$x=p\sec heta $ and $y=q an heta $ thenA.${{x}^{2}}-{{y}^{2}}={{p}^{2}}{{q}^{2}}$ B.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=pq$ C.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}={{p}^{2}}{{q}^{2}}$D.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=\dfrac{1}{{{p}^{2}}{{q}^{2}}}$

Question

$x=p\sec 	heta $ and $y=q	an 	heta $ thenA.${{x}^{2}}-{{y}^{2}}={{p}^{2}}{{q}^{2}}$ B.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=pq$ C.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}={{p}^{2}}{{q}^{2}}$D.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=\dfrac{1}{{{p}^{2}}{{q}^{2}}}$

Vedantu Content Team · Accepted Answer

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.

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

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$x=p\sec \theta $ and $y=q\tan \theta $ thenA.${{x}^{2}}-{{y}^{2}}={{p}^{2}}{{q}^{2}}$ B.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=pq$ C.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}={{p}^{2}}{{q}^{2}}$D.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=\dfrac{1}{{{p}^{2}}{{q}^{2}}}$

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