Math, asked by hiralgarg7, 2 months ago

find the focus, directrix and length of latus rectum of parabola:y²=2√3x​

Answers

Answered by pulakmath007
20

SOLUTION

TO DETERMINE

The focus, directrix and length of latus rectum of parabola : y² = 2√3x

EVALUATION

Here the given equation of the parabola

 \sf{ {y}^{2}  = 2 \sqrt{3}x }

Comparing with y² = 4ax we get

 \displaystyle \sf{4a = 2 \sqrt{3} }

 \displaystyle \sf{ \implies \: a =  \frac{ \sqrt{3} }{2} }

Hence the required coordinates of focus

 \displaystyle \sf{  = ( a   \:, \: 0)}

 \displaystyle \sf{  =   \bigg( \frac{ \sqrt{3} }{2} \: , \: 0 \bigg) }

The required equation of directrix

 \displaystyle \sf{  x +  \frac{ \sqrt{3} }{2} \:  =  \: 0 }

Length of the latus rectum

= 4a unit

 =  \sf{2 \sqrt{3}  \:  \:  \: unit}

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