Let chord $A B$ of the parabola $y^{2} = 4 x$ subtend $90^{\circ}$ at the vertex. If the coordinates of $A$ and $B$ are $(t_{1}^{2}, 2 t_{1})$ and $(t_{2}^{2}, 2 t_{2})$ respectively, then

t_{1} t_{2} = - 1

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t_{1} t_{2} = - 2

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t_{1} t_{2} = - 4

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t_{1} t_{2} = - 8

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Solution

The correct option is C $t_{1} t_{2} = - 4$
Here, $O (0, 0)$ is the vertex.
Slope of $(O A) \times$ Slope of $(O B) = - 1$
$\Rightarrow \frac{2 t_{1} - 0}{t_{1}^{2} - 0} \times \frac{2 t_{2} - 0}{t_{2}^{2} - 0} = - 1 \Rightarrow t_{1} t_{2} = - 4$

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Let chord AB of the parabola y2=4x subtend 90∘ at the vertex. If the coordinates of A and B are (t21,2t1) and (t22,2t2) respectively, then

The correct option is C t1t2=−4Here, O(0,0) is the vertex. Slope of (OA) × Slope of (OB)=−1 ⇒2t1−0t21−0×2t2−0t22−0=−1⇒t1t2=−4

Let chord $A B$ of the parabola $y^{2} = 4 x$ subtend $90^{\circ}$ at the vertex. If the coordinates of $A$ and $B$ are $(t_{1}^{2}, 2 t_{1})$ and $(t_{2}^{2}, 2 t_{2})$ respectively, then

The correct option is C $t_{1} t_{2} = - 4$
Here, $O (0, 0)$ is the vertex.
Slope of $(O A) \times$ Slope of $(O B) = - 1$
$\Rightarrow \frac{2 t_{1} - 0}{t_{1}^{2} - 0} \times \frac{2 t_{2} - 0}{t_{2}^{2} - 0} = - 1 \Rightarrow t_{1} t_{2} = - 4$