Question

The tangent at the point $(4\cos \varphi ,\frac{16}{\sqrt{11}}\sin \varphi )$ to the ellipse $16x^2+11y^2=256$ is also a tangent to the circle $x^2+y^2-2x=15$. Find the absolute value of $\varphi$ in degrees.

• A. $30$
• B. $45$
• C. $60$
• D. $90$

Answer 1

The correct option is C $60$

The equation of the ellipse is $\frac{x^2}{16}+\frac{11y^2}{256}=1.$
Tangent at $(4\cos \varphi ,\frac{16}{\sqrt{11}}\sin \varphi )$ is given by,

$\frac{x}{16}4\cos \varphi +\frac{11}{256}y\cdot \frac{16}{\sqrt{11}}\sin \varphi =1.$

$i.e.\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$ or $4\cos \varphi x+\sqrt{11}\sin \varphi y=16$

If it is a tangent to the circle $x^2+y^2-2x-15=0,$ then perpendicular distance from the centre $(1,0)$ is equal to radius $\sqrt{(1+15)}=4$
or $\frac{4\cos \varphi -16}{\sqrt{(16{\cos }^2\varphi +11{\sin }^2\varphi )}}=4$ or ${(\cos \varphi -4)}^2=16{\cos }^2\varphi +11(1-{\cos }^2\varphi )$ or ${\cos }^2\varphi -8\cos \varphi +16=5{\cos }^2\varphi +11$ or $4{\cos }^2\varphi +8\cos \varphi -5=0$ or $(2\cos \varphi +5)(2\cos \varphi -1)=0$

$\therefore \cos \varphi =\frac{1}{2}\therefore \varphi =\pm {60}^{\circ }$