Show that the line $x - y + 4 = 0$ is a tangent to the ellipse $x^2 + 3y^2 = 12$. Find the coordinates of the point of contact.
Equation of given line is $\;$ $x - y + 4 = 0$
i.e. $\;$ $y = x + 4$ $\;\;\; \cdots \; (1)$
Slope of given line $= m = 1$
Intercept of given line $= c = 4$
Equation of ellipse is $\;$ $x^2 + 3y^2 = 12$
i.e. $\;$ $\dfrac{x^2}{12} + \dfrac{y^2}{12 / 3} = 1$
i.e. $\;$ $\dfrac{x^2}{12} + \dfrac{y^2}{4} = 1$ $\;\;\; \cdots \; (2)$
The major axis of the ellipse given by equation $(2)$ is along the X axis.
$\therefore$ $\;$ Comparing equation $(2)$ with the tandard equation of ellipse $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\;$ gives
$a^2 = 12$ $\;$ and $\;$ $b^2 = 4$
Condition that the given line is a tangent to the ellipse is $\;$ $c^2 = a^2 m^2 + b^2$ $\;\;\; \cdots \; (3)$
Substituting the values of $c$, $a^2$, $m$ and $b^2$ in equation $(3)$ we have,
$\left(4\right)^2 = 12 \times \left(1\right)^2 + 4$
i.e. $\;$ $16 = 16$ $\;$ which is true.
$\therefore$ $\;$ The line given by equation $(1)$ is a tangent to the ellipse given by equation $(2)$.
The point of contact of the tangent with the ellipse is $\;$ $\left(\dfrac{-a^2 m}{c}, \dfrac{b^2}{c}\right)$
i.e. $\;$ $\left(\dfrac{-12 \times 1}{4}, \dfrac{4}{4}\right)$ $\;\;$ i.e. $\;$ $\left(-3, 1\right)$