Question

# Focus of the parabola $4{y^2} - 6x - 4y = 5$ is,a. $(\dfrac{ - 8}{5},2)$b. $(\dfrac{ - 5}{8},\dfrac{1}{2})$c. $(\dfrac{1}{2},\dfrac{5}{8})$d. $\dfrac{5}{8},\dfrac{ - 1}{2})$

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Hint: Reduce the given equation to the standard form of that conic and then compare ${x_0}{\text{, }}{y_0}$ and a with the standard equation of parabola.Now, find the required solution by putting values.

As, we know that the standard equation of parabola is ${\left( {y - {y_0}} \right)^2} = 4a\left( {x - {x_0}} \right)$.

In which,

$\Rightarrow {\text{Vertex = }}\left( {{x_0},{y_0}} \right)$ and,

$\Rightarrow {\text{Focus of parabola is }}\left( {{x_0} + a,{y_0}} \right)$

Given Equation of parabola is $4{y^2} - 6x - 4y = 5$

First we have to convert given equation to the standard equation of parabola,

Taking 6x to RHS of the given equation it becomes,

$\Rightarrow 4{y^2} - 4y = 6x + 5$

Adding both sides $4*\dfrac{1}{4}$ equation becomes,

$\Rightarrow 4\left( {{y^2} - y} \right) + {\text{4*}}\dfrac{1}{4} = 6x + 5{\text{ + 4*}}\dfrac{1}{4}$

Taking 4 common in LHS and 6 common in RHS, equation becomes,

$\Rightarrow 4\left( {{y^2} - y + \dfrac{1}{4}} \right) = 6\left( {x + 1} \right)$

Taking 4 to the denominator of RHS, equation becomes,

$\Rightarrow {\left( {y - \dfrac{1}{2}} \right)^2} = \dfrac{3}{2}\left( {x + 1} \right){\text{ }}\left( 1 \right)$

Comparing equation 1with standard equation of parabola we get,

$\Rightarrow {x_0} = - 1,{\text{ }}{y_0} = \dfrac{1}{2}{\text{ and }}a = \dfrac{3}{8}$

So, focus of the parabola in equation 1 will be,

$\Rightarrow {\text{focus}} = \left( { - 1 + \dfrac{3}{8},\dfrac{1}{2}} \right) = \left( {\dfrac{{ - 5}}{8},\dfrac{1}{2}} \right)$

Hence the correct option for the question will be (b)

NOTE: - Understand the diagram properly and a good command over formulas will be an added advantage to get the right answer.