Question

The center of the conic section $14{{x}^{2}}-4xy+11{{y}^{2}}-44x-58y+71=0$ is:
(a). $\left( 2,3 \right)$
(b). $\left( 2,-3 \right)$
(c). $\left( -2,3 \right)$
(d). $\left( -2,-3 \right)$

Answer 1

Hint: Compare the given equation with the general form of a conic section:
$a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$
Find the values of the coefficients. Then solve the following equations to find out the center:
$\begin{aligned} & a{{x}_{1}}+h{{y}_{1}}+g=0 \\\ & h{{x}_{1}}+b{{y}_{1}}+f=0 \\\ \end{aligned}$

Complete step-by-step answer:
If a point moves in a plane in such a way that its distance from a fixed point always bears a constant ratio to its distance from a fixed straight line, then the locus of the moving point is called a conic section or simply a conic.
We know that the general form of a conic section is:
$a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0.....(1)$
Let the center of the conic be at $\left( {{x}_{1}},{{y}_{1}} \right)$.
Then the center of the conic will satisfy the following equations:
$\begin{aligned} & a{{x}_{1}}+h{{y}_{1}}+g=0 \\\ & h{{x}_{1}}+b{{y}_{1}}+f=0 \\\ \end{aligned}$
We can find out the center of the conic by solving these two equations.
Now the given equation is:
$14{{x}^{2}}-4xy+11{{y}^{2}}-44x-58y+71=0.....(2)$
Now if we compare the coefficients of the variables from (1) and (2) we will get,
$a=14$
$b=11$
$2h=-4\Rightarrow h=-2$
$2g=-44\Rightarrow g=-22$
$2f=-58\Rightarrow f=-29$
$c=71$
Let the center of the conic be at $\left( {{x}_{1}},{{y}_{1}} \right)$.
The center will satisfy the following equations:
$\begin{aligned} & a{{x}_{1}}+h{{y}_{1}}+g=0 \\\ & h{{x}_{1}}+b{{y}_{1}}+f=0 \\\ \end{aligned}$
Let us put the values of the coefficients,
$\begin{aligned} & 14{{x}_{1}}-2{{y}_{1}}-22=0.......(3) \\\ & -2{{x}_{1}}+11{{y}_{1}}-29=0.......(4) \\\ \end{aligned}$
We will solve these to equations by cancelling out ${{x}_{1}}$ from (1) and (2)
To cancel out ${{x}_{1}}$ we have to make the coefficients the same.
Let us multiply equation (4) by 7, we will get:
$-14{{x}_{1}}+77{{y}_{1}}-203=0......(5)$
Now by adding equation (3) and (4), we will get:
$14{{x}_{1}}-2{{y}_{1}}-22-14{{x}_{1}}+77{{y}_{1}}-203=0$
By cancelling out the same terms we will get:
$\begin{aligned} & \Rightarrow -2{{y}_{1}}-22+77{{y}_{1}}-203=0 \\\ & \Rightarrow 75{{y}_{1}}-225=0 \\\ & \Rightarrow 75{{y}_{1}}=225 \\\ & \Rightarrow {{y}_{1}}=3 \\\ \end{aligned}$
Now let us put the value of ${{y}_{1}}$ in (3) to find out the value of ${{x}_{1}}$:
$\begin{aligned} & 14{{x}_{1}}-\left( 2\times 3 \right)-22=0 \\\ & \Rightarrow 14{{x}_{1}}-6-22=0 \\\ & \Rightarrow 14{{x}_{1}}-28=0 \\\ & \Rightarrow 14{{x}_{1}}=28 \\\ & \Rightarrow {{x}_{1}}=2 \\\ \end{aligned}$
Therefore, ${{x}_{1}}=2,{{y}_{1}}=3$
Hence the center of the given conic is $\left( 2,3 \right)$.
Therefore option (a) is correct.

Note: We can also find out center of a conic by the following formula:
${{x}_{1}}=\dfrac{hf-bg}{ab-{{h}^{2}}},{{y}_{1}}=\dfrac{gh-af}{ab-{{h}^{2}}}$