Solveeit Logo
Login

Question

Question: The equation \(16{x^2} - 3{y^2} - 32x + 12y - 44 = 0\) represents a hyperbola. A.The length of who...

The equation 16x2−3y2−32x+12y−44=016{x^2} - 3{y^2} - 32x + 12y - 44 = 0 represents a hyperbola.
A.The length of whose transverse axis is434\sqrt 3
B.The length of whose conjugate axis is 4.
C.Whose centre is (-1, 2)
D.Whose eccentricity is 193\sqrt {\dfrac{{19}}{3}}

Explanation

Solution

We will first convert the given equation into a standard equation of a hyperbola i.e.,
(x−h)2a2−(y−k)2b2=1\dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 with its centre at (– h, – k). We will then calculate the value of its length of transverse axis by the formula 2a, length of conjugate axis by the formula 2b, coordinate of centre: (– h, – k) and the eccentricity of the hyperbola using the formula: e = 1+b2a2\sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} and then check all the obtained values with the given options if they are correct or not.

Complete step-by-step answer:
We are given an equation: 16x2−3y2−32x+12y−44=016{x^2} - 3{y^2} - 32x + 12y - 44 = 0and it represents a hyperbola.

Let us simplify the given equation into the terms of a standard hyperbola.
Let us re – arrange the equation as:
⇒16x2−32x−3y2+12y=44\Rightarrow 16{x^2} - 32x - 3{y^2} + 12y = 44
Taking 16 common from the terms having x and -3 common from terms with y, we get
⇒16(x2−2x)−3(y2−4y)=44\Rightarrow 16\left( {{x^2} - 2x} \right) - 3\left( {{y^2} - 4y} \right) = 44
Now, adding and subtracting 1 from the first bracket and adding and subtracting 4 from the second bracket, we get
⇒16(x2−2x+1−1)−3(y2−4y+4−4)=44\Rightarrow 16\left( {{x^2} - 2x + 1 - 1} \right) - 3\left( {{y^2} - 4y + 4 - 4} \right) = 44
Rearranging the terms, we get
⇒16(x2−2x+1)−16−3(y2−4y+4)+12=44\Rightarrow 16\left( {{x^2} - 2x + 1} \right) - 16 - 3\left( {{y^2} - 4y + 4} \right) + 12 = 44
Now, we can see that both the brackets are the expansion of perfect squares of the form: (a−b)2=a2−2ab+b2{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}
Therefore, we can write
⇒16(x−1)2−3(y−2)2−4=44\Rightarrow 16{\left( {x - 1} \right)^2} - 3{\left( {y - 2} \right)^2} - 4 = 44
⇒16(x−1)2−3(y−2)2=48\Rightarrow 16{\left( {x - 1} \right)^2} - 3{\left( {y - 2} \right)^2} = 48
Dividing both sides by 48, we get
⇒16(x−1)248−3(y−2)248=4848 ⇒(x−1)23−(y−2)216=1  \Rightarrow \dfrac{{16{{\left( {x - 1} \right)}^2}}}{{48}} - \dfrac{{3{{\left( {y - 2} \right)}^2}}}{{48}} = \dfrac{{48}}{{48}} \\\ \Rightarrow \dfrac{{{{\left( {x - 1} \right)}^2}}}{3} - \dfrac{{{{\left( {y - 2} \right)}^2}}}{{16}} = 1 \\\
This is the standard form of a hyperbola. Comparing this equation with (x−h)2a2−(y−k)2b2=1\dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1having its centre as (– h, – k), we get
h = 1 and k = 2, therefore, the centre of this hyperbola will be (1, 2)
therefore, option (C) is incorrect.
Now, we know that the transverse length of a hyperbola is 2a. here, a = 3\sqrt 3 . Therefore, its transverse length will be 23\sqrt 3 .
Therefore, option (A) is incorrect.
Similarly, length of the conjugate axis is given by 2b. here, b = 4. Hence, the length of the conjugate axis will be 8.
Therefore, option(B) is incorrect.
Now, we have the formula of the eccentricity of the hyperbola as: e = 1+b2a2\sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}}
Therefore, e=1+163=3+163=193e = \sqrt {1 + \dfrac{{16}}{3}} = \sqrt {\dfrac{{3 + 16}}{3}} = \sqrt {\dfrac{{19}}{3}}
Hence, option (D) is correct.

Note: In such questions, you can also verify the options by checking them individually instead of reducing their values from the obtained standard equation of hyperbola. You may get confused while solving for the standard equation as we generally use the equation of hyperbola having its centre at origin.