NIMCET 2015 — Mathematics PYQ

Q: How do I solve NIMCET 2015 Mathematics questions?

Practice NIMCET 2015 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of NIMCET Mathematics questions?

NIMCET Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are NIMCET previous year papers available for free?

Yes. Mathem Solvex provides free access to NIMCET previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

NIMCET 2015 — Mathematics PYQ

NIMCET | Mathematics | 2015

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

Choose the correct answer:

Explanation

1. Find the Derivative $\frac{dy}{dx}$

To find the slope $m$ , we differentiate the parametric equations with respect to $t$ :

Given $x = a \cos 2t$ :

\frac{dx}{dt} = -2a \sin 2t = -4a \sin t \cos t

Given $y = 2\sqrt{2}a \sin t$ :

\frac{dy}{dt} = 2\sqrt{2}a \cos t

The slope $m$ is given by:

m = \frac{dy/dt}{dx/dt} = \frac{2\sqrt{2}a \cos t}{-4a \sin t \cos t}

m = \frac{2\sqrt{2}}{-4 \sin t} = -\frac{\sqrt{2}}{2 \sin t} = -\frac{1}{\sqrt{2} \sin t}

From this, we can find $\sin t$ in terms of $m$ :

\sin t = -\frac{1}{\sqrt{2}m}

2. Express Coordinates in Terms of $m$

Now we find $x$ and $y$ using $\sin t = -\frac{1}{\sqrt{2}m}$ .

For $y$ :

y = 2\sqrt{2}a \sin t = 2\sqrt{2}a \left( -\frac{1}{\sqrt{2}m} \right) = -\frac{2a}{m}

For $x$ :

Use the identity $\cos 2t = 1 - 2\sin^2 t$ :

x = a(1 - 2\sin^2 t) = a \left[ 1 - 2 \left( -\frac{1}{\sqrt{2}m} \right)^2 \right]

x = a \left[ 1 - 2 \left( \frac{1}{2m^2} \right) \right] = a \left( 1 - \frac{1}{m^2} \right)

x = a - \frac{a}{m^2}

3. Equation of the Tangent

Using the point-slope form $y - y_1 = m(x - x_1)$ :

y - \left( -\frac{2a}{m} \right) = m \left( x - \left( a - \frac{a}{m^2} \right) \right)

y + \frac{2a}{m} = mx - ma + \frac{ma}{m^2}

y + \frac{2a}{m} = mx - ma + \frac{a}{m}

Rearranging to solve for $y$ :

y = mx - ma + \frac{a}{m} - \frac{2a}{m}

y = mx - ma - \frac{a}{m}

y = mx - a \left( m + \frac{1}{m} \right)

Correct Option: (b)