The equations of the tangents to the ellipse | vedclass

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(A) The given equation of the ellipse is $9x^2 + 16y^2 = 144$,which can be written as $\frac{x^2}{16} + \frac{y^2}{9} = 1$. Here,$a^2 = 16$ and $b^2 =

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$y = 3, x + y = 5$

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(A) The given equation of the ellipse is $9x^2 + 16y^2 = 144$,which can be written as $\frac{x^2}{16} + \frac{y^2}{9} = 1$. Here,$a^2 = 16$ and $b^2 = 9$. The equation of a tangent to the ellipse with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$,which is $y = mx \pm \sqrt{16m^2 + 9}$. Since the tangent passes through $(2, 3)$,we have $3 = 2m \pm \sqrt{16m^2 + 9}$. Rearranging gives $3 - 2m = \pm \sqrt{16m^2 + 9}$. Squaring both sides: $(3 - 2m)^2 = 16m^2 + 9$. $9 - 12m + 4m^2 = 16m^2 + 9$. $12m^2 + 12m = 0$. $12m(m + 1) = 0$. Thus,$m = 0$ or $m = -1$. For $m = 0$,the tangent is $y - 3 = 0(x - 2) \implies y = 3$. For $m = -1$,the tangent is $y - 3 = -1(x - 2) \implies y - 3 = -x + 2 \implies x + y = 5$. Therefore,the equations of the tangents are $y = 3$ and $x + y = 5$.

The equations of the tangents to the ellipse $9x^2 + 16y^2 = 144$ which pass through the point $(2, 3)$ are

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