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  • The locus of the point of intersection of two normals to the parabola , which are at

The locus of the point of intersection of two normals to the parabola \mathrm{x^2=8 y}, which are at right angles to each other, is

Option: 1

\mathrm{x^2=2(y-6)}


Option: 2

\mathrm{x^2=2(y+6)}


Option: 3

\mathrm{x^2=-2(y-6)}


Option: 4

\mathrm{\text { None of these }}


Answers (1)

best_answer

Given parabola is \mathrm{x^2=8 y}                      \mathrm{....(i)}

Let \mathrm{\left(4 t_1 2 t_1^2\right)} and \mathrm{Q\left(4 t_2, 2 t_2^2\right)} be two points on the parabola (i)

Normal at \mathrm{P, Q} are \mathrm{ y-2 t_1^2=-\frac{1}{t_1}\left(x-4 t_1\right) }   \mathrm{....(ii)}  and \mathrm{ y-2 t_2^2=-\frac{1}{t_2}\left(x-4 t_2\right) }     \mathrm{....(iii)}
(ii)-(iii) gives \mathrm{ 2\left(t_2^2-t_1^2\right)=x\left(\frac{1}{t_2}-\frac{1}{t_1}\right)=x \frac{t_1-t_2}{t_1 t_2} }   \mathrm{\therefore x=-2 t_1 t_2\left(t_2+t_1\right)}           \mathrm{.....(iv)}


                                                    \mathrm{y=2 t_1^2-\frac{1}{t_1}\left(-2 t_1 t_2\left(t_2+t_1\right)-4 t_1\right)=2 t_1^2+2 t_2\left(t_1+t_2\right)+4 \quad \Rightarrow}
From    (ii),

\mathrm{ y=2 t_1^2+2 t_1 t_2+2 t_2^2+4 }     \mathrm{.....(v)}

Since normals (ii) and (iii) are at right angles, \mathrm{\therefore} \mathrm{ \left(-\frac{1}{t_1}\right)\left(-\frac{1}{t_2}\right)=-1 \Rightarrow t_1 t_2=-1 }

\mathrm{\therefore \quad}  From (iv), \mathrm{x=2\left(t_1+t_2\right)} and from (v) \mathrm{y=2 t_1^2-2+2 t_2^2+4 \Rightarrow y=2\left[t_1^2+t_2^2+1\right]}

\mathrm{ =2\left[\left(t_1+t_2\right)^2-2 t_1 t_2+1\right] }

\mathrm{\Rightarrow \quad y^2 2\left[\left(t_1+t_2\right)^2+2+1\right]=2\left[\left(t_1+t_2\right)^2+3 \Rightarrow y=2\left[\frac{x^2}{4}+3\right]=\frac{x^2}{2}+6 \Rightarrow x^2=2(y-6)\right.,} which is the required locus.

Posted by

Gautam harsolia

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