The normal at the point (at12, 2at1) on the parabola, cuts the parabola again at the point whose parameter is

Question

The normal at the point (at12, 2at1) on the parabola, cuts the parabola again at the point whose parameter is (A) t2 = t1 - (2/t1) (B) t2 = t1 + (2/t1) (C) t2 = - ( t1 + (2/t1) ) (D) None of these

Tardigrade Admin · Accepted Answer

Let the normal at ât1â cuts the parabola again at the point 't2'. the equation of the normal at (at12, 2at1) is y + t1x = 2at1 + at13 Since it passes through the point 't2' i.e (at22, 2at2) ∴ 2at2 + at1t2 2 = 2 at1 + at1 3 ⇒ 2a(t1 - t2 ) + at1(t1 2 - t2 2) = 0 ⇒ 2 + t1 (t1 + t2) = 0 (∵ t1 - t2 ≠0) ⇒ 2 + t1 2 + t1 t2 = 0 ⇒ t1 t2 = -(t1 2 + 2 ) ⇒ t2 = - ( t1 + (2/t1))

Q. The normal at the point $(a t_{1}^{2}, 2 a t_{1})$ on the parabola, cuts the parabola again at the point whose parameter is}

$t_{2} = t_{1} - \frac{2}{t _{1}}$

$t_{2} = t_{1} + \frac{2}{t _{1}}$

$t_{2} = - (t_{1} + \frac{2}{t _{1}})$

None of these

Q. The normal at the point (at1​2,2at1​) on the parabola, cuts the parabola again at the point whose parameter is}

t2​=t1​−t1​2​

t2​=t1​+t1​2​

t2​=−(t1​+t1​2​)

None of these

Q. The normal at the point $(a t_{1}^{2}, 2 a t_{1})$ on the parabola, cuts the parabola again at the point whose parameter is}

$t_{2} = t_{1} - \frac{2}{t _{1}}$

$t_{2} = t_{1} + \frac{2}{t _{1}}$

$t_{2} = - (t_{1} + \frac{2}{t _{1}})$