Question

The $x+y=2$ line meets the ellipse $x^2+2y^2=10$ in the points $A\&B$ . The point of intersection of tangents to the ellipse at this point is

• A. $\left(5,\frac{5}{2}\right)$
• B. $\left(\frac{5}{2},5\right)$
• C. $\left(\frac{5}{2},\frac{5}{2}\right)$
• D. $\left(5,5\right)$

Answer 1

The correct option is A $\left(5,\frac{5}{2}\right)$

The line $x+y=2$ meets the ellipse $x^2+2y^2=10$ in points A and B.

1) Equation of line is $x+y=2$

$\therefore x=2-y$

Put this value in equation of ellipse, we get,

${(2-y)}^2+2y^2=10$

$\therefore 4-4y+y^2+2y^2=10$

$\therefore 3y^2-4y+4=10$

$\therefore 3y^2-4y-6=0$

$\therefore y=\frac{-(-4)\pm \sqrt{{(-4)}^2-4\times 3\times -6}}{2\times 3}$

$\therefore y=\frac{4\pm \sqrt{16+72}}{6}$

$\therefore y=\frac{4\pm \sqrt{88}}{6}$

$\therefore y=\frac{4\pm 2\sqrt{22}}{6}$

$\therefore y=\frac{2\pm \sqrt{22}}{3}$

case 1) For $y=\frac{2+\sqrt{22}}{3}$,

$x=2-\frac{2+\sqrt{22}}{3}$

$\therefore x=\frac{6-2-\sqrt{22}}{3}$

$\therefore x=\frac{4-\sqrt{22}}{3}$

Thus, coordinates of point A are $(\frac{4-\sqrt{22}}{3},\frac{2+\sqrt{22}}{3})$

Case 2) For $y=\frac{2-\sqrt{22}}{3}$

$x=2-\frac{2-\sqrt{22}}{3}$

$\therefore x=\frac{6-2+\sqrt{22}}{3}$

$\therefore x=\frac{4+\sqrt{22}}{3}$

Thus, coordinates of point B are $(\frac{4+\sqrt{22}}{3},\frac{2-\sqrt{22}}{3})$

2) Equation of ellipse is $x^2+2y^2=10$

Dividing both sides by 10, we get,

$\frac{x^2}{10}+\frac{y^2}{5}=1$

$\frac{x^2}{{\left(\sqrt{10}\right)}^2}+\frac{y^2}{{\left(\sqrt{5}\right)}^2}=1$

$\therefore a=\sqrt{10}$ and $b=\sqrt{5}$

2) Equation of tangent to the ellipse from point A is given by,

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

$\therefore \frac{x}{10}\left(\frac{4-\sqrt{22}}{3}\right)+\frac{y}{5}\left(\frac{2+\sqrt{22}}{3}\right)=1$

$\therefore \frac{x(4-\sqrt{22})}{30}+\frac{y(2+\sqrt{22})}{15}=1$

$\therefore \frac{x(4-\sqrt{22})}{30}+\frac{y(4+2\sqrt{22})}{30}=1$

$\therefore x(4-\sqrt{22})+y(4+2\sqrt{22})=30$ (1)

Equation of tangent to the ellipse from point B is given by,

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

$\therefore \frac{x}{10}\left(\frac{4+\sqrt{22}}{3}\right)+\frac{y}{5}\left(\frac{2-\sqrt{22}}{3}\right)=1$

$\therefore \frac{x(4+\sqrt{22})}{30}+\frac{y(2-\sqrt{22})}{15}=1$

$\therefore \frac{x(4+\sqrt{22})}{30}+\frac{y(4-2\sqrt{22})}{30}=1$

$\therefore x(4+\sqrt{22})+y(4-2\sqrt{22})=30$ (2)

Adding equations (1) and (2), we get,

$8x+8y=60$

$\therefore 2x+2y=15$ (3)

Subtract equation (1) from (2), we get,

$\therefore 2\sqrt{22}x-4\sqrt{22}y=0$

$\therefore x-2y=0$ (4)

Adding equations (3) and (4), we get,

$3x=15$

$\therefore x=5$

Put this value in equation (4), we get,

$\therefore 5-2y=0$

$\therefore 2y=5$

$\therefore y=\frac{5}{2}$

Thus, coordinates of point of intersection of tangents is $(5,\frac{5}{2})$