The equations of the tangents to the ellipse | vedclass

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(A) For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the equation of a tangent with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.Here, $a^2 =

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$x - \sqrt{3}y \pm \sqrt{43} = 0$

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(A) For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the equation of a tangent with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.Here, $a^2 = 16$, $b^2 = 9$, and the inclination is $30^\circ$, so $m = 	an(30^\circ) = \frac{1}{\sqrt{3}}$.Substituting these values into the formula:$y = \frac{1}{\sqrt{3}}x \pm \sqrt{16(\frac{1}{\sqrt{3}})^2 + 9}$$y = \frac{x}{\sqrt{3}} \pm \sqrt{\frac{16}{3} + 9}$$y = \frac{x}{\sqrt{3}} \pm \sqrt{\frac{16 + 27}{3}}$$y = \frac{x}{\sqrt{3}} \pm \sqrt{\frac{43}{3}}$Multiply by $\sqrt{3}$:$\sqrt{3}y = x \pm \sqrt{43}$$x - \sqrt{3}y \pm \sqrt{43} = 0$.

The equations of the tangents to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ making an inclination of $30^\circ$ with the major axis are

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