Let an ellipse x^2a^2+y^2b^2=1

Question

Let an ellipse $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$

Accepted Answer

Since $a$\frac{{x}\^{2}}{{a}\^{2}}+\frac{{y}\^{2}}{{b}\^{2}}=1,$

we use the eccentricity formula

$e=\sqrt{1-\frac{{a}\^{2}}{{b}\^{2}}}.$

Given

$e=\frac{\sqrt{5}}{3},$

so

${e}\^{2}=\frac{5}{9}.$

Hence,

$1-\frac{{a}\^{2}}{{b}\^{2}}=\frac{5}{9}$

which gives

$\frac{{a}\^{2}}{{b}\^{2}}=1-\frac{5}{9}=\frac{4}{9}.$

Therefore,

${a}\^{2}=\frac{4}{9}{b}\^{2}.$

Now the ellipse passes through $(4,3)$, so it satisfies the equation:

$\frac{{4}\^{2}}{{a}\^{2}}+\frac{{3}\^{2}}{{b}\^{2}}=1.$

That is,

$\frac{16}{{a}\^{2}}+\frac{9}{{b}\^{2}}=1.$

Substitute ${a}\^{2}=\frac{4}{9}{b}\^{2}$:

$\frac{16}{\frac{4}{9}{b}\^{2}}+\frac{9}{{b}\^{2}}=1.$

Simplifying,

$16\cdot \frac{9}{4{b}\^{2}}+\frac{9}{{b}\^{2}}=1$

$\frac{36}{{b}\^{2}}+\frac{9}{{b}\^{2}}=1$

$\frac{45}{{b}\^{2}}=1$

so

${b}\^{2}=45.$

Then

${a}\^{2}=\frac{4}{9}\cdot 45=20.$

For an ellipse with major axis along the $y$-axis, the length of latus rectum is

$\frac{2{a}\^{2}}{b}.$

So,

$=\frac{2\cdot 20}{\sqrt{45}}=\frac{40}{3\sqrt{5}}.$

Rationalising,

$\frac{40}{3\sqrt{5}}\cdot \frac{\sqrt{5}}{\sqrt{5}}=\frac{8\sqrt{5}}{3}.$

Therefore, the correct answer is

$

Solutions · 1