The equation of the tangent at any point of the curve x=a \cos 2 t, y=2 \sqrt{2} a \sin t, with m as its slope, is:

Correct Answer is : y=mx−a(m+1m)

NIMCET 2023 Question Paper with solutions

Question : 34 of 120

Marks: +1, -0

The equation of the tangent at any point of the curve

x=a \cos 2 t, y=2 {√{2}} a sin \;t

, with

m

as its slope, is:

Solution:

Equation of tangent, with slope

m

x\;=a \cos 2 t

\;{d x}/{d t}\;=-2 a sin \;2 t

y\;=2 {√{2}} a sin \;t

\;{d y}/{d t}\;=2 {√{2}} a \cos t

\;{d y}/{d x}\;=\;{d y \/ \dt}/{d x \/ \dt}=\;{2 {√{2}} a \cos t}/{-2 a sin \;2 t}

\;=\;{{√{2}} \cos t}/{-2 sin \;t \cos t}

\;=\;{-1}/{{√{2}} sin \;t}=m \;\text " given "\;

sin \;t\;=\;{-1}/{{√{2}} m}

sin \;t\;=\;{-1}/{{√{2}} m} ⇒ sin \;t=\;{-1}/{{√{2}} m}

Then,

x\;=a \cos 2 t

\;=a (1-2 sin \;^2 t)

\;=a (1-2 × (\;{-1}/{{√{2}} m}) ^2)

y\;=2 {√{2}} a sin \;t

\;=2 {√{2}} a × (\;{-1}/{{√{2}} m})

\;y=\;{-2 a}/{m}

\;x=a (1-\;{1}/{m^2})

Then equation of tangent

y-y_1\;=m (x-x_1)

y+\;{2 a}/{m}\;=m (x-a (1-\;{1}/{m^2}) )

y+\;{2 a}/{m}\;=m x-a m+\;{a}/{m}

y\;=m x-a (m+\;{1}/{m})

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