If the chord joining the points P_1(x_1,y_1) and P_2(x_2,y_2) on the parabola y^2=12x subt…

Question

If the chord joining the points ${P}_{1}({x}_{1},{y}_{1})$ and ${P}_{2}({x}_{2},{y}_{2})$ on the parabola ${y}^{2}=12x$ subtends a right angle at the vertex of the parabola, then ${x}_{1}{x}_{2}-{y}_{1}{y}_{2}$ is equal to

Accepted Answer

For the parabola

${y}^{2}=12x,$

compare it with ${y}^{2}=4ax$. We get

$4a=12\Rightarrow a=3.$

The parametric coordinates of a point on the parabola are

$(a{t}^{2},2at)=(3{t}^{2},6t).$

Thus, take

${P}_{1}=(3{t}_{2}^{1},6{t}_{1}),\, {P}_{2}=(3{t}_{2}^{2},6{t}_{2}).$

The chord subtends a right angle at the vertex $O(0,0)$. Hence,

$\overset{	o \leftarrow }{O{P}_{1}}\cdot \overset{	o \leftarrow }{O{P}_{2}}=0.$

So,

$(3{t}_{2}^{1})(3{t}_{2}^{2})+(6{t}_{1})(6{t}_{2})=0,$

$9{t}_{2}^{1}{t}_{2}^{2}+36{t}_{1}{t}_{2}=0.$

Since the points are not the vertex,

${t}_{1}{t}_{2}=-4.$

Now,

$\begin{gathered} {x}_{1}{x}_{2}-{y}_{1}{y}_{2} =(3{t}_{2}^{1})(3{t}_{2}^{2})-(6{t}_{1})(6{t}_{2}) \ =9({t}_{1}{t}_{2}{)}^{2}-36{t}_{1}{t}_{2} \ =9(-4{)}^{2}-36(-4) \ =144+144 \ =288. \end{gathered}$

Hence, the correct answer is

$288$

Option B.

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