Question

The latus rectum of the conic 3x2 + 4y2 - 6x + 8y - 5 = 0 is
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Answer 1

Answer:

6+8y-6x+8y-5=16y-1-6x=0

Answer 2

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

$\green{\tt{\therefore{Latus\:rectum(LL')=\frac{3}{2}}}}$

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

$\green {\underline \bold{Given : }} \\ \tt{ : \implies Eqn \: of \: conic =3{x}^{2}+ {4y}^{2} -6x+8y-5=0} \\ \\ \red {\underline \bold{To \: Find: }} \\ \tt {: \implies Length \: of \: latus \: rectum (LL')=?}$

• According to given question :

$\tt{: \implies 3 {x}^{2} + 4 {y}^{2} - 6x + 8y - 5 = 0} \\ \\ \tt{: \implies 3 {x}^{2} - 6x +3 - 3 + 4 {y}^{2} + 8y + 4 - 4 - 5 = 0} \\ \\ \tt{: \implies (\sqrt{3} x - \sqrt{3})^{2} + {(2y + 2)}^{2} = 12} \\ \\ \tt{: \implies \frac{(x - 1)^{2}}{ \frac{12}{3}} + \frac{ {(y + 1)}^{2} }{ \frac{12}{4} } = 1} \\ \\ \tt{: \implies \frac{(x - 1)^{2} }{4} + \frac{ {(y + 1)}^{2} }{3} = 1}$

$\text{So, \: it \: is \: in \: the \: form \: of} \\ \tt{\to \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} \\ \\ \bold{Where : } \\ \tt{\circ \: {a}^{2} = 4} \\ \\ \tt{\circ \: {b}^{2} = 3} \\ \\ \bold{As \: we \: know \: that} \\ \tt{ : \implies Latus \: rectum = \frac{2 {b}^{2} }{a} } \\ \\ \text{Putting \: given \: values} \\ \tt{ : \implies Latus \: rectum = \frac{2 \times 3}{4} } \\ \\ \green{\tt{ : \implies Latus \: rectum = \frac{3 }{2} }}$