Question

Find the value of \[{x^2}\; + {\rm{ }}4{y^2}\;\] if \[2ycos\theta = xsin\theta \] and \[2x{\rm{ }}sec\theta - ycosec\theta = 3\]
A. \[4\]
B. \[ - 4\]
C. \[ \pm 4\]
D. None of these

Answer 1

Hint: The power of algebra allows us to solve for the unknown variable to determine what the other variables would equal. This is known as the Quadratic equation. The quadratic equation is a type of equation that has two solutions, which is what we will do here. The solution will give you one variable, and then you can use that variable with the original equation to solve for another variable and so on until you have found all of them.

Formula used: The following trigonometry formula will be useful for this question
\[\begin{array}{l}\cos ec\theta = \dfrac{1}{{\sin \theta }}\\\sec \theta = \dfrac{1}{{\cos \theta }}\end{array}\]

Complete step-by-step solution: Firstly we will square both the side
\[\begin{array}{l}2ycos\theta = xsin\theta \\4{y^2}co{s^2}\theta = {x^2}si{n^2}\theta \end{array}\]
Now, we will solve for the equation \[2x{\rm{ }}sec\theta - ycosec\theta = 3\]
\[\begin{array}{l}2xsec\theta -ycosec\theta = 3\\\dfrac{{2x}}{{cos\theta }}-\dfrac{y}{{sin\theta }} = 3\\2xsin\theta -ycos\theta = 3sin\theta cos{\rm{ }}\theta \end{array}\] …(i)
Putting the value of \[ycos\theta = \dfrac{{xsin\theta }}{2}\]
\[\begin{array}{l}2xsin\theta -\dfrac{{x\sin \theta }}{2} = 3sin\theta cos\theta \\\begin{array}{*{20}{l}}{3xsin\theta = 6sin\theta cos\theta }\\{x = 2cos\theta }\end{array}\end{array}\] …(ii)
From (i) and (ii)
\[\begin{array}{l}{x^2}\; + {\rm{ }}4{y^2}\; = {\rm{ }}{\left( {2cos\theta } \right)^2}\; + {\rm{ }}\left[ {\dfrac{{\left( {{x^2}\;si{n^2}\theta } \right)}}{{co{s^2}\theta }}} \right]\\\begin{array}{*{20}{l}}{{x^2}\; + {\rm{ }}4{y^2} = {\rm{ }}4{\rm{ }}cos2\theta {\rm{ }} + {\rm{ }}\dfrac{{\left( {4{\rm{ }}co{s^2}\theta } \right)si{n^2}\theta }}{{co{s^2}\theta }}}\\{{x^2}\; + {\rm{ }}4{y^2} = {\rm{ }}4\left( {co{s^2}\theta {\rm{ }} + {\rm{ }}si{n^2}\theta } \right)}\\{{x^2}\; + {\rm{ }}4{y^2} = {\rm{ }}4}\end{array}\end{array}\]
Hence, the value of \[{x^2} + 4{y^2}\] is \[4\]
Therefore, the option (A) is correct.

Additional Information: There are three main types of quadratic equations: standard form, factored form and vertex form. The standard form is the most common form of a quadratic equation, and it is the form that is typically used when solving problems. Factored form is a less common form of a quadratic equation, but it can be useful when trying to find the roots of the equation. Vertex form is the least common form of a quadratic equation, but it can be helpful when graphing the equation.

Note : The vertex form of a quadratic equation is the least common form of a quadratic equation, but it can be helpful when graphing the equation. The vertex is the point on the graph of the equation where the curve changes direction.