Find the length of the normal chord which subtends an angle of 90∘ at the vertex of the parabola y2=4x.

If the normal at Pt1 meets the parabola at Qt2, then t2=−t1−2t1 ….(i)Since the normal chord subtends an angle of 90∘ at the vertex, thent1t2=−4From Eq. (i), we get t12+t1t2+2=0⇒ t12−4+2=0⇒ t12−2=0Also, t22=−t1−2t12 =t12+4t12+4 =2+2+4=8Therefore, PQ2 =a2t12−t222+4a2t1−t22 =1⋅(2−8)2+4(2+22)2 =36+72=108⇒ PQ=63.

Find the length of the normal chord which subtends an angle of 90∘ at the vertex of the parabola y2=4x.

If the normal at Pt1 meets the parabola at Qt2, then t2=−t1−2t1 ….(i)Since the normal chord subtends an angle of 90∘ at the vertex, thent1t2=−4From Eq. (i), we get t12+t1t2+2=0⇒ t12−4+2=0⇒ t12−2=0Also, t22=−t1−2t12 =t12+4t12+4 =2+2+4=8Therefore, PQ2 =a2t12−t222+4a2t1−t22 =1⋅(2−8)2+4(2+22)2 =36+72=108⇒ PQ=63.

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Find the length of the normal chord which subtends an angle of $90^{\circ}$ at the vertex of the parabola $y^{2} = 4 x$ .

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$P Q = - 6 \sqrt{3}$

$P Q = 4 \sqrt{3}$

$P Q = 6 \sqrt{3}$

$P Q = 6 \sqrt{2}$

answer is A.

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Detailed Solution

If the normal at $P (t_{1})$ meets the parabola at $Q (t_{2})$ , then

$t_{2} = - t_{1} - \frac{2}{t_{1}}$ ….(i)

Since the normal chord subtends an angle of $90^{\circ}$ at the vertex, then

$t_{1} t_{2} = - 4$

From Eq. (i), we get

$\begin{array}{l} t_{1}^{2} + t_{1} t_{2} + 2 = 0 \\ \Rightarrow t_{1}^{2} - 4 + 2 = 0 \\ \Rightarrow t_{1}^{2} - 2 = 0 \end{array}$

Also, $t_{2}^{2} = {(- t_{1} - \frac{2}{t_{1}})}^{2}$

$\begin{array}{l} = t_{1}^{2} + \frac{4}{t_{1}^{2}} + 4 \\ = 2 + 2 + 4 = 8 \end{array}$

Therefore,

$\begin{array}{l} P Q^{2} = a^{2} {(t_{1}^{2} - t_{2}^{2})}^{2} + 4 a^{2} {(t_{1} - t_{2})}^{2} \\ = 1 \cdot (2 - 8)^{2} + 4 (\sqrt{2} + 2 \sqrt{2})^{2} \\ = 36 + 72 = 108 \end{array}$