Let the length of the focal chord PQ of the parabola y^2=12 x be 15 units.

Question

Let the length of the focal chord PQ of the parabola $y^2=12 x$ be 15 units. If the distance of $$\mathrm{PQ}$$ from the origin is $$\mathrm{p}$$, then $$10 \mathrm{p}^2$$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 72

Accepted Answer

$\begin{gathered} AB=15\Rightarrow (3({t}^{2}-\frac{1}{{t}^{2}}))+{(6(t+\frac{1}{t}))}^{2}=225 \ \Rightarrow 9({t}^{2}-\frac{1}{{t}^{2}})+36{(t+\frac{1}{t})}^{2}=225 \end{gathered}$

$\begin{gathered} \Rightarrow {(t+\frac{1}{t})}^{2}[\, (t-\frac{1}{t})+4]=25 \ {(t+\frac{1}{t})}^{2}{(t+\frac{1}{t})}^{2}=25\Rightarrow {(t+\frac{1}{t})}^{4}=25 \ \Rightarrow t+\frac{1}{t}=\pm \sqrt{5}\Rightarrow (t-\frac{1}{t})=\pm 1 \end{gathered}$

Equation of $AB:(y-6t)=(\frac{2t}{{t}^{2}-1})(x-3{t}^{2})$

$\Rightarrow$ Distance from $y-6t=mx-3m{t}^{2}$

$\Rightarrow p=\frac{|3m{t}^{2}-6t|}{\sqrt{1+{m}^{2}}}=\frac{|(\frac{6t}{{t}^{2}-1})|}{\sqrt{5}}=\frac{6}{\sqrt{5}}$

$[m=\frac{2t}{{t}^{2}-1}=\frac{}{t-\frac{1}{t}}=\pm 2\Rightarrow {m}^{2}=4]$

$\Rightarrow \, 10{p}^{2}=\frac{10	imes 36}{5}\Rightarrow 72$

Solutions · 1