Question

If the tangent at the point $\left(4\cos \varphi ,\frac{16}{\sqrt{11}}\sin \varphi \right)$ to the ellipse $16x^2+11y^2=256$ is also a tangent to the circle $x^2+y^2-2x=15$, then the value of $\varphi$ is:

• A. $\pm \frac{\pi }{2}$
• B. $\pm \frac{\pi }{4}$
• C. $\pm \frac{\pi }{3}$
• D. $\pm \frac{\pi }{6}$

Answer 1

The correct option is C $\pm \frac{\pi }{3}$

$4\times 16cos\varphi x+11\times \frac{16}{\sqrt{11}}\times sin\varphi =256$

$4cos\varphi x+\sqrt{11}sin\varphi y=16$------(1)

$(x-1{)}^2+y^2=16$

$(x-1{)}^2+{\left(\frac{16-4cos\varphi x}{sin\varphi \sqrt{11}}\right)}^2=16$ (From $e{q}^n$(1))

$11si{n}^2\varphi x^2+11si{n}^2\varphi (-2x)+256+16co{s}^2\varphi x^2-128cos\varphi x=15\times 11si{n}^2\varphi$

$\because$ line is tangent to the circle

$\therefore D=0$ (of the quadratic $e{q}^n$)

$(-128cos\varphi -22si{n}^2\varphi {)}^2-4\times (16co{s}^2\varphi +11si{n}^2\varphi )\times (256-165si{n}^2\varphi )=0$

$4co{s}^2\varphi +8cos\varphi -5=0$

$cos\varphi =+\frac{1}{2},cos\varphi =-2-5$ (not possible)

$\therefore \varphi =\frac{\pi }{3}$ and

$\varphi =-\frac{\pi }{3}$