Question

$AB,AC$ are tangents to a parabola $y^2=4ax.p_1,p_2$ and $p_3$ are the lengths of the perpendiculars from $A,B$ and $C$ respectively on any tangent to the curve, then $p_2,p_1,p_3$ are in

• A. A.P.
• B. G.P.
• C. H.P.
• D. None of these

Answer 1

The correct option is B G.P.

Let $B=(a{t}_1^2,2a{t}_1)$ and $C=(a{t}_2^2,2a{t}_2)$

Then, $A=(a{t}_1{t}_2,a(t_1+t_2\left)\right)$

Let, $y=mx+\frac{a}{m}$ be any tangent.

Then $p_2=\frac{|2a{t}_1-ma{t}_1^2-\frac{a}{m}|}{\sqrt{1+m^2}}=|\frac{a}{m}|\frac{(m{t}_1-1{)}^2}{\sqrt{1+m^2}}$

$p_3=\frac{|2a{t}_2-ma{t}_2^2-\frac{a}{m}|}{\sqrt{1+m^2}}=|\frac{a}{m}|\frac{(m{t}_2-1{)}^2}{\sqrt{1+m^2}}$

$p_1=\frac{|am{t}_1{t}_2-a(t_1+t_2)-\frac{a}{m}|}{\sqrt{1+m^2}}=|\frac{a}{m}|\frac{(m{t}_1-1)(m{t}_2-1)}{\sqrt{1+m^2}}$

$\Rightarrow p_2{p}_3=\sqrt{p_1}$

So, $p_2,p_1,p_3$ are in G.P.

Option B is correct.