Let A be a point on the x-axis. Common tangen | vedclass

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Question

$y = mx +\frac{4}{ m }$

$\frac{\left|\frac{4}{ m }ight|}{\sqrt{1+ m ^2}}=2 \sqrt{2} 	herefore m =\pm 1$

$y=\pm x \pm 4$. Point of contact on

Vedclass · Accepted Answer

$72$

Vedclass · Answer

$y = mx +\frac{4}{ m }$

$\frac{\left|\frac{4}{ m }ight|}{\sqrt{1+ m ^2}}=2 \sqrt{2} 	herefore m =\pm 1$

$y=\pm x \pm 4$. Point of contact on parabola

Let $m=1,\left(\frac{a}{m^2}, \frac{2 a}{m}ight)$

$R (4,8)$

Point of contact on circle $Q (-2,2)$

$	herefore( QR )^2=36+36=72$

Let $A$ be a point on the $x$-axis. Common tangents are drawn from $A$ to the curves $x^2+y^2=8$ and $y^2= 16x.$ If one of these tangents touches the two curves at $Q$ and $R$, then $( QR )^2$ is equal to

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