Parabola Question 3
Question 3 - 25 January - Shift 1
The distance of the point $(6,-2 \sqrt{2})$ from the common tangent $y=mx+c, m>0$, of the curves $x=2 y^{2}$ and $x=1+y^{2}$ is
(1) $\frac{1}{3}$
(2) 5
(3) $\frac{14}{3}$
(4) $5 \sqrt{3}$
Show Answer
Answer: (2)
Solution:
Formula: Equation of the Tangent at any Point, Distance between point and line
For
$y^{2}=\frac{x}{2}, T: y=mx+\frac{1}{8 m}$
For tangent to $y^{2}+1=x$
$\Rightarrow(mx+\frac{1}{8 m})^{2}+1=x$
$D=0 \Rightarrow m=\frac{1}{2 \sqrt{2}}$
$\therefore T: x-2 \sqrt{2} y+1=0$
$d=|\frac{6+8+1}{\sqrt{9}}|=5$
