The equation of the tangent to the ellipse x^ | vedclass

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(C) The given ellipse is $x^2 + 16y^2 = 16$,which can be written as $\frac{x^2}{16} + \frac{y^2}{1} = 1$.Here,$a^2 = 16$ and $b^2 = 1$.The slope of th

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$\sqrt{3}x - y \pm 7 = 0$

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(C) The given ellipse is $x^2 + 16y^2 = 16$,which can be written as $\frac{x^2}{16} + \frac{y^2}{1} = 1$.Here,$a^2 = 16$ and $b^2 = 1$.The slope of the tangent $m = 	an(60^\circ) = \sqrt{3}$.The equation of a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with slope $m$ is given by $y = mx \pm \sqrt{a^2m^2 + b^2}$.Substituting the values: $y = \sqrt{3}x \pm \sqrt{16(\sqrt{3})^2 + 1}$.$y = \sqrt{3}x \pm \sqrt{16(3) + 1}$.$y = \sqrt{3}x \pm \sqrt{48 + 1}$.$y = \sqrt{3}x \pm \sqrt{49}$.$y = \sqrt{3}x \pm 7$,which can be rewritten as $\sqrt{3}x - y \pm 7 = 0$.

The equation of the tangent to the ellipse $x^2 + 16y^2 = 16$ making an angle of $60^\circ$ with the $x$-axis is

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