Question

If a chord of solpe $m$ of the circle $x^2+y^2=4$ touches the parabola $y^2=4x$, then
(where [.] represents G.I.F)

• A. least value of $\left[m^2\right]=0$
• B. least value of $\left[m\right]=0$
• C. $m\in (-\infty ,-\sqrt{\frac{\sqrt{2}-1}{2}})\cup (\sqrt{\frac{\sqrt{2}-1}{2}},\infty )$
• D. $m\in (-\sqrt{\frac{\sqrt{2}-1}{2}},\sqrt{\frac{\sqrt{2}-1}{2}})$

Answer 1

Answer: (A), (C)

The correct options are
A least value of $\left[m^2\right]=0$
C $m\in (-\infty ,-\sqrt{\frac{\sqrt{2}-1}{2}})\cup (\sqrt{\frac{\sqrt{2}-1}{2}},\infty )$

Equation of tangent to the parabola with slope $m$ will be
$y=mx+\frac{1}{m}$
$\because$ this will be the chord of the given circle
$\therefore$ length of perpendicular from $(0,0)$ will be less than radius
$\therefore \left|\frac{1}{m\sqrt{1+m^2}}\right|<2\Rightarrow 4m^4+4m^2-1>0\Rightarrow (m^2-\frac{\sqrt{2}-1}{2})(m^2+\frac{\sqrt{2}+1}{2})>0\Rightarrow m^2-\frac{\sqrt{2}-1}{2}>0\Rightarrow m\in (-\infty ,-\sqrt{\frac{\sqrt{2}-1}{2}})\cup (\sqrt{\frac{\sqrt{2}-1}{2}},\infty )$