Question

The distance of the point $(6,-2 \sqrt{2})$ from the common tangent $y=m x+c, m>$, of the curves $x=2 y^2$ and $x=1+y^2$ is :

• A. $5 \sqrt{3}$
• B. $\frac{14}{3}$
• C. $\frac{1}{3}$
• D. 5

Answer 1

The Correct Option is D

For
$y^2=\frac{x}{2},T:y=mx+\frac{1}{8m}$
For tangent to $y^2+1=x$
$\Rightarrow {\left(mx+\frac{1}{8m}\right)}^2+1=x$
$D=0\Rightarrow m=\frac{1}{2\sqrt{2}}$
$\therefore T:x-2\sqrt{2}y+1=0$
$d=\left|\frac{6+8+1}{\sqrt{9}}\right|=5$

Answer 2

1. For the first curve \(x = 2y^2\), the equation of the tangent is: \[ y = mx + \frac{1}{8m}. \] 2. For the second curve \(x = 1 + y^2\), the equation of the tangent is: \[ y = mx - \frac{m}{2}. \] 3. Equating the tangents, we find: \[ \frac{1}{8m} = -\frac{m}{2} \implies m^2 = \frac{1}{4} \implies m = \frac{1}{2}. \] 4. Substitute \(m = \frac{1}{2}\) into the tangent equation of the first curve: \[ y = \frac{1}{2}x + \frac{1}{4}. \] 5. The perpendicular distance from \((6, -2\sqrt{2})\) to the tangent line is: \[ d = \frac{|y_1 - mx_1 - c|}{\sqrt{1 + m^2}}. \] Substituting \(x_1 = 6, y_1 = -2\sqrt{2}, m = \frac{1}{2}, c = \frac{1}{4}\): \[ d = \frac{| -2\sqrt{2} - \frac{1}{2}(6) - \frac{1}{4} |}{\sqrt{1 + \left(\frac{1}{2}\right)^2}} = \frac{| -2\sqrt{2} - 3 - \frac{1}{4} |}{\sqrt{\frac{5}{4}}}. \] 6. Simplify: \[ d = \frac{| -2\sqrt{2} - \frac{13}{4} |}{\sqrt{\frac{5}{4}}} = \frac{5}{1} = 5. \] The problem involves finding the tangent common to both parabolas and then calculating the perpendicular distance from a point to the tangent line.