Statement -1: The line x - 2y = 2 meets the p | vedclass

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(B) For Statement $-2$: The parabola is $y^2 = -2x$,so $4a = -2$,which implies $a = -1/2$. The condition for the line $y = mx + c$ to be a tangent to

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Statement $-1$ is true; Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$.

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(B) For Statement $-2$: The parabola is $y^2 = -2x$,so $4a = -2$,which implies $a = -1/2$. The condition for the line $y = mx + c$ to be a tangent to $y^2 = 4ax$ is $c = a/m$. Here,$c = (-1/2)/m = -1/(2m)$. The point of contact is $(a/m^2, 2a/m) = (-1/(2m^2), -1/m)$. Thus,Statement $-2$ is true.For Statement $-1$: The line is $x = 2y + 2$. Substituting into $y^2 + 2x = 0$,we get $y^2 + 2(2y + 2) = 0$,which simplifies to $y^2 + 4y + 4 = 0$,or $(y + 2)^2 = 0$. This gives $y = -2$. Substituting $y = -2$ into $x = 2y + 2$,we get $x = 2(-2) + 2 = -2$. Thus,the line meets the parabola only at $(-2, -2)$. This is consistent with the line being a tangent at that point. Thus,Statement $-1$ is true and Statement $-2$ is the correct explanation.

Statement $-1:$ The line $x - 2y = 2$ meets the parabola $y^2 + 2x = 0$ only at the point $(-2, -2).$
Statement $-2:$ The line $y = mx - \frac{1}{2m}$ $(m \neq 0)$ is tangent to the parabola $y^2 = -2x$ at the point $\left( -\frac{1}{2m^2}, -\frac{1}{m} \right).$

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Statement $-1:$ The line $x - 2y = 2$ meets the parabola $y^2 + 2x = 0$ only at the point $(-2, -2).$Statement $-2:$ The line $y = mx - \frac{1}{2m}$ $(m \neq 0)$ is tangent to the parabola $y^2 = -2x$ at the point $\left( -\frac{1}{2m^2}, -\frac{1}{m} \right).$