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Mathematics Question on Parabola

The normal at the point (at12, 2at1)(at_1^2 ,\, 2at_1) on the parabola meets the parabola again in the point (at12, 2at2)(at_1^2 ,\, 2at_2) , then

A

t2=−t1+2t1 t_{2} = -t_{1} +\frac{2}{t_{1}}

B

t2=−t1−2t1 t_{2} = -t_{1} -\frac{2}{t_{1}}

C

t2=t1−2t1 t_{2} = t_{1} -\frac{2}{t_{1}}

D

t2=t1+2t1 t_{2} = t_{1} +\frac{2}{t_{1}}

Answer

t2=−t1−2t1 t_{2} = -t_{1} -\frac{2}{t_{1}}

Explanation

Solution

Equation of the normal at point (at12, 2at1)(at_1^2 ,\, 2at_1) on parabola is
y=−t1x+2at1+at13y = -t_{1}x + 2at_{1} + at_{1}^{3}
It also passes through (at12, 2at2)\left(at_{1}^{2} ,\, 2at_{2}\right)
So, 2at2=−t1(at22)+2at1+at132at_{2} = -t_{1}\left(at^{2}_{2}\right)+2at_{1}+at^{3}_{1}
⇒2t2−2t1=−t1(t22−t12)\Rightarrow 2t_{2}-2t_{1} = -t_{1}\left(t^{2}_{2}-t^{2}_{1}\right)
⇒t1+t2=−2t1\Rightarrow t_{1}+t_{2} = \frac{-2}{t_{1}}
⇒t2=−t1−2t1\Rightarrow t_{2} = -t_{1} -\frac{2}{t_{1}}