PA and PB are the tangents drawn to y2 = 4ax from point P. These tangents meet the y-axis at the points A1 and B1 respectively. If the area of triangle PA1B1 is 2 sq. units, then locus of P is:

P≡at1t2,at2+t2 t, y=x+at12 at1t2 at1+t2 1 0 at1 1 0 at2 1=4 ⇒A1=0,at1,B1=0,at2⇒at1t2⋅at1−t2=4⇒at1−t2=4/h⇒a2t1+t22-4t1t2=4h2⇒k2-4ah=16h2⇒y2−4axx2=16

PA and PB are the tangents drawn to y2 = 4ax from point P. These tangents meet the y-axis at the points A1 and B1 respectively. If the area of triangle PA1B1 is 2 sq. units, then locus of P is:

P≡at1t2,at2+t2 t, y=x+at12 at1t2 at1+t2 1 0 at1 1 0 at2 1=4 ⇒A1=0,at1,B1=0,at2⇒at1t2⋅at1−t2=4⇒at1−t2=4/h⇒a2t1+t22-4t1t2=4h2⇒k2-4ah=16h2⇒y2−4axx2=16

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PA and PB are the tangents drawn to y² = 4ax from point P. These tangents meet the y-axis at the points A₁ and B₁ respectively. If the area of triangle PA₁B₁ is 2 sq. units, then locus of P is:

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$(y^{2} - 4 ax) x^{2} = 8$

$(y^{2} + 4 ax) x^{2} = 16$

$(y^{2} - 4 ax) x^{2} = 16$

$(y^{2} + 4 ax) x^{2} = 8$

answer is D.

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Detailed Solution

$P \equiv ({at}_{1} t_{2}, a (t_{2} + t_{2})) t, y = x + {at}_{1}^{2} |\begin{array}{l} {at}_{1} t_{2} a (t_{1} + t_{2}) 1 \\ 0 {at}_{1} 1 \\ 0 {at}_{2} 1 \end{array}| = 4 \begin{array}{l} \Rightarrow A_{1} = (0, {at}_{1}), B_{1} = (0, {at}_{2}) \\ \Rightarrow {at}_{1} t_{2} \cdot a (t_{1} - t_{2}) = 4 \\ \Rightarrow a (t_{1} - t_{2}) = 4 / h \Rightarrow a^{2} [{(t_{1} + t_{2})}^{2} - 4 t_{1} t_{2}] = {(\frac{4}{h})}^{2} \\ \Rightarrow k^{2} - 4 ah = \frac{16}{h^{2}} \\ \Rightarrow (y^{2} - 4 ax) x^{2} = 16 \end{array}$