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  • AB, AC are tangents to a parabola <img alt="\mathrm{y^2=4 a x, p_1, p_2, p_3}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7By%5E2%3D4%20a%20x%2C%20p_1%2C%20p_2%2C%20p_3%7D"

AB, AC are tangents to a parabola \mathrm{y^2=4 a x, p_1, p_2, p_3}are the lengths of the perpendiculars from A,B,C on any tangent to the curve, then \mathrm{p_2, p_1, p_3} are in 

Option: 1

AP


Option: 2

GP


Option: 3

HP


Option: 4

None of these 


Answers (1)

best_answer

Let, \mathrm{B \equiv\left(a t_1^2, 2 a t_1\right), C \equiv\left(a t_2^2, 2 a t_2\right)}

then, \mathrm{A \equiv\left(a t_1 t_2, a\left(t_1+t_2\right)\right)}

Let any tangent to \mathrm{y^2=4 a x} is 

                              \mathrm{t y=x+a t^2}

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

\therefore    p_1=\text { length of perpendicular from } A \text { on Eq. (i) }

\begin{aligned} & =\frac{\left|a t_1 t_2-t a\left(t_1+t_2\right)+a t^2\right|}{\sqrt{\left(1+t^2\right)}} \\ \\& =\frac{\left|a\left(t-t_1\right)\left(t-t_2\right)\right|}{\sqrt{\left(1+t^2\right)}} \end{aligned}

\mathrm{p_2=\text { length of perpendicular from } B \text { on Eq. (i) }}

=\frac{\left|a t_1^2-t\left(2 a t_1\right)+a t^2\right|}{\sqrt{\left(1+t^2\right)}}=\frac{\left|a\left(t-t_1\right)^2\right|}{\sqrt{\left(1+t^2\right)}}

and p_3 length of perpendicular from C on Eq. (i)

\begin{aligned} & =\frac{\left|a t_2^2-\left(2 a t_2\right) t+a t^2\right|}{\sqrt{\left(1+t^2\right)}} \\ & =\frac{\left|a\left(t-t_2\right)^2\right|}{\sqrt{\left(1+t^2\right)}} \end{aligned}

\\\text{It is clear that}\: p_2 p_3=p_1^2. \\\text{Hence}, p_2, p_1, p_3, are\: in \: GP

Posted by

seema garhwal

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