HyperbolaMedium
Question
Let O be the origin, and P and Q be two points on the rectangular hyperbola $xy = 12$ such that the mid point of the line segment PQ is $\left( \frac{1}{2}, - \frac{1}{2} \right)$. Then the area of the triangle OPQ equals :
Options
A.$\frac{3}{2}$
B.$\frac{5}{2}$
C.$\frac{7}{2}$
D.$\frac{9}{2}$
Solution
Sol. [Key Concept: Chord with given midpoint ($T = S_{1}$) gives $P,Q$ on $xy = 12$; area of $OPQ$ by the shoelace formula.]
Equation of chord whose midpoint is $\left( \frac{1}{2},\frac{- 1}{2} \right)$
$$\begin{matrix} & \frac{x}{x_{1}} + \frac{y}{y_{1}} = 2 \\ & 2x - 2y = 2 \\ & x - y = 1 \end{matrix}$$
solving with $xy = 12$
$$P( - 3, - 4),Q(4,3)$$
Area $( \bigtriangleup OPQ) = \frac{1}{2}| - 3 \times 3 - ( - 4)(4)| = \frac{7}{2}$
Create a free account to view solution
View Solution FreeMore Hyperbola Questions
AB is a double ordinate of the hyperbola = 1 such that ᐃAOB (where ′O′ is the origin) is an equilatera...The equation 16x2 − 3y2 − 32x + 12y − 44 = 0 represents a hyperbola-...The tangents to the hyperbola x2 − y2 = 3 are parallel to the straight line 2x + y + 8 = 0 at the following points...The ellipse = 1 and the hyperbola = 1 have in common -...For the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{3} = 1$ the incorrect statement is :...