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 :
Solution
$$
\begin{aligned}
& y^2=\frac{x}{2} \Rightarrow \text { tangent } y=m x+\frac{1}{8 m} \\\\
& y^2=x-1 \Rightarrow \text { tangent } y=m(x-1)+\frac{1}{4 m} \\\\
& \text { For common tangent } \frac{1}{8 m}=-m+\frac{1}{4 m} \\\\
& \Rightarrow 1=-8 m^2+2 \\\\
& \because m>0 \Rightarrow m=\frac{1}{2 \sqrt{2}} \\\\
& \Rightarrow \text { Common tangent is } y=\frac{x}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}} \\\\
& \Rightarrow x-2 \sqrt{2} y+1=0
\end{aligned}
$$<br/><br/>
Distance of point $(6,-2 \sqrt{2})$ from common tangent $=5$
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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