Statement - 1 : For all non-zero values of m, | vedclass

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(A) For the parabola $y^2 = 4ax$,the equation of the tangent with slope $m$ is $y = mx + a/m$.Here,$y^2 = -4x$,so $4a = -4$,which implies $a = -1$.Sub

Vedclass · Accepted Answer

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

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(A) For the parabola $y^2 = 4ax$,the equation of the tangent with slope $m$ is $y = mx + a/m$.Here,$y^2 = -4x$,so $4a = -4$,which implies $a = -1$.Substituting $a = -1$ into the tangent equation,we get $y = mx - 1/m$.Thus,Statement $- 1$ is true.For the parabola $y^2 = -4x$,the tangent at point $(x_1, y_1)$ is $yy_1 = -2(x + x_1)$.The $x$-intercept is found by setting $y = 0$,which gives $0 = -2(x + x_1)$,so $x = -x_1$.Since the parabola $y^2 = -4x$ lies in the region $x \leq 0$,the $x$-coordinate of any point on the parabola is non-positive $(x_1 \leq 0)$.Therefore,the $x$-intercept $x = -x_1$ is non-negative $(x \geq 0)$.Statement $- 2$ is true and it explains why the tangent behaves as described.

Statement $- 1 :$ For all non-zero values of $m$,$y = mx - 1/m$ is always a tangent to the parabola $y^2 = -4x$.
Statement $- 2 :$ Every tangent to the parabola $y^2 = -4x$ touches its axis at a point whose $x$-coordinate is non-negative.

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Statement $- 1 :$ For all non-zero values of $m$,$y = mx - 1/m$ is always a tangent to the parabola $y^2 = -4x$.Statement $- 2 :$ Every tangent to the parabola $y^2 = -4x$ touches its axis at a point whose $x$-coordinate is non-negative.