Let a circle of radius 4 be concentric to the ellipse 15 x^2+19 y^2=285.

Question

Let a circle of radius 4 be concentric to the ellipse $15 x^{2}+19 y^{2}=285$. Then the common tangents are inclined to the minor axis of the ellipse at the angle :

Accepted Answer

We have, equation of ellipse : $15{x}^{2}+19{y}^{2}=285$

or $\frac{{x}^{2}}{19}+\frac{{y}^{2}}{15}=1$

Let the coordinate of center of circle be $(0,0)$.

Equation of circle is ${x}^{2}+{y}^{2}=16$

Equation of tangent of ellipse is

$\begin{gathered} y=mx\pm \sqrt{19{m}^{2}+15} \ mx-y\pm \sqrt{19{m}^{2}+15}=0 \end{gathered}$

It is also tangent to the circle ${x}^{2}+{y}^{2}=16$

Perpendicular distance from center of circle to tangent $=4$

$\frac{|0-0\pm \sqrt{19{m}^{2}+15}|}{\sqrt{{m}^{2}+1}}=4$

On squaring both side, we get

$\begin{gathered} 19{m}^{2}+15=16{m}^{2}+16 \ 3{m}^{2}=1 \ {m}^{2}=\frac{1}{3} \ m=\pm \frac{1}{\sqrt{3}} \end{gathered}$

$	an 	heta =\frac{1}{\sqrt{3}}\Rightarrow 	heta =\frac{\pi }{6}X$$\frac{\pi }{3}$ with $Y$-axis

Hence, the common tangents are inclined to the minor axis of the ellipse at an angle of $\frac{\pi }{3}$.

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