%%

The distance of the point $$(6,-2\sqrt2)$$ from the common tangent $$\mathrm{y=mx+c,m > 0}$$, of the curves $$x=2y^2$$ and $$x=1+y^2$$ is :

JEE · Math · previous-year question

  1. A.$$\frac{1}{3}$$
  2. B.5correct
  3. C.$$\frac{14}{3}$$
  4. D.5$$\sqrt3$$

Answer

B. 5

Explanation

$$ \begin{aligned} & y^2=\frac{x}{2} \Rightarrow \text { tangent } y=m x+\frac{1}{8 m} \\\\ & y^2=x-1 \Rightarrow \text { tangent } y=m(x-1)+\frac{1}{4 m} \\\\ & \text { For common tangent } \frac{1}{8 m}=-m+\frac{1}{4 m} \\\\ & \Rightarrow 1=-8 m^2+2 \\\\ & \because m>0 \Rightarrow m=\frac{1}{2 \sqrt{2}} \\\\ & \Rightarrow \text { Common tangent is } y=\frac{x}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}} \\\\ & \Rightarrow x-2 \sqrt{2} y+1=0 \end{aligned} $$ Distance of point $(6,-2 \sqrt{2})$ from common tangent $=5$

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions