Question

If P and Q are the points $\left( at_{1}^{2},2at_{1} \right)$ and $\left( at_{2}^{2},2at_{2} \right)$and normals at P and Q meet on the parabola $y^{2} = 4ax,$ then $t_{1},t_{2}$ equals

• A. 2
• B. -1
• C. -2
• D. -4

Answer 1

Answer: (A)

2

Answer 2

Normal at $'t_{1}'$is $y = - t_{1}x + 2at_{1} + at_{1}^{3}$…..(1)

& normal at $'t_{2}$' is $y = - t_{2}x + 2at_{2} + at_{2}^{3}$…….(2)

solving $(1)\&(2)$

we get $\left( 2a + a\left( t_{1}^{2} + t_{2}^{2} + t_{1}t_{2} \right), - at_{1}t_{2}\left( t_{1} + t_{2} \right) \right)$

which is lie on $y^{2} = 4ax$

∴ $a^{2}t_{1}^{2}t_{2}^{2}\left( t_{1} + t_{2} \right)^{2} = 8a^{2} + 4a^{2}\left( t_{1}^{2} + t_{2}^{2} + t_{1}t_{2} \right)$

⇒ $t_{1}^{2}t_{2}^{2}\left( t_{1} + t_{2} \right)^{2} = 8 + 4\left( t_{1}^{2} + t_{2}^{2} + t_{1}t_{2} \right)$

= $8 + 4\left\{ \left( t_{1} + t_{2} \right)^{2} - t_{1}t_{2} \right\}$

⇒ $\left( t_{1} + t_{2} \right)^{2}\{ t_{1}^{2}t_{2}^{2} - 4\rbrack = - 4\left( t_{1}t_{2} - 2 \right)$

∴ $t_{1}t_{2} - 2 = 0$

⇒ $t_{1}t_{2} = 2$