Question

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

Answer 1

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, $L=\frac{3}{5}\Rightarrow 5L=3$