Let O be the origin, and P and Q be two points on the rectangular hyperbola xy = 12 such t…

Question

Let O be the origin, and P and Q be two points on the rectangular hyperbola $xy = 12$ such that the midpoint of the line segment PQ is $\left( \frac{1}{2}, -\frac{1}{2} \right)$. Then the area of the triangle OPQ equals :

Accepted Answer

Let

$P({x}_{1},{y}_{1}),\, Q({x}_{2},{y}_{2})$

be two points on the rectangular hyperbola

$xy=12$

So, both points satisfy

${x}_{1}{y}_{1}=12\, \, {x}_{2}{y}_{2}=12$

Also, the midpoint of $PQ$ is given as

$(\frac{1}{2},-\frac{1}{2})$

Using the midpoint formula,

$\frac{{x}_{1}+{x}_{2}}{2}=\frac{1}{2},\, \frac{{y}_{1}+{y}_{2}}{2}=-\frac{1}{2}$

Hence,

${x}_{1}+{x}_{2}=1,\, {y}_{1}+{y}_{2}=-1$

Now we need the area of triangle $OPQ$, where $O=(0,0)$.

The area of triangle formed by points $O(0,0)$, $P({x}_{1},{y}_{1})$, $Q({x}_{2},{y}_{2})$ is

$=\frac{1}{2}|{x}_{1}{y}_{2}-{x}_{2}{y}_{1}|$

So our aim is to find $|{x}_{1}{y}_{2}-{x}_{2}{y}_{1}|$.

Since ${x}_{1}$ and ${x}_{2}$ are the $x$-coordinates of points on the hyperbola, and their sum is $1$, let us use

${y}_{1}=\frac{12}{{x}_{1}},\, {y}_{2}=\frac{12}{{x}_{2}}$

Now using

${y}_{1}+{y}_{2}=-1$

we get

$\frac{12}{{x}_{1}}+\frac{12}{{x}_{2}}=-1$

$12(\frac{{x}_{1}+{x}_{2}}{{x}_{1}{x}_{2}})=-1$

Since ${x}_{1}+{x}_{2}=1$,

$12(\frac{1}{{x}_{1}{x}_{2}})=-1$

So,

${x}_{1}{x}_{2}=-12$

Thus ${x}_{1}$ and ${x}_{2}$ are roots of the quadratic equation

${t}^{2}-({x}_{1}+{x}_{2})t+{x}_{1}{x}_{2}=0$

${t}^{2}-t-12=0$

$(t-4)(t+3)=0$

Hence,

${x}_{1}=4,\, {x}_{2}=-3$

or vice versa.

Then corresponding $y$-coordinates are

${y}_{1}=\frac{12}{4}=3,\, {y}_{2}=\frac{12}{-3}=-4$

So the points are

$P(4,3),\, Q(-3,-4)$

Now area of triangle $OPQ$ is

$=\frac{1}{2}|4(-4)-(-3)(3)|$

$=\frac{1}{2}|-16+9|$

$=\frac{1}{2}\cdot 7$

$=\frac{7}{2}$

Therefore, the required area is

$\frac{7}{2}$

So, the correct option is:

Solutions · 1