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The shortest distance between the parabolas $\mathrm{y^2=4 x\: and \: y^2-2 x+6=0}$ is

Option: 1
$\sqrt{2}$

Option: 2
$\sqrt{5}$

Option: 3
$\sqrt{3}$

Option: 4
$\text{None of these}$

Answers (1)

best_answer

Shortest distance between two curves exist along the common normal, we need the equation of normals to both the curves.
Now equation of normal to the curve $\mathrm{y^2=4 x \: at \: \left(t^2, 2 t\right)}$ is taken as

$\mathrm{y-y_1=-\left(\frac{1}{\frac{d y}{d x}}\right)\left(x-x_1\right) \text { at }(x, y)=\left(t^2, 2 t\right) }$

$\mathrm{\therefore \quad(y-2 t) =-t\left(x-t^2\right) }$

$\mathrm{\Rightarrow y+t x-2 t-t^3=0}$

Similarly normal to $\mathrm{ y^2-2 x+6=0 at \left(\frac{1}{2} t^2+3, t\right)}$ is

$\mathrm{ y+t(x-3)-t-\frac{1}{2} t^3=0 }$

Equating (i) and (ii), we get

$\mathrm{ \quad-2 t-t^3=-4 t-\frac{1}{2} t^3 \Rightarrow 2 t=t^3-\frac{1}{2} t^3 \Rightarrow 2 t=\frac{1}{2} t^3 }$

$\mathrm{ \Rightarrow t\left(2-\frac{1}{2} t^2\right)=0 \Rightarrow t=0, t= \pm 2 }$

$\mathrm{ \therefore \quad \text { Points on the parabolas are }\left(t^2, 2 t\right)=(4,4) }$

$\mathrm{ \text { and }\left(\frac{1}{2} t^2+3, t\right)=(5,2) }$

$\mathrm{ \therefore \quad \text{Points on the parabolas are} \left(t^2, 2 t\right)=(4,4) }$

$\mathrm{ and \left(\frac{1}{2} t^2+3, t\right)=(5,2) }$

$\mathrm{ \therefore \text{Shortest distance} =\sqrt{(5-4)^2+(4-2)^2}=\sqrt{5} }$

Hence option 2 is correct.

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