If the line ax + y = c touches both the curve | vedclass

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(B) The equation of the tangent to the parabola $y^2 = 4ax$ is $y = mx + \frac{a}{m}$. Here,$4a = 4\sqrt{2}$,so $a = \sqrt{2}$. Thus,the tangent is $y

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$\sqrt{2}$

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(B) The equation of the tangent to the parabola $y^2 = 4ax$ is $y = mx + \frac{a}{m}$. Here,$4a = 4\sqrt{2}$,so $a = \sqrt{2}$. Thus,the tangent is $y = mx + \frac{\sqrt{2}}{m}$.This line can be rewritten as $mx - y + \frac{\sqrt{2}}{m} = 0$.Since this line is also tangent to the circle $x^2 + y^2 = 1$,the perpendicular distance from the center $(0, 0)$ to the line must equal the radius $r = 1$.Using the distance formula: $\left| \frac{\frac{\sqrt{2}}{m}}{\sqrt{m^2 + (-1)^2}} \right| = 1$.Squaring both sides: $\frac{2}{m^2(m^2 + 1)} = 1 \Rightarrow m^4 + m^2 - 2 = 0$.Let $t = m^2$,then $t^2 + t - 2 = 0 \Rightarrow (t + 2)(t - 1) = 0$. Since $m^2 > 0$,we have $m^2 = 1$,so $m = \pm 1$.Substituting $m = 1$ into the tangent equation: $y = x + \sqrt{2} \Rightarrow x - y + \sqrt{2} = 0$. Comparing with $ax + y = c$,we get $a = -1$ and $c = -\sqrt{2}$,so $|c| = \sqrt{2}$.Substituting $m = -1$: $y = -x - \sqrt{2} \Rightarrow x + y = -\sqrt{2}$. Comparing with $ax + y = c$,we get $a = 1$ and $c = -\sqrt{2}$,so $|c| = \sqrt{2}$.

If the line $ax + y = c$ touches both the curves $x^2 + y^2 = 1$ and $y^2 = 4\sqrt{2}x$,then $|c|$ is equal to

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