The equations of the tangents to the ellipse 9x^{2} + 16y^{2} = 144 from the point (2, 3) are

#### Options

y = 3, x = 5

x = 2, y = 3

x = 3, y = 2

x + y = 5, y = 3

#### Solution

\[ x + y = 5 , y = 3\]

\[9 x^2 + 16 y^2 = 144\]

\[ \Rightarrow \frac{x^2}{16} + \frac{y^2}{9} = 1\]

Equation of the tangent in case of an ellipse is given by

\[y = mx + \sqrt{a^2 m^2 + b^2}\]

\[ \Rightarrow y = mx + \sqrt{16 m^2 + 9} . . . (1) \]

\[\text{ Substituting }x=2\text{ and }y=3,\text{ we get: }\]

\[3 = 2m \pm \sqrt{16 m^2 + 9}\]

\[ \Rightarrow 3 - 2m = \sqrt{16 m^2 + 9}\]

On squaring both sides, we get:

\[ \left( 3 - 2m \right)^2 = \left( 16 m^2 + 9 \right)\]

\[ \Rightarrow 9 + 4 m^2 - 12m = \left( 16 m^2 + 9 \right)\]

\[ \Rightarrow 12 m^2 + 12m = 0\]

\[ \Rightarrow 12m\left( m + 1 \right) = 0\]

\[ \Rightarrow m = 0, - 1\]

Substituting values ofmin eq. (1), we get:

\[\text{ For }m = 0, y = 3\]

\[\text{ For }m = - 1, y = - x + 5\text{ or }x + y = 5\]