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Normals are drawn at points \mathrm{A, B \: and\: C} on the parabola \mathrm{y^2=4 x} which intersect at \mathrm{P(h, k)}. The locus of the point \mathrm{P} if the slope of the line joining the feet of two of them is 2 , is
 

Option: 1

\mathrm{x+y=1}
 


Option: 2

\mathrm{x-y=3}
 


Option: 3

\mathrm{y^2=2(x-1)}
 


Option: 4

\mathrm{y^2=2\left(x-\frac{1}{2}\right)}


Answers (1)

best_answer

The equation of normal at \mathrm{\left(a t^2, 2 a t\right)} is

\mathrm{ y+t x=2 a t+a t^3 }          ....(i)

As (i) passes through \mathrm{P(h, k)}, so

\mathrm{ a t^3+t(2 a-h)-k=0 }     .........(ii)

Here, \mathrm{ a=1 }

\mathrm{ \therefore t_1+t_2+t_3=0 }...........(iii)

\mathrm{ \text { Also, } \frac{2}{t_1+t_2}=2 }

\mathrm{ \Rightarrow t_1+t_2=1 }   ........(iv)

From (iii) and (iv), we get \mathrm{t_3=-1}

Putting \mathrm{t_3=-1} in (ii), we get

\mathrm{ -1-1(2-h)-k=0 \Rightarrow-1-2+h-k=0 }

\mathrm{ \therefore \quad \text { Locus of } P(h, k) \text {, is } x-y=3 }

Hence option 2 is correct.
 

Posted by

Kuldeep Maurya

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