Question

Let the equation of a circle and a parabola be ${{x}^{2}}+{{y}^{2}}-4x-6=0$ and ${{y}^{2}}=9x$ respectively. Then
[a] (1,-1) is a point on the common chord of contact.
[b] The equation of the common chord is y+1=0
[c] The length of the common chord is 6
[d] None of these.

Answer 1

Hint:Draw a rough diagram of the curves. Find the coordinates of the points of intersection of these curves. Use the fact that the equation of the line passing through points $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$. Hence find the equation of the common chord. Use the fact that the distance between the points $P\left( {{x}_{1}},{{y}_{1}} \right)$ and $Q\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $PQ=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$. Hence find the length of the common chord.

Complete step-by-step answer:

AB is the common chord.
Finding the coordinates of A and B:
A and B are the points of intersection of the curves ${{x}^{2}}+{{y}^{2}}-4x-6=0$ and ${{y}^{2}}=9x$.
Hence we solve these equations simultaneously to get the coordinates of A and B.
Substituting the value of ${{y}^{2}}$ from the equation of the parabola in the equation of the circle, we get
${{x}^{2}}+9x-4x-6=0\Rightarrow {{x}^{2}}+5x-6=0$
We use the method of splitting the middle term to solve the quadratic equation.
We have 5 = 6-1 and $6\times 1=6$
Hence, we have
${{x}^{2}}+5x-6=0\Rightarrow {{x}^{2}}+6x-x-6=0$
Taking x common from the first two terms and -1 common from the last two terms, we get
$x\left( x+6 \right)-1\left( x+6 \right)=0$
Taking x+6 common from both the terms, we get
$\left( x+6 \right)\left( x-1 \right)=0$
Using zero product property, we have
$x+6=0\text{ or }x-1=0$
Hence, we have
$x=-6\text{ or }x=1$
Since from the equation of the parabola, we have
${{y}^{2}}=9x\Rightarrow 9x\ge 0\Rightarrow x\ge 0$
Hence, we have x= -6 is rejected.
Hence, we have x =1
When x= 1, we have
${{y}^{2}}=9\Rightarrow y=\pm 3$
Hence, we have
$B\equiv \left( 1,3 \right)$ and $A\equiv \left( 1,-3 \right)$
Now, we know that the equation of the line passing through points $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$.
Hence, we have
$\left( y-{{y}_{1}} \right)\left( {{x}_{2}}-{{x}_{1}} \right)=\left( x-{{x}_{1}} \right)\left( {{y}_{2}}-{{y}_{1}} \right)$
Hence the equation fo the line AB is
$\begin{align}
  & \left( y-3 \right)\left( 1-1 \right)=\left( x-1 \right)\left( -3-3 \right) \\
 & \Rightarrow -6\left( x-1 \right)=0 \\
 & \Rightarrow x-1=0 \\
\end{align}$
Hence the equation of the line AB is x-1= 0
Hence option [b] is incorrect.
Clearly, the point (1,-1) lies on AB.
Hence option [a] is correct.
Now, we know that the distance between the points $P\left( {{x}_{1}},{{y}_{1}} \right)$ and $Q\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $PQ=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$.
Hence, we have
$AB=\sqrt{{{\left( 1-1 \right)}^{2}}+{{\left( -3-3 \right)}^{2}}}=6$
Hence, the length of the common chord is 6.
Hence option [c] is correct.
Hence options [a] and [c] are correct.

Note: [1] If using the equations of two second-degree curves, you can eliminate the quadratic terms, then the resulting equation is the equation of the common chord. In the above case, it was not possible. This, however, is possible in case of two ellipses or two circles, or a circle and an ellipse or two parabolas etc.