The equation of the tangent at any point of the curvex = a cos 2t, y = 2 \sqrt2 a sin t , with ‘m’ as its slope is;

Correct Answer is : y = mx - a (m+1m)

NIMCET 2015 Question Paper with solutions

Question : 18 of 120

Marks: +1, -0

The equation of the tangent at any point of the curvex = a cos 2t, y = 2

√2

a sin t , with ‘m’ as its slope is;

Solution:

We have,

x=a\cos 2t

and

y=2√2a\sin t

{dx}/{dt}=-2a\sin2t

and

{dy}/{dt}=2√2a\cos t

∴{dy}/{dt}={2√2a\cos t}/{-2a\sin 2t}

={-√2\cos t}/{\sin^2t}

⇒{dy}/dx={-√2\cos t}/{2\sin t\cos t}={-1}/{√2\sint}

It is given that, m is the slope of tangent

∴m=-1/{√2\sin t}

⇒√2\sint={-1}/m

Squaring both sides

⇒2\sin^2t={-1}/m^2

⇒1-\cos 2t=1/m^2

⇒1-1/m^2=\cos2t

{m^2-1}/{m^2}=\cos2t

Equation of tangent of curve is

y-2√{2}a\sint=m(x-a\cos2t)

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

⇒=mx-ma({m^{2}-1}/{m^{2}})-{2a}/{m}

⇒y=mx-a({m^{2}-1+2}/{m})

⇒y=mx-a(m+{1}/{m})

Go to Question:

Prev Question

Next Question