Question

Question 12. Eliminate t between the equations :
(i) p tan t + q sec t = x,
p'tan t + q' sec t = y.​

Answer 1

$\textbf{Given:}$

$\mathrm{p\,tan\,t+q\,sec\,t=x}$

$\mathrm{p'\,tan\,t+q'\,sec\,t=y}$

$\textbf{To find:}$

$\text{Eliminate t from the given equations}$

$\textbf{Solution:}$

$\text{Consider,}$

$\mathrm{p\,tan\,t+q\,sec\,t-x=0}$

$\mathrm{p'\,tan\,t+q'\,sec\,t-y=0}$

$\text{Using Cross-mutiplication rule, we get}$

$\mathrm{\dfrac{tan\,t}{-qy+q'x}=\dfrac{sec\,t}{-xp'+py}=\dfrac{1}{pq'-p'q}}$

$\implies\mathrm{sec\,t=\dfrac{py-p'x}{pq'-p'q}}$

$\mathrm{tan\,t=\dfrac{q'x-qy}{pq'-p'q}}$

$\text{But}$

$\mathrm{sec^2t-tan^2t=1}$

$\mathrm{(\dfrac{py-p'x}{pq'-p'q})^2-(\dfrac{q'x-qy}{pq'-p'q})^2=1}$

$\mathrm{\dfrac{(py-p'x)^2}{(pq'-p'q)^2}-\dfrac{(q'x-qy)^2}{(pq'-p'q)^2}=1}$

$\implies\boxed{\mathrm{\bf\,(py-p'x)^2-(q'x-qy)^2=(pq'-p'q)^2}}$

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Answer 2

hope this will help you.