Question

Find the equations to the tangents to the ellipse $4x^2+3y^2=5$ which are parallel to the straight line $y=3x+7$.
Find also the coordinates of the points of contact of the tangents which are inclined at ${60}^o$ to the axis of $x$.

Answer 1

Given ellipse ${4x}^2+{3y}^2=5⟹\frac{x^2}{{\left(\sqrt{\frac{5}{4}}\right)}^2}+\frac{y^2}{{\left(\sqrt{\frac{5}{3}}\right)}^2}=1$

Slope of a tangent at a point $(x_1,y_1)$ is $\frac{-b^2{x}_1}{y_1{a}^2}$

$\frac{-\left(\frac{5}{3}\right)x_1}{\left(\frac{5}{4}\right)y_1}=3$

$⟹-4x_1=9y_1$

$⟹x_1=\frac{-9}{4}{y}_1⟶1$

we know that point $(x_1,y_1)$ lies on ellipse

$4x_1^2+3y_1^2=5$

$⟹4\left(\frac{81}{16}{y}_1^2\right)+3y_1^2=5$

$⟹y_1^2=\frac{5\times 4}{93}$

$⟹y_1=\pm \sqrt{\frac{20}{93}}$

$x_1=\mp \frac{9}{4}\sqrt{\frac{20}{93}}$

Equation of tangents are $4x{x}_1+3y{y}_1=5$

$\mp 4x\cdot \frac{9}{4}\sqrt{\frac{20}{93}}+3y(\pm \sqrt{\frac{20}{93}})=5$

$⟹\mp 9x\pm 3y=5\sqrt{\frac{93}{20}}$

b) Given $\frac{{-4x}_1}{{3y}_1}=\sqrt{3}⟹x_1=\frac{-3\sqrt{3}{y}_1}{4}$

$4x_1^2+3y_1^2=5⟹(\frac{27}{4}+3)$

$y_1^2=5⟹y_1=\pm \sqrt{\frac{20}{39}}$

$x_1=\mp \frac{3}{4}\sqrt{\frac{20}{13}}$

Required points of contacts are $(\frac{-3}{4}\sqrt{\frac{20}{39}},\sqrt{\frac{20}{13}}),(\frac{+3}{4}\sqrt{\frac{20}{13}},-\sqrt{\frac{20}{39}})$