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If the normals drawn from a point to the parabola \mathrm{x^2=4 y} cuts the line \mathrm{y=2} in points whose abscissa are in A.P., then the slopes of the tangents at the three conormal points are in

Option: 1

A.P.
 


Option: 2

G.P.
 


Option: 3

H.P.
 


Option: 4

none of these


Answers (1)

best_answer

Any point on \mathrm{x^2=4 y} is \mathrm{\left(2 t, t^2\right)}
Normal at this point is \mathrm{y-t^2=-\frac{1}{t}(x-2 t)}

Or \mathrm{x+t y=2 t+t^3}                       ...........(1)

It is drawn from \mathrm{A\left(x_1, y_1\right).} The condition is

\mathrm{ x_1+t y_1=2 t+t^3 }

Or \mathrm{t^3+t\left(2-y_1\right)-x_1=0}

Let its roots be \mathrm{t_1, t_2, t_3}

\mathrm{ \therefore \quad t_1+t_2+t_3=0 }

The intersection of (1) with \mathrm{y=2} is given by

\mathrm{ \begin{aligned} & x+2 t=2 t+t^3 \\ & t^3=x \end{aligned} }                                   ...........(2)

The abscissa of three points of intersection are

\mathrm{ t_1^3, t_2^3, t_3^3 }

They are in A.P.

\mathrm{\Rightarrow \quad 2 t_2^3=t_1^3+t_3^3}

Or \mathrm{3 t_2^3=t_1^3+t_2^3+t_3^2}                                   ...........(3)

As \mathrm{t_1+t_2+t_3=0 \Rightarrow\left(t_1+t_2+t_3\right)^3=0}

Or \mathrm{t_1^3+t_2^3+t_3^3=3 t_1 t_2 t_3}                       ...........(4)

From (3) and (4)

\mathrm{ 3 t_2^3=3 t_1 t_2 t_3 }

Or \mathrm{t_2^2=t_1 t_3}

Or \mathrm{\frac{1}{t_1} \frac{1}{t_3}=\frac{1}{t_2^2}}

But \mathrm{\frac{1}{t_1}, \frac{1}{t_2}, \frac{1}{t_3}} are slopes of tangents at three conormal points.

Therefore, they are in G.P.

The answer is (b)

Posted by

Pankaj

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