NIMCET 2018 Question Paper with solutions

Let the equation of line y - mx + c ... (i)
If Eq. (i) is tangent to the hyperbola ${\mathit{x}}^{29}-\frac{{\mathit{y}}^2}{4}$ = 1
∴ c = ± $\sqrt{9{\mathit{m}}^2-4}$ [a = 3 , b = 2]
So, equation of line (i) is
y = mx ± $\sqrt{9{\mathit{m}}^2-}$ ... (ii)
It is also tangent to the circle ${\mathit{x}}^2+{\mathit{y}}^2-8\mathit{x}=0$ here, centre
c - (4, 0) and radius (r) = 4
Perpendicular distance from centre to the tangent = Radius
∴ $\frac{\left|4\mathit{m}\pm \sqrt{9{\mathit{m}}^2-4}\right|}{\sqrt{1+{\mathit{m}}^2}}$ = 4
${\left(4\mathit{m}\pm \sqrt{9{\mathit{m}}^2-}\right)}^2$ = 16 $\left(1+{\mathit{m}}^2\right)$ ... (iii) [squaring both sides]
By solving Eq. (iii), we get
m = $\frac{2}{\sqrt{5}}$
Put the value of m = $\frac{2}{\sqrt{5}}$ in eq . (i)
∴ y = $\frac{2}{\sqrt{5}}\mathit{x}$± $\sqrt{9\times \frac{4}{5}}$ - 4
= $\frac{2}{\sqrt{5}}\mathit{x}\pm \frac{\sqrt{16}}{5}$ = $\frac{2\mathit{x}}{\sqrt{5}}\pm \frac{4}{\sqrt{5}}$
⇒ $\sqrt{5}$y = 2x ± 4
⇒ 2x ± 4 - $\sqrt{5}$y = 0
⇒ 2x - $\mathit{v}5\mathit{y}$ + 4 = 0