Question

The normal at the point $\left(b{t}_1^2,2b{t}_1\right)$ on a parabola meets the parabola again in the point $\left(b{t}_2^2,2b{t}_2\right),$ then :

• A. $t_2=-t_1-\frac{2}{t_1}$
• B. $t_2=-t_1+\frac{2}{t_1}$
• C. $t_2=t_1-\frac{2}{t_1}$
• D. $t_2=t_1+\frac{2}{t_1}$

Answer 1

Answer: (A)

Key Idea : Equation of normal at a point $\left(a{t}^2,2at\right)$ on $a$ perabola $y^2=4ax$ is $y=-tx+2at+b{t}^3$
Equation of the normal at point $\left(b{t}_1^2,2b{t}_1\right)$ on parabola is
$y=-t_{1}x+2b{t}_1+b{t}_1^3$
It also passes through $\left(b{t}_2^2,2b{t}_2\right),$ then
$2b{t}_2=-t_1\cdot b{t}_2^2+2b{t}_1+b{t}_1^3$
$2t_2-2t_1=-t_1\left(t_2^2-t_1^2\right)$
$=-t_1\left(t_2+t_1\right)\left(t_2-t_1\right)$
$\Rightarrow 2=-t_1\left(t_2+t_1\right)$
$\Rightarrow t_2+t_1=-\frac{2}{t_1}$
$\Rightarrow t_2=-t_1-\frac{2}{t_1}$
Note :
1. If a normal drawn at a point $t_1$ on the parabola meets the parabola again at a point $t_2$, then
$t_2=-t_1-\frac{2}{t_1}$
2. Three normals can be drawn at a point on a parabola.
3. Foot of the normals drawn from a point are called conormal points.