\sqrt{6}}}{\frac{sin\theta }{\sqrt{3}}}=-1$

$\Rightarrow \frac{cos\theta }{\sqrt{6}}\times \frac{\sqrt{3}}{sin\theta }=1$

$\Rightarrow tan\theta =\frac{1}{\sqrt{2}}$

$\therefore P(2,1)$

$E{q}^n$ of normal PQ

$y-1=1(x-2)$

$y-1=1(x-2)$

$x-y=1$ ___ (iv)

$x+y=7$ ___ (ii)

$\therefore Q(4,3)$

$\therefore$ minimum diatance = $PQ=\sqrt{4+4}=2\sqrt{2}$

$=\frac{K}{\sqrt{2}}=2\sqrt{2}$

$\therefore K=4$