If any tangent drawn to the ellipse frac{x^2} | vedclass

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(C) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$,where $a^2 = 16$ and $b^2 = 9$.Any tangent to the ellipse is given by $y = mx

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$3 \leq \alpha \leq 4$

Vedclass · Answer

(C) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$,where $a^2 = 16$ and $b^2 = 9$.Any tangent to the ellipse is given by $y = mx \pm \sqrt{a^2m^2 + b^2}$,which is $y = mx \pm \sqrt{16m^2 + 9}$.This line is a tangent to the circle $x^2 + y^2 = \alpha^2$ if the perpendicular distance from the center $(0, 0)$ to the line is equal to the radius $\alpha$.The distance $d = \frac{|m(0) - 0 \pm \sqrt{16m^2 + 9}|}{\sqrt{m^2 + 1}} = \alpha$.Squaring both sides,we get $\alpha^2 = \frac{16m^2 + 9}{m^2 + 1}$.Let $t = m^2$,where $t \geq 0$. Then $\alpha^2 = \frac{16t + 9}{t + 1} = \frac{16(t + 1) - 7}{t + 1} = 16 - \frac{7}{t + 1}$.Since $t \geq 0$,the value of $t + 1$ ranges from $1$ to $\infty$.Thus,$\frac{7}{t + 1}$ ranges from $0$ to $7$.Therefore,$\alpha^2$ ranges from $16 - 7 = 9$ to $16 - 0 = 16$.So,$9 \leq \alpha^2 \leq 16$,which implies $3 \leq \alpha \leq 4$ (considering $\alpha > 0$).

List-$I$	List-$II$
$(i)$ The equation of the polar of $(-5, 1)$ with respect to $C$	$(A)$ $y = 0$
$(ii)$ The equation of the tangent at $(8, 0)$ to $C$	$(B)$ $y = 6$
$(iii)$ The equation of the normal at $(2, 6)$ to $C$	$(C)$ $x + y = 7$
$(iv)$ The equation of the diameter of $C$ through $(8, 12)$	$(D)$ $13x + 5y = 98$
	$(E)$ $x = 8$

If any tangent drawn to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ touches one of the circles $x^2 + y^2 = \alpha^2$,then the range of $\alpha$ is

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