NIMCET 2018 — Mathematics PYQ

Calculation

The equation of the circle can be written as (x-4)2+y2=42.
Comparing with the general form of a circle, we have center O(4,0) and radius r=4.
The equation of the given hyperbola can be written as $\frac{{\mathrm{x}}^2}{3^2}-\frac{{\mathrm{y}}^2}{2^2}=1.$
Comparing with the general form of a hyperbola, we have a=3 and b=2.
The equation of the tangent to this hyperbola will have the form:
$$\mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2-{\mathrm{b}}^2}$$

$$\Rightarrow \mathrm{y}=\mathrm{mx}\pm \sqrt{9{\mathrm{m}}^2-4}$$

Since this line is a tangent to the circle as well, we must have:Distance from the center O(4,0) of the circle to the tangent y=mx$\pm \sqrt{9{\mathrm{m}}^2-4}=$radius (r=4) of the circle.
Using the formula for th
$$\frac{\left|\mathrm{m}(4)+(-1)(0)\pm \sqrt{9{\mathrm{m}}^2-4}\right|}{\sqrt{{\mathrm{m}}^2+(-1{)}^2}}=4$$

$$\Rightarrow \left|\begin{array}{c}4\text{m}\pm \sqrt{9{\text{m}}^2-4}\end{array}\right|=4\sqrt{{\text{m}}^2+1}$$

On squaring both sides, we get:
$\Rightarrow 16{\mathrm{m}}^2\pm 8$m$\sqrt{9}{\mathrm{m}}^2-4+9{\mathrm{m}}^2-4=16{\mathrm{m}}^2+16$
$\Rightarrow \pm 8$m$\sqrt{9}{\mathrm{m}}^2-4=20-9{\mathrm{m}}^2$
Squaring again, we get

$\Rightarrow 64m^2(9m^2-4)=400+81m^4-360m^2$  

$\Rightarrow 576m^4-256m^2=400+81m^4-360m^2$  

$\Rightarrow 495m^4+104m^2-400=0$  

$\Rightarrow m^2=\frac{-104\pm \sqrt{104^2-4(495)(-400)}}{2\times 495}$  

$\Rightarrow m^2=\frac{-104\pm \sqrt{823216}}{990}$  

$\Rightarrow m^2=\frac{-104+896}{990}$  

$\Rightarrow m^2=\frac{4}{5}$  

\text{Discarding the negative value of } $m^2$.  

\text{Since the slope is positive, we get:}  

$\Rightarrow m=\frac{2}{\sqrt{5}}$  

\therefore \text{Equation of the tangent will be:}  

$y=mx\pm \sqrt{9m^2-4}$  

$y=\frac{2}{\sqrt{5}}x\pm \sqrt{9\left(\frac{4}{5}\right)-4}$  

$y=\frac{2}{\sqrt{5}}x\pm \frac{4}{\sqrt{5}}$  

$2x-\sqrt{5}y+4=0$