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

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(D) The equation of the tangent to the parabola $y^2 = 16x$ (where $a=4$) is $y = mx + \frac{4}{m}$.This line is also a tangent to the circle $x^2 + y

Vedclass · Accepted Answer

$72$

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(D) The equation of the tangent to the parabola $y^2 = 16x$ (where $a=4$) is $y = mx + \frac{4}{m}$.This line is also a tangent to the circle $x^2 + y^2 = 8$ (radius $r = 2\sqrt{2}$).The perpendicular distance from the center $(0,0)$ to the line $mx - y + \frac{4}{m} = 0$ is equal to the radius:$\frac{|4/m|}{\sqrt{m^2 + 1}} = 2\sqrt{2} \implies \frac{16}{m^2(m^2+1)} = 8 \implies m^2(m^2+1) = 2$.Let $m^2 = t$,then $t^2 + t - 2 = 0 \implies (t+2)(t-1) = 0$. Since $t > 0$,$t = 1$,so $m = \pm 1$.Taking $m = 1$,the tangent is $y = x + 4$.The point of contact $R$ on the parabola $y^2 = 16x$ is $(\frac{a}{m^2}, \frac{2a}{m}) = (4, 8)$.The point of contact $Q$ on the circle $x^2 + y^2 = 8$ is given by the foot of the perpendicular from the origin to the line $x - y + 4 = 0$,which is $(-2, 2)$.Then $(QR)^2 = (4 - (-2))^2 + (8 - 2)^2 = 6^2 + 6^2 = 36 + 36 = 72$.

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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