A tangent having slope m = -1/2 to the ellips | vedclass

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(A) The equation of the ellipse is $3x^2 + 4y^2 = 12$, which can be written as $\frac{x^2}{4} + \frac{y^2}{3} = 1$. Here $a^2 = 4$ and $b^2 = 3$.The e

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$4 	ext{ sq. units}$

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(A) The equation of the ellipse is $3x^2 + 4y^2 = 12$, which can be written as $\frac{x^2}{4} + \frac{y^2}{3} = 1$. Here $a^2 = 4$ and $b^2 = 3$.The equation of a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.Given $m = -1/2$, the equation becomes $y = -\frac{1}{2}x \pm \sqrt{4(-\frac{1}{2})^2 + 3} = -\frac{1}{2}x \pm \sqrt{1 + 3} = -\frac{1}{2}x \pm 2$.For the $X$-intercept $(A)$, set $y = 0$: $0 = -\frac{1}{2}x \pm 2 \implies x = \pm 4$. So $A = (\pm 4, 0)$.For the $Y$-intercept $(B)$, set $x = 0$: $y = \pm 2$. So $B = (0, \pm 2)$.The area of $\Delta AOB = \frac{1}{2} 	imes |x_A| 	imes |y_B| = \frac{1}{2} 	imes 4 	imes 2 = 4 	ext{ sq. units}$.

$A$ tangent having slope $m = -1/2$ to the ellipse $3x^2 + 4y^2 = 12$ intersects the $X$-axis and $Y$-axis at the points $A$ and $B$ respectively. If $O$ is the origin, then the area of $\Delta AOB$ is ...

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