# $x=p\sec \theta $ and $y=q\tan \theta $ then

A.${{x}^{2}}-{{y}^{2}}={{p}^{2}}{{q}^{2}}$

B.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=pq$

C.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}={{p}^{2}}{{q}^{2}}$

D.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=\dfrac{1}{{{p}^{2}}{{q}^{2}}}$

Last updated date: 25th Mar 2023

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Answer

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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.

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