Statement 1: y = mx - frac{1}{m} is always a | vedclass

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(D) For the parabola $y^2 = -4ax$,the equation of the tangent with slope $m$ is $y = mx + \frac{a}{m}$.Here,$4a = 4$,so $a = 1$. The parabola is $y^2

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Statement $1$ is true,Statement $2$ is true,Statement $2$ is not a correct explanation of Statement $1$.

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(D) For the parabola $y^2 = -4ax$,the equation of the tangent with slope $m$ is $y = mx + \frac{a}{m}$.Here,$4a = 4$,so $a = 1$. The parabola is $y^2 = -4x$,so it opens to the left. The standard form is $y^2 = -4ax$ with $a=1$.The tangent equation is $y = mx - \frac{1}{m}$. Thus,Statement $1$ is true.For the tangent $y = mx - \frac{1}{m}$,the intersection with the axis $(y=0)$ is $0 = mx - \frac{1}{m}$,which gives $x = \frac{1}{m^2}$.Since $m^2 > 0$ for all $m 
eq 0$,$x = \frac{1}{m^2} > 0$. Thus,the abscissa is always positive (non-negative). Statement $2$ is true.Statement $2$ describes a property of the tangent,but it is not the logical derivation or explanation for the specific form of the tangent equation given in Statement $1$. Therefore,Statement $2$ is not the correct explanation for Statement $1$.

Statement $1$: $y = mx - \frac{1}{m}$ is always a tangent to the parabola $y^2 = -4x$ for all non-zero values of $m$.
Statement $2$: Every tangent to the parabola $y^2 = -4x$ will meet its axis at a point whose abscissa is non-negative.

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Statement $1$: $y = mx - \frac{1}{m}$ is always a tangent to the parabola $y^2 = -4x$ for all non-zero values of $m$.Statement $2$: Every tangent to the parabola $y^2 = -4x$ will meet its axis at a point whose abscissa is non-negative.