Question

length of latus rectum of parabola y^2-8y-4x=0 iis

Answer 1

Answer:

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Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Latus\:rectum(LL')=4\:units}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green {\underline \bold{Given : }} \\ \tt{ : \implies Eqn \: of \: parabola = {y}^{2} -8y-4x=0} \\ \\ \red {\underline \bold{to \: find: }} \\ \tt {: \implies Length \: of \: latus \: rectum (LL')=?}$

• According to given question :

$\tt{: \implies {y}^{2} - 8y - 4x = 0} \\ \\ \tt{: \implies {y}^{2} - 8y + 16 - 16- 4x = 0} \\ \\ \tt {: \implies {(y - 4)}^{2} = 4x + 16 } \\ \\ \tt{: \implies {(y - 4)}^{2} = 4(x + 4)} \\ \\ \text{So, \: it \: is \: in \: the \: form \: of} \\ \tt{: \implies Y^{2} = 4aX} \\ \\ \bold{where :} \\ \tt{\circ \: a = 1} \\ \\ \bold{As \: we \: know \: that} \\ \tt{: \implies Latus \: rectum = 4a} \\ \\ \green{\tt{: \implies Latus \: rectum = 4 \: units}}$