Question

if a point (x,y)=(tana+sina,tana-sina) then the locus of (x,y) is

Answer 1

$\textbf{Given:}$

$\text{The moving point (x,y) with}$

$x=tanA+sinA\;\text{and}\;y=tanA-sinA$

$\textbf{To find: Locus of (x,y)}$

$x=tanA+sinA$

$y=tanA-sinA$

$\text{Then,}\;\;x^2-y^2$

$=(tanA+sinA)^2-(tanA-sinA)^2$

$=tan^2A+sin^2A+2\,tanA\,sinA-tan^2A-sin^2A+2\,tanA\,sinA$

$=4\,tanA\,sinA$..........(1)

$xy=(tanA+sinA)(tanA-sinA)$

$xy=tan^2A-sin^2A$

$xy=\dfrac{sin^2A}{cos^2A}-sin^2A$

$xy=sin^2A(\dfrac{1}{cos^2A}-1)$

$xy=sin^2A(sec^2A-1)$

$xy=sin^2A\;tan^2A$.......(2)

$(1)\implies\,(x^2-y^2)^2=16\,sin^2A\,tan^2A$

$\implies\bf\,(x^2-y^2)^2=16xy$

$\therefore\textbf{The locus of (x,y) is}$

$\bf\,(x^2-y^2)^2=16xy$

Find more:

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