Find the point on the curve y^{2}= 4x,which is nearest to the point (2,-8).

 

 

 

 
 
 
 
 

Answers (1)

given the curve of the form y^{2}= 4x,which is a means to the point \left ( 2,-8 \right ) and let p\left ( x,y \right ) be the point on the curve.
             \because y^{2}= 4x
             then \because \frac{y^{2}}{4}= x
\therefore p\left ( x,y \right ) will \: be\: p\left ( \frac{y^{2}}{4},y \right )
Now the distance between point A & P is given by
AP= \sqrt{\left ( x-2 \right )^{2}+\left ( y+8 \right )^{2}}
         AP= \sqrt{\left ( \frac{y^{2}}{4} -2\right )^{2}+\left ( y+8 \right )^{2}}
\Rightarrow \sqrt{\frac{y^{2}}{16}-y^{2}+4+y^{2}+16y+64}\Rightarrow \sqrt{\frac{y^{4}}{16}+16y+68}and Let Zz= \frac{y^{4}}{16}+16y+68-(i)
\frac{dz}{dy}= \frac{1}{16}\times 4y^{3}+16
= \frac{y^{3}}{4}+16
For the maximum or minimum value of Z, we have
                          \frac{dz}{dy}= 0
\Rightarrow \frac{y^{3}}{4}+16= 0 \: \Rightarrow y^{3}+64= 0
\Rightarrow \left ( y+4 \right )\left ( y^{2} -4y+16\right )= 0 \left ( \because y^{2}-4y+16= 0\: gives\:imaginary\: value \: of\: y \right )\Rightarrow y= -4
Now,\frac{d^{2}z}{dy^{2}}= \frac{1}{4}\times 3y^{2}= \frac{3}{4}y^{2}
At y= -4
\frac{d^{2}Z}{dy^{2}}= \frac{3}{4}\left ( -4 \right )^{2}= 12> 0
Thus, z is maximum when y= -4
substituting  y= -4  in the equation of the curve y^{2}= 4x
we have x= 4
Hence, the point(4,-4) on the curve y^{2}= 4x is nearest to the point(12,-8)
 

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