NIMCET 2008 — Mathematics PYQ

1. Equation of Tangent to the Parabola

The given parabola is $y^2=4ax$ where $a=1$.

The equation of any tangent to the parabola $y^2=4x$ in slope form ($m$) is:

Rearranging into general form:

2. Condition for Tangency to the Circle

The given circle is $(x-3{)}^2+y^2=9$.

• Center: $C(3,0)$

• Radius: $r=3$

For the line $m^{2}x-my+1=0$ to be tangent to the circle, the perpendicular distance from the center $(3,0)$ to the line must equal the radius ($3$).

The distance formula is $d=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$.

3. Solving for $m$

Squaring both sides:

4. Selecting the Correct Tangent

The problem specifies the tangent is above the x-axis.

• If we use $m=\frac{1}{\sqrt{3}}$:

  $$y=\frac{1}{\sqrt{3}}x+\frac{1}{1/\sqrt{3}}⟹y=\frac{x}{\sqrt{3}}+\sqrt{3}$$

  Multiplying by $\sqrt{3}$:

  $$\sqrt{3}y=x+3$$

• If we use $m=-\frac{1}{\sqrt{3}}$:

  $$\sqrt{3}y=-x-3⟹\sqrt{3}y=-(x+3)$$

Since the tangent is "above the x-axis," we look at the y-intercept. For $x=0$, the first equation gives $y=\sqrt{3}$ (positive/above) and the second gives $y=-\sqrt{3}$ (negative/below).

Therefore, the required equation is $\sqrt{3}y=x+3$.

Correct Option: (c) $\sqrt{3}y=x+3$