Q.A tangent having slope – 1/2 to the ellipse $3x^2 + 4y^2 = 12$ intersects the X and Y axes in the points A and B respectively. If O is the origin, find the area of the triangle.

Differentiating the standard ellipse equation gives the slope $dy/dx=-\dfrac{b^2x}{a^2y}$ at any point, leading to the point-form tangent at $(x_1,y_1)$: $\dfrac{xx_1}{a^2}+\dfrac{yy_1}{b^2}=1$ — again the familiar "$x^2\to xx_1,\,y^2\to yy_1$" substitution seen for the parabola. Using the eccentric-angle parametrisation instead gives the equally useful form $\dfrac{x\cos\theta_1}{a}+\dfrac{y\sin\theta_1}{b}=1$.

For a tangent of a given slope $m$, matching $y=mx+c$ against the point-form tangent gives the condition of tangency: $c^2=a^2m^2+b^2$ — so the tangent can always be written $y=mx\pm\sqrt{a^2m^2+b^2}$ for ANY real slope $m$ (unlike the hyperbola, where a similar condition restricts which slopes are possible). The point of contact is $\left(-\dfrac{a^2m}{c},\dfrac{b^2}{c}\right)$. …