Answer

Verified

433.5k+ views

Hint: Out of the given options the right-hand side of the equation is always independent of $\theta $. So, we have to perform some calculations to make left-hand side equations independent of $\theta $ too. Thus, we need to make equations based on $\theta $ and solve it accordingly to remove $\theta $.

Complete step-by-step answer:

Here, we have $x=p\sec \theta $ and $y=q\tan \theta $

Considering each option’s nature, we can see that the right-hand side of the equations are independent of $\theta $. So, we have to form equations based on $\theta $ to eliminate it from these equations.

Now, from the given equations, for $x=p\sec \theta $, we get

$\Rightarrow x=p\sec \theta $

On cross-multiplication, we get

$\Rightarrow \sec \theta =\dfrac{x}{p}...\text{ }\left( 1 \right)$

Similarly, from $y=q\tan \theta $, we have

\[\begin{align}

& \Rightarrow y=q\tan \theta \\

& \Rightarrow \tan \theta =\dfrac{y}{q}...\text{ }\left( 2 \right) \\

\end{align}\]

Since, we need to make the equations independent of $\theta $, we can use trigonometric identities, i.e.,

${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $

And, substituting values of $\tan \theta $ and $\sec \theta $ from equation (1) and (2), we get

$\begin{align}

& \Rightarrow {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \\

& \Rightarrow {{\left( \dfrac{x}{p} \right)}^{2}}=1+{{\left( \dfrac{y}{q} \right)}^{2}} \\

\end{align}$

By transposing the variables, we get

\[\begin{align}

& \Rightarrow {{\left( \dfrac{x}{p} \right)}^{2}}=1+{{\left( \dfrac{y}{q} \right)}^{2}} \\

& \Rightarrow {{\left( \dfrac{x}{p} \right)}^{2}}-{{\left( \dfrac{y}{q} \right)}^{2}}=1 \\

& \Rightarrow \dfrac{{{x}^{2}}}{{{p}^{2}}}-\dfrac{{{y}^{2}}}{{{q}^{2}}}=1 \\

\end{align}\]

Taking LCM on LHS of the equation, we get

\[\begin{align}

& \Rightarrow \dfrac{{{x}^{2}}}{{{p}^{2}}}-\dfrac{{{y}^{2}}}{{{q}^{2}}}=1 \\

& \Rightarrow \dfrac{{{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}}{{{p}^{2}}{{q}^{2}}}=1 \\

\end{align}\]

On cross-multiplication, we get

\[\begin{align}

& \Rightarrow \dfrac{{{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}}{{{p}^{2}}{{q}^{2}}}=1 \\

& \Rightarrow {{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=1\left( {{p}^{2}}{{q}^{2}} \right) \\

& \Rightarrow {{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}={{p}^{2}}{{q}^{2}} \\

\end{align}\]

Hence, on solving the given equations, we get \[{{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}={{p}^{2}}{{q}^{2}}\], i.e., option [C] is correct.

Note: The easiest way to solve this kind of problem is by substituting the values of $x$ and $y$ on LHS of every option available and checking it with the values of RHS of that particular option. This method is very useful if the given equations of $x$ and $y$ are very complex.

Complete step-by-step answer:

Here, we have $x=p\sec \theta $ and $y=q\tan \theta $

Considering each option’s nature, we can see that the right-hand side of the equations are independent of $\theta $. So, we have to form equations based on $\theta $ to eliminate it from these equations.

Now, from the given equations, for $x=p\sec \theta $, we get

$\Rightarrow x=p\sec \theta $

On cross-multiplication, we get

$\Rightarrow \sec \theta =\dfrac{x}{p}...\text{ }\left( 1 \right)$

Similarly, from $y=q\tan \theta $, we have

\[\begin{align}

& \Rightarrow y=q\tan \theta \\

& \Rightarrow \tan \theta =\dfrac{y}{q}...\text{ }\left( 2 \right) \\

\end{align}\]

Since, we need to make the equations independent of $\theta $, we can use trigonometric identities, i.e.,

${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $

And, substituting values of $\tan \theta $ and $\sec \theta $ from equation (1) and (2), we get

$\begin{align}

& \Rightarrow {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \\

& \Rightarrow {{\left( \dfrac{x}{p} \right)}^{2}}=1+{{\left( \dfrac{y}{q} \right)}^{2}} \\

\end{align}$

By transposing the variables, we get

\[\begin{align}

& \Rightarrow {{\left( \dfrac{x}{p} \right)}^{2}}=1+{{\left( \dfrac{y}{q} \right)}^{2}} \\

& \Rightarrow {{\left( \dfrac{x}{p} \right)}^{2}}-{{\left( \dfrac{y}{q} \right)}^{2}}=1 \\

& \Rightarrow \dfrac{{{x}^{2}}}{{{p}^{2}}}-\dfrac{{{y}^{2}}}{{{q}^{2}}}=1 \\

\end{align}\]

Taking LCM on LHS of the equation, we get

\[\begin{align}

& \Rightarrow \dfrac{{{x}^{2}}}{{{p}^{2}}}-\dfrac{{{y}^{2}}}{{{q}^{2}}}=1 \\

& \Rightarrow \dfrac{{{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}}{{{p}^{2}}{{q}^{2}}}=1 \\

\end{align}\]

On cross-multiplication, we get

\[\begin{align}

& \Rightarrow \dfrac{{{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}}{{{p}^{2}}{{q}^{2}}}=1 \\

& \Rightarrow {{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=1\left( {{p}^{2}}{{q}^{2}} \right) \\

& \Rightarrow {{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}={{p}^{2}}{{q}^{2}} \\

\end{align}\]

Hence, on solving the given equations, we get \[{{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}={{p}^{2}}{{q}^{2}}\], i.e., option [C] is correct.

Note: The easiest way to solve this kind of problem is by substituting the values of $x$ and $y$ on LHS of every option available and checking it with the values of RHS of that particular option. This method is very useful if the given equations of $x$ and $y$ are very complex.

Recently Updated Pages

Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

The branch of science which deals with nature and natural class 10 physics CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Trending doubts

What type of defect is shown by NaCl in a Stoichiometric class 12 chemistry CBSE

Difference Between Plant Cell and Animal Cell

Distinguish between tetrahedral voids and octahedral class 12 chemistry CBSE

Derive an expression for electric potential at point class 12 physics CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE