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Statement-1: The line $x-2 y=2$ meets the parabola, $y^2+2 x=0$ only at the point $(-2,-2)$.
Statement-2: The line $y=m x-\frac{1}{2 m}(m \neq 0)$ is tangent to the parabola, $y^2=-2 x$ at the point $\left(-\frac{1}{2 m^2},-\frac{1}{m}\right)$
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Statement-2: The line $y=m x-\frac{1}{2 m}(m \neq 0)$ is tangent to the parabola, $y^2=-2 x$ at the point $\left(-\frac{1}{2 m^2},-\frac{1}{m}\right)$
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The correct answer is:
Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for statement-1.
Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for statement-1.
Both statements are true and statement-2 is the correct explanation of statement-1
$\therefore \quad$ The straight line $y=m x+\frac{a}{m}$ is always a tangent to the parabola $y^2=4 a x$ for any value of $m$.
The co-ordinates of point of contact
$$
\left(\frac{a}{m^2}, \frac{2 a}{m}\right)
$$
$\therefore \quad$ The straight line $y=m x+\frac{a}{m}$ is always a tangent to the parabola $y^2=4 a x$ for any value of $m$.
The co-ordinates of point of contact
$$
\left(\frac{a}{m^2}, \frac{2 a}{m}\right)
$$
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