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# Engineering

If the line ax+y=c , touches both the curves x^{2}+y^{2}=1

and y^{2}=4\sqrt2x , then |c| is equal to :

  • Option 1)

    2

  • Option 2)

    \frac{1}{\sqrt2}

  • Option 3)

    \frac{1}{2}

  • Option 4)

    \sqrt2

 

Answers (1)
V Vakul

line ax+y=c

curve eqns x^{2}+y^{2}=1 & y^{2}=4\sqrt2x

A tangent to the parabola 

y^{2}=4\sqrt2x  can be taken as (y^{2}=4ax)

y=mx+\frac{\sqrt2}{m}   ....................(1)   (y=mx+\frac{a}{m})

 

It will also touch the given circle x^{2}+y^{2}=1

if (\frac{\sqrt2}{m})^{2}=1.(m^{2}+1)

=> m^{2}(m^{2}+1)=2                                   (eqn c^{2}=r^{2}(m^{2}+1) of line touching the circle)

=>m^{4}+m^{2}-2=0

=>m=\pm 1

\therefore c=\sqrt2

So, option (4) is correct.

 


Option 1)

2

Option 2)

\frac{1}{\sqrt2}

Option 3)

\frac{1}{2}

Option 4)

\sqrt2

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