NIMCET 2023 — Mathematics PYQ

Step 1: Find the derivative $\frac{dy}{dx}$ (Slope of the tangent)

The given parametric equations of the curve are:

$$x=a\cos 2t$$

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

Differentiating $x$ and $y$ with respect to $t$:

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

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

Now, find the slope of the tangent $m$:

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

Simplifying the expression:

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

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

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

Step 2: Express the point $(x,y)$ in terms of $m$

We know that:

$$\cos 2t=1-2{\sin }^{2}t$$

Substitute $\sin t=-\frac{1}{\sqrt{2}m}$ into the parametric coordinates:

• For $x$:

  $$x=a\cos 2t=a(1-2{\sin }^{2}t)$$

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

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

• For $y$:

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

  $$y=-\frac{2a}{m}$$

Step 3: Write the equation of the tangent

The point-slope form of the equation of a line passing through $(x_1,y_1)$ with slope $m$ is:

$$y-y_1=m(x-x_1)$$

Substitute $(x_1,y_1)=\left(a\left(1-\frac{1}{m^2}\right),-\frac{2a}{m}\right)$:

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

$$y+\frac{2a}{m}=mx-am\left(1-\frac{1}{m^2}\right)$$

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

Rearranging the terms to isolate $y$:

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

$$y=mx-am-\frac{a}{m}$$

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

Correct Answer:

Correct Option: B