Let a focal chord $12 x+5 y-27=0$ of the parabola $\mathrm{y}^2=\mathrm{kx}$ intersect the parabola at the…
Let a focal chord $12 x+5 y-27=0$ of the parabola $\mathrm{y}^2=\mathrm{kx}$ intersect the parabola at the points $\mathrm{P}$ and $\mathrm{P}^{\prime}$. If $\mathrm{S}$ is the focus of this parabola, then $9\left(\mathrm{SP}+\mathrm{SP}^1\right)=$
$27$
$108$
16 SP.SP $^{\prime}$
$4 \mathrm{SP}. \mathrm{SP}^{\prime}$
Solution
Given that equation of parabola $y^2=\mathrm{km}$ then focus is $\left(\frac{k}{4}, 0\right)$ which is lies on focal chord. $\therefore 12 \times \frac{k}{4}+0-27=0 \Rightarrow k=9$, focus s $\left(\frac{9}{4}, 0\right)$
On solving the equation $y^2=9 x$ and $12 x+5 y-27=0$ We get $P(1,3)$ and $P^{\prime}\left(\frac{81}{16}, \frac{-27}{4}\right)$.
$\begin{aligned}
& \therefore S P=\sqrt{\left(\frac{9}{4}-1\right)^2+3^2}=\frac{13}{4} \\
& S P^{\prime}=\sqrt{\left(\frac{81}{16}-\frac{9}{4}\right)^2+\left(\frac{27}{4}\right)^2}=\frac{117}{16}
\end{aligned}$
Now, $g\left(S P+S P^{\prime}\right)=\frac{1521}{1}=4 S P . S P^{\prime}$.