Find the equations of the tangents to the ell | vedclass

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(A) The given ellipse is $3x^{2} + 4y^{2} = 12$,which can be written as $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$. Here,$a^{2} = 4$ and $b^{2} = 3$. The

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$x - 2y \pm 4 = 0$

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(A) The given ellipse is $3x^{2} + 4y^{2} = 12$,which can be written as $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$. Here,$a^{2} = 4$ and $b^{2} = 3$. The slope of the line $y + 2x = 4$ is $m_{1} = -2$. Since the tangent is perpendicular to this line,its slope $m$ must satisfy $m 	imes (-2) = -1$,so $m = \frac{1}{2}$. The equation of a tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^{2}m^{2} + b^{2}}$. Substituting the values,$y = \frac{1}{2}x \pm \sqrt{4(\frac{1}{4}) + 3}$. $y = \frac{1}{2}x \pm \sqrt{1 + 3} = \frac{1}{2}x \pm 2$. Multiplying by $2$,we get $2y = x \pm 4$,which simplifies to $x - 2y \pm 4 = 0$.

Find the equations of the tangents to the ellipse $3x^{2} + 4y^{2} = 12$ which are perpendicular to the line $y + 2x = 4$.

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