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\mathrm{A} is a point on the parabola \mathrm{y^2=4 a x}. The normal at \mathrm{A} cuts the parabola again at \mathrm{B}. If \mathrm{AB} subtends a right angle at the vertex of the parabola, find the slope of \mathrm{AB}.
 

Option: 1

1
 


Option: 2

\pm \sqrt3


Option: 3

\pm \sqrt5


Option: 4

\pm \sqrt2


Answers (1)

best_answer

Let \mathrm{\mathrm{A}=\left(a t_1{ }^2, 2 a t_1\right)}
Let the normal at \mathrm{A} cut the parabola at \mathrm{B} and let \mathrm{B=\left(a t_2^2, 2 a t_2\right).}Then we know that

\mathrm{ t_2=-t_1-\frac{2}{t_1} }

Equation of \mathrm{ A B \: is \: y\left(t_1+t_2\right)=2\left(x+a t_1 t_2\right) }
The combined equation of the lines \mathrm{ O A \: and \: O B } is obtained by making the equation of the parabola homogeneous with the help of \mathrm{ A B }

we write \mathrm{ y^2-4 a x(1)=0 \quad\left[\because \frac{y\left(t_1+t_2\right)-2 x}{2 a t_1 t_2}=1\right] }

\mathrm{ \therefore \text{Combined equation of} \: O A\: and \: O B\: \: is, }

\mathrm{ y^2-4 a x\left[\frac{y\left(t_1+t_2\right)}{2 a t_1 t_2}-\frac{2 x}{2 a t_1 t_2}\right]=0 }

Since \mathrm{ O A\: and \: O B } are perpendicular,

\mathrm{ coeff. of \: x^2+ \: coeff. \: of \: y^2=0 }

\mathrm{ \Rightarrow \frac{8 a}{2 a t_1 t_2}+1=0 \Rightarrow t_1 t_2=-4 }

\mathrm{ \Rightarrow t_2=-\frac{4}{t_1} }

\mathrm{ \therefore-t_1-\frac{2}{t_1}=-\frac{4}{t_1} }

\mathrm{ \Rightarrow t_1^2=2 \quad \therefore t_1= \pm \sqrt{2} }

\mathrm{ Slope\: of\: A B=\frac{2}{t_1+t_2}=\frac{2}{\frac{-2}{t_1}}=-t_1= \pm \sqrt{2} }

Hence option 4 is correct.






 

Posted by

Pankaj Sanodiya

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