The distance of the point (6, - 2 2 ) from the common tangent y =...

Given,

Equation of the curves $x = 2{y}^{2} and x = 1 + {y}^{2}$

Now for parabola ${y}^{2} = \dfrac{x}{2}$, equation of tangent is given by $y = mx + \dfrac{1}{8m}$

Which is also tangent to parabola ${y}^{2} + 1 = x$,

So, $y = mx + \dfrac{1}{8m}$ will satisfy the curve ${y}^{2} + 1 = x$,

Hence, ${\left(mx + \dfrac{1}{8m}\right)}^{2} + 1 = x$

$⇒ \left({m}^{2}{x}^{2} + \dfrac{1}{64{m}^{2}} + \dfrac{x}{4}\right) + 1 = x$

$⇒ {m}^{2}{x}^{2} - \dfrac{3x}{4} + 1 + \dfrac{1}{64{m}^{2}} = 0$

Now making discriminant zero we get,

$D = 0 ⇒ {\left(\dfrac{3}{4}\right)}^{2} - 4 \times {m}^{2} \times \left(1 + \dfrac{1}{64{m}^{2}}\right)$

$⇒ \left(\dfrac{9}{16}\right) - \left(4{m}^{2} + \dfrac{1}{16}\right) = 0$

$⇒ m = \dfrac{1}{2\sqrt{2}}$

Hence, the equation of tangent is $x - 2\sqrt{2}y + 1 = 0$

Now distance of point \left(6, - 2\sqrt{2}\right)$from the tangent will be,$d = \left|\dfrac{6 + 8 + 1}{\sqrt{9}}\right| = 5$