Explanation:
B Given, equation of normal \(\mathrm{mx}-\mathrm{y}+\mathrm{c}=0\) Parabola \(\mathrm{y}^2=16 \mathrm{x}\) Comparing with general equation of parabola i.e.,
\(y^2 =4 a x\)
\(4 a =16\)
\(a =\frac{16}{4}=4\)
Let the coordinates of point \(P\) on parabola be \(\left(4 t^2, 8 t\right)\)
\(\because\) Slope of tangent at point \(\mathrm{P}=\frac{1}{\mathrm{t}}\)
Also, slope of tangent \(=-\frac{1}{\text { Slope of normal }}=-\frac{1}{\mathrm{~m}}\)
From equation (i) and (ii),
\(\frac{1}{t} =-\frac{1}{m}\)
\(t =-m\)
\(\therefore\) Coordinate of point is \(\mathrm{P}\left(4 \mathrm{~m}^2,-8 \mathrm{~m}\right)\).
Focus of parabola is \((a, 0)\) or \((4,0)\)
Given, focal distance \(=40\)
Using distance formula,
\(\sqrt{\left(4 m^2-4\right)^2+(-8 m-0)^2}=40\)
\(\sqrt{\left(\mathrm{m}^2-1\right)^2+(-2 \mathrm{~m})^2}=10\)
\(\sqrt{\left(\mathrm{m}^2-1\right)^2+4 \mathrm{~m}^2}=10 \Rightarrow \sqrt{\left(\mathrm{m}^2+1\right)^2}=10\)
\(\mathrm{~m}^2+1=10 \text { or } \mathrm{m}^2=9 \text { or } \mathrm{m}= \pm 3\)
Therefore, coordinate of point \(\mathrm{P}\)
\(\left(4 \times( \pm 3)^2,-8( \pm 3)\right)=(36, \mp 24)\)
\(\because\) Point \(\mathrm{P}\) lies on line \(\mathrm{mx}-\mathrm{y}+\mathrm{c}=0\)
Then it must satisfy equation of line
\(( \pm 3)(36)-(\mp 24)+c =0\)
\(\pm(3 \times 36+24)+\mathrm{c} =0\)
\(\mathrm{c} = \pm 132\)
\(|\mathrm{c}| =132\)
