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  • From a variable point P, on a fixed normal to the parabola   ,

From a variable point P, on a fixed normal to the parabola  \mathrm{y^2=4 a x} , two other normals are drawn to the parabola. The chord joining the feet of these two normals is:

Option: 1

inclined at 60^{\circ} to a fixed line
 


Option: 2

 parallel to x-axis.
 


Option: 3

 perpendicular to a fixed line
 


Option: 4

 parallel to a fixed line with non-zero stope


Answers (1)

 The equation of the normal at the point \mathrm{\mathrm{at}^2, 2 \mathrm{at} } is \mathrm{ \mathrm{y}-2 \mathrm{at}=-\mathrm{t}\left(\mathrm{x}-\mathrm{at}{ }^2\right) }
If the variable point (h, k) lies on it, then a t^3-t(h-2 a)-k=0
If one of the normals is fixed, then one of the roots, say \mathrm{t_1 } is known. From (1), we have
\mathrm{\begin{aligned} & \mathrm{t}_1+\mathrm{t}_2+\mathrm{t}_3=0 \\ & \Rightarrow \quad \mathrm{t}_2+\mathrm{t}_3=-\mathrm{t}_1 \\ & \end{aligned} }
Slope of the chord through the feet of the normals at \mathrm{\left(a t_2^2, 2 \mathrm{at}_2\right) and \left(\mathrm{at}_3^2, 2 \mathrm{at}_3\right) } is equal to
\mathrm{\frac{2 a t_2-2 a t_3}{a t_2^2-a t_3^2}=\frac{2}{t_2+t_3}=-\frac{2}{t_1} }
which is a fixed number.
Hence the chord is parallel to the line, say
\mathrm{y=-\frac{2}{t_1}(x-a) }
Which is a fixed line.

Posted by

Sumit Saini

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