Question

If the length of latusrectum of the parabola $5y^2-3x-8y-1=0$ is $L$ then $5L=$

Answer 1

Answer: 3

The given parabola is $5y^2-3x-8y-1=0$
$\Rightarrow 5y^2-8y=3x+1$
$\Rightarrow y^2-\frac{8}{5}y=\frac{3}{5}x+\frac{1}{5}$
$\Rightarrow y^2-\frac{8}{5}y+\frac{16}{25}=\frac{3}{5}x+\frac{1}{5}+\frac{16}{25}$
$\Rightarrow {\left(y-\frac{4}{5}\right)}^2=\frac{3}{5}x+\frac{21}{25}$
$\Rightarrow {\left(y-\frac{4}{5}\right)}^2=\frac{3}{5}\left(x+\frac{7}{5}\right)$
The above equation is of the form ${\left(y-k\right)}^2=4a\left(x-h\right)$
Length of latusrectum is $4a=\frac{3}{5}$
Hence, $L=\frac{3}{5}\Rightarrow 5L=3$