Question

A tangent to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ intersects the tangent at the end of the major axis at,the points $P$ and $Q$. If the circle on $PQ$ as diameter passes through $R$, then $R$ may be

• A. $(0,\sqrt{5})$
• B. $(\sqrt{5},0)$
• C. $(3,2)$
• D. $(0,0)$

Answer 1

The correct option is B $(\sqrt{5},0)$

The given equation of ellipse is,

$\frac{x^2}{9}+\frac{y^2}{4}=1$

$\therefore a=3$ and $b=2$

The equation of tangent to the ellipse is given by,

$\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$

$\therefore \frac{x\left(3cos\theta \right)}{9}+\frac{y\left(2sin\theta \right)}{4}=1$

$\therefore \frac{xcos\theta }{3}+\frac{ysin\theta }{2}=1$

To find coordinates of point Q, put $x=3$ in above equation.

$\therefore \frac{3cos\theta }{3}+\frac{ysin\theta }{2}=1$

$\therefore cos\theta +\frac{ysin\theta }{2}=1$

$\therefore \frac{ysin\theta }{2}=1-cos\theta$

$\therefore y=\frac{2(1-cos\theta )}{sin\theta }$

$\therefore y=\frac{2\left(2{sin}^2\left(\frac{\theta }{2}\right)\right)}{2sin\left(\frac{\theta }{2}\right)cos\left(\frac{\theta }{2}\right)}$

$\therefore y=2tan\left(\frac{\theta }{2}\right)$

Thus, coordinates of point Q are $Q(3,2tan\left(\frac{\theta }{2}\right))$

Similarly, to find coordinates of point P, put $x=-3$ in equation of tangent.

$\therefore \frac{-3cos\theta }{3}+\frac{ysin\theta }{2}=1$

$\therefore -cos\theta +\frac{ysin\theta }{2}=1$

$\therefore \frac{ysin\theta }{2}=1+cos\theta$

$\therefore y=\frac{2(1+cos\theta )}{sin\theta }$

$\therefore y=\frac{2\left(2{cos}^2\left(\frac{\theta }{2}\right)\right)}{2sin\left(\frac{\theta }{2}\right)cos\left(\frac{\theta }{2}\right)}$

$\therefore y=2cot\left(\frac{\theta }{2}\right)$

Thus, coordinates of point P are $P(-3,2cot\left(\frac{\theta }{2}\right))$

These two points are end-points of diameter of circle. Thus, equation of circle is given by,

$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

$\therefore (x+3)(x-3)+(y-2cot\frac{\theta }{2})(y-2tan\frac{\theta }{2})=0$

$\therefore x^2-9+y^2-2y(tan\frac{\theta }{2}+cot\frac{\theta }{2})+4=0$

$\therefore x^2+y^2-2y(tan\frac{\theta }{2}+cot\frac{\theta }{2})-5=0$

To find point of intersection of this circle with x-axis, put $y=0$ in above equation.

$\therefore x^2-5=0$

$\therefore x^2=5$

$\therefore x=\pm \sqrt{5}$

Thus, coordinates are $(\sqrt{5},0)$ and $(-\sqrt{5},0)$

Thus, correct option is option (B)