Question

Find the equations to the tangents to the ellipse 4x² + 3y² = 5 which are parallel to the straight line y = 3x + 7. Also find points of tangency.

Answer 1

The equations of the tangents are $y = 3x + \frac{\sqrt{465}}{6}$ and $y = 3x - \frac{\sqrt{465}}{6}$. The points of tangency are $\left(-\frac{3\sqrt{465}}{62}, \frac{2\sqrt{465}}{93}\right)$ and $\left(\frac{3\sqrt{465}}{62}, -\frac{2\sqrt{465}}{93}\right)$.

Answer 2

The standard equation of an ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The given ellipse $4x^2 + 3y^2 = 5$ can be rewritten as $\frac{x^2}{5/4} + \frac{y^2}{5/3} = 1$, so $a^2 = 5/4$ and $b^2 = 5/3$. The given line $y = 3x + 7$ has a slope $m=3$. Tangents parallel to this line will also have a slope $m=3$. The equation of a tangent to the ellipse with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$. Substituting the values of $a^2$, $b^2$, and $m$, we find the equations of the tangents: $y = 3x \pm \sqrt{\frac{5}{4}(3^2) + \frac{5}{3}} = 3x \pm \sqrt{\frac{45}{4} + \frac{5}{3}} = 3x \pm \sqrt{\frac{135+20}{12}} = 3x \pm \sqrt{\frac{155}{12}} = 3x \pm \frac{\sqrt{155 \times 3}}{\sqrt{12 \times 3}} = 3x \pm \frac{\sqrt{465}}{6}$. The points of tangency $(x_0, y_0)$ for a tangent $y = mx + c$ to the ellipse are given by $x_0 = -\frac{a^2m}{c}$ and $y_0 = \frac{b^2}{c}$. For $c = \frac{\sqrt{465}}{6}$, $x_0 = -\frac{5/4 \times 3}{\sqrt{465}/6} = -\frac{15/4}{\sqrt{465}/6} = -\frac{15}{4} \times \frac{6}{\sqrt{465}} = -\frac{90}{4\sqrt{465}} = -\frac{45}{2\sqrt{465}} = -\frac{45\sqrt{465}}{2 \times 465} = -\frac{45\sqrt{465}}{930} = -\frac{3\sqrt{465}}{62}$. And $y_0 = \frac{5/3}{\sqrt{465}/6} = \frac{5}{3} \times \frac{6}{\sqrt{465}} = \frac{30}{3\sqrt{465}} = \frac{10}{\sqrt{465}} = \frac{10\sqrt{465}}{465} = \frac{2\sqrt{465}}{93}$. For $c = -\frac{\sqrt{465}}{6}$, the points of tangency are $\left(\frac{3\sqrt{465}}{62}, -\frac{2\sqrt{465}}{93}\right)$.