Question Number 57578 by rahul 19 last updated on 07/Apr/19
$${Minimum}\:{distance}\:{between}\:{curves} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{x}\:{and}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{31}=\mathrm{0}\:{is}\:? \\ $$
Answered by mr W last updated on 07/Apr/19
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{31}=\mathrm{0} \\ $$$$\left({x}−\mathrm{6}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{circle}\:{with}\:{radius}\:\sqrt{\mathrm{5}}\:{and}\:{center}\:\left(\mathrm{6},\mathrm{0}\right). \\ $$$$ \\ $$$${eqn}.\:{of}\:{circle}\:{with}\:{radius}\:{r}\:{and}\:{center}\:\left(\mathrm{6},\mathrm{0}\right): \\ $$$$\left({x}−\mathrm{6}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${such}\:{that}\:{the}\:{circle}\:{touches}\:{the}\:{curve} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{x} \\ $$$$\Rightarrow\left({x}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{4}{x}={r}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{8}{x}+\left(\mathrm{36}−{r}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\Delta=\mathrm{64}−\mathrm{4}\left(\mathrm{36}−{r}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{r}=\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$ \\ $$$${Minimum}\:{distance}\:{between}\:{curves} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{x}\:{and}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{31}=\mathrm{0}\:{is}\: \\ $$$$\mathrm{2}\sqrt{\mathrm{5}}−\sqrt{\mathrm{5}}=\sqrt{\mathrm{5}} \\ $$
Commented by mr W last updated on 07/Apr/19
Commented by rahul 19 last updated on 08/Apr/19
thank you sir.
Answered by MJS last updated on 07/Apr/19
![the center of the circle is ((6),(0) ); r=(√5) ⇒ the point of the parabola with minimum distance from this center is the same as the point with the minimum distance of the circle. ∣ [(x),((2(√x))) ]− [(6),(0) ]∣^2 =x^2 −8x+36 (d/dx)[x^2 −8x+36]=2x−8=0 ⇒ x=4 distance=(√(x^2 −8x+36))−(√5)=(√(20))−(√5)=(√5)](Q57582.png)
$$\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{is}\:\begin{pmatrix}{\mathrm{6}}\\{\mathrm{0}}\end{pmatrix};\:{r}=\sqrt{\mathrm{5}} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{the}\:\mathrm{parabola}\:\mathrm{with}\:\mathrm{minimum} \\ $$$$\mathrm{distance}\:\mathrm{from}\:\mathrm{this}\:\mathrm{center}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{the} \\ $$$$\mathrm{point}\:\mathrm{with}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{circle}. \\ $$$$\mid\begin{bmatrix}{{x}}\\{\mathrm{2}\sqrt{{x}}}\end{bmatrix}−\begin{bmatrix}{\mathrm{6}}\\{\mathrm{0}}\end{bmatrix}\mid^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{36} \\ $$$$\frac{{d}}{{dx}}\left[{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{36}\right]=\mathrm{2}{x}−\mathrm{8}=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{4} \\ $$$$\mathrm{distance}=\sqrt{{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{36}}−\sqrt{\mathrm{5}}=\sqrt{\mathrm{20}}−\sqrt{\mathrm{5}}=\sqrt{\mathrm{5}} \\ $$
Commented by rahul 19 last updated on 08/Apr/19
thank you sir.