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

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(A) 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

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

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(A) 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 is $m = 	an 60^{\circ} = \sqrt{3}$. The equation of the 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,we get $y = \sqrt{3}x \pm \sqrt{16(\sqrt{3})^2+1}$. $y = \sqrt{3}x \pm \sqrt{16(3)+1} = \sqrt{3}x \pm \sqrt{48+1} = \sqrt{3}x \pm \sqrt{49}$. $y = \sqrt{3}x \pm 7$. Thus,the equations are $\sqrt{3}x-y+7=0$ or $\sqrt{3}x-y-7=0$. Comparing with the given options,$\sqrt{3}x-y+7=0$ is the correct choice.

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

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