Question

All chords of a curve $3{{x}^{2}}-{{y}^{2}}-2x+4y=0$ which subtend a right angle at origin, pass through a fixed point which is:
A. (-1, 2)
B. (2, 1)
C. (1, -2)
D. (-2, 1)

Answer 1

Hint: Here we will find the fixed points from where the chord of a curve passes. First, we will use the equation of the chord which is a straight line to homogenize the equation of the curve and then we will find the constant term. Now we will solve the equation for finding the points which are fixed. A chord is a pointy which has both its endpoints on the curve.

Complete step by step answer:
Let the equation of the curve be $3{{x}^{2}}-{{y}^{2}}-2x+4y=0$. Equation of a straight line is given by $y=mx+c$, where $m$ is the slope and $c$ is the y-intercept. Since the chords of the curve is a straight line, then the equation of the chord is given by,
$y=mx+c$
So, find $OA\times OB$, where A and B are points of intersection of chord and circle, we use homogenisation.
$\Rightarrow 3{{x}^{2}}-{{y}^{2}}-2\left( x+2y \right)\left( y-\dfrac{mx}{c} \right)=0$
Further we can write it as,
$\begin{align}
  & \Rightarrow c3{{x}^{2}}-c{{y}^{2}}-2xy-2m{{x}^{2}}-4{{y}^{2}}+4mxy=0 \\
 & \Rightarrow 3c{{x}^{2}}-c{{y}^{2}}-2xy-2m{{x}^{2}}-4{{y}^{2}}+4mxy=0 \\
 & \Rightarrow {{x}^{2}}\left( 3c-2m \right)-{{y}^{2}}\left( c+4 \right)+xy\left( 4m-2 \right)=0 \\
\end{align}$
Therefore OA and OB are at right angles.
$\begin{align}
  & {{m}_{1}}{{m}_{2}}=\dfrac{a}{b}=-1 \\
 & \Rightarrow \dfrac{3c-2m}{-\left( c+4 \right)}=-1 \\
\end{align}$
So,
$\begin{align}
  & \Rightarrow 3c-2m=c+4 \\
 & \Rightarrow 2c=2\left( m+2 \right) \\
 & \Rightarrow c=\left( m+2 \right) \\
\end{align}$
Hence the chord is, $y=\left( mx+m+2 \right)$. This equation always satisfies (-1, 2) irrespective of the value of m. hence the fixed point is (-1, 2).

So, the correct answer is “Option A”.

Note: We should know that if the chords of the curve are at right angles, then they are said to be perpendicular. Three or more lines are said to be current if they all pass through a common point. Thus the common point is known as the point of concurrency. Homogenisation is the process of making the degree of all the terms of the equations equal.