Question

If the tangents to the parabola $y^2=4ax$ at $(x_1,y_1),(x_2,y_2)$ intersect at $(x_3,y_3),$ then

• A. $x_1,x_3,x_2$ are in $A.P.$
• B. $x_1,x_3,x_2$ are in $G.P.$
• C. $y_1,y_3,y_2$ are in $A.P.$
• D. $y_1,y_3,y_2$ are in $G.P.$

Answer 1

Answer: (B), (C)

The correct options are
B $x_1,x_3,x_2$ are in $G.P.$
C $y_1,y_3,y_2$ are in $A.P.$

Let $(x_1,y_1)=(a{t}_1^2,2a{t}_1)$
$(x_2,y_2)=(a{t}_2^2,2a{t}_2)$
$(x_3,y_3)=(a{t}_1{t}_2,a(t_1+t_2\left)\right)$
$\therefore x_1\cdot x_2=a{t}_1^2\cdot a{t}_2^2=(a{t}_1{t}_2{)}^2=x_3^2$
So, $x_1,x_3,x_2$ are in $G.P.$
and
$y_3=a(t_1+t_2)=\frac{1}{2}(2a{t}_1+2a{t}_2)=\frac{1}{2}(y_1+y_2)$
So, $y_1,y_3,y_2$ are in $A.P.$