Match the conics in Column I with the stateme | vedclass

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(A) $(p)$ The line $hx+ky=1$ touches $x^2+y^2=4$ if the perpendicular distance from the origin $(0,0)$ to the line equals the radius $2$.
$\frac{|h(0

Vedclass · Accepted Answer

$A-p, B-s, t, C-r, D-q, s$

Vedclass · Answer

(A) $(p)$ The line $hx+ky=1$ touches $x^2+y^2=4$ if the perpendicular distance from the origin $(0,0)$ to the line equals the radius $2$.
$\frac{|h(0)+k(0)-1|}{\sqrt{h^2+k^2}}=2 \Rightarrow \frac{1}{\sqrt{h^2+k^2}}=2 \Rightarrow h^2+k^2=\frac{1}{4}$. This is a circle.
$(q)$ $|z+2|-|z-2|=\pm 3$. This represents the difference of distances from two fixed points $(\pm 2, 0)$ being a constant $3$. Since $3 
$(r)$ $x=\sqrt{3}\left(\frac{1-t^2}{1+t^2}ight), y=\frac{2 t}{1+t^2}$. Let $t=	an 	heta$. Then $x=\sqrt{3}\cos 2	heta$ and $y=\sin 2	heta$. Thus,$\frac{x^2}{3}+y^2=1$,which is an ellipse.
$(s)$ Eccentricity $e=1$ for a parabola,and $e>1$ for a hyperbola. Thus,$1 \leq e 
$(t)$ Let $z=x+iy$. $\operatorname{Re}(z+1)^2 = \operatorname{Re}((x+1+iy)^2) = (x+1)^2-y^2$. Given $(x+1)^2-y^2 = x^2+y^2+1 \Rightarrow x^2+2x+1-y^2 = x^2+y^2+1 \Rightarrow 2x = 2y^2 \Rightarrow x=y^2$. This is a parabola.
Matching: $A-p, B-s, t, C-r, D-q, s$.

Column $I$	Column $II$
$(A)$ Circle	$(p)$ The locus of the point $(h, k)$ for which the line $h x+k y=1$ touches the circle $x^2+y^2=4$
$(B)$ Parabola	$(q)$ Points $z$ in the complex plane satisfying $\|z+2\|-\|z-2\|= \pm 3$
$(C)$ Ellipse	$(r)$ Points of the conic have parametric representation $x=\sqrt{3}\left(\frac{1-t^2}{1+t^2}\right), y=\frac{2 t}{1+t^2}$
$(D)$ Hyperbola	$(s)$ The eccentricity of the conic lies in the interval $1 \leq x < \infty$
	$(t)$ Points $z$ in the complex plane satisfying $\operatorname{Re}(z+1)^2=\|z\|^2+1$

List-$I$	List-$II$
$(A)$ $\left[\frac{p}{2}\left(t+\frac{1}{t}\right), \frac{q}{2}\left(t-\frac{1}{t}\right)\right]$	$(I)$ parabola
$(B)$ $(p+q \cos \theta, r+q \sin \theta)$	$(II)$ circle
$(C)$ $(p+\lambda^2, q-\lambda)$	$(III)$ ellipse
	$(IV)$ hyperbola

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