Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Find the length of the latus rectum, the eccentricity and the coordinates of the foci of the ellipse.
$5{{x}^{2}}+4{{y}^{2}}=1$.

seo-qna
Last updated date: 24th Jul 2024
Total views: 450.9k
Views today: 7.50k
Answer
VerifiedVerified
450.9k+ views
Hint: First convert the given equation in the $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ form. Then use the formulae ${{e}^{2}}=1-\dfrac{{{a}^{2}}}{{{b}^{2}}}$, \[F\left( 0,\pm be \right)\] and \[L.R.=\dfrac{2{{a}^{2}}}{b}\] to find the values of eccentricity, foci and latus rectum respectively.

Complete step-by-step solution -
To solve the above problem we will write the given equation first,
$5{{x}^{2}}+4{{y}^{2}}=1$
As the given equation is of ellipse therefore we have to convert the equation it its standard form which is shown below,
Standard Form of Ellipse:
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
Therefore we can write given equations by shifting the constants in the denominator as shown below,
$\dfrac{{{x}^{2}}}{\dfrac{1}{5}}+\dfrac{{{y}^{2}}}{\dfrac{1}{4}}=1$
Now the denominators can be expressed in terms of square as,
$\therefore \dfrac{{{x}^{2}}}{{{\left( \dfrac{1}{\sqrt{5}} \right)}^{2}}}+\dfrac{{{y}^{2}}}{{{\left( \dfrac{1}{2} \right)}^{2}}}=1$
Now if we compare the above equation with standard form we will get,
$a=\dfrac{1}{\sqrt{5}}$, and $b=\dfrac{1}{2}$ ……………………………………………… (1)
As, b > a therefore the ellipse is vertical and for well understanding we have shown it below,
As shown in figure A and B are the two foci of the ellipse.
As the ellipse is vertical therefore we have to use all the formula of Vertical ellipse.
Now we will find the required terminologies one by one,
Eccentricity:
The formula for eccentricity of vertical ellipse is given by,
${{e}^{2}}=1-\dfrac{{{a}^{2}}}{{{b}^{2}}}$
Now if we put the values of equation (1) in the above formula we will get,
$\therefore {{e}^{2}}=1-\dfrac{{{\left( \dfrac{1}{\sqrt{5}} \right)}^{2}}}{{{\left( \dfrac{1}{2} \right)}^{2}}}$
$\therefore {{e}^{2}}=1-\dfrac{\dfrac{1}{5}}{\dfrac{1}{4}}$
$\therefore {{e}^{2}}=1-\dfrac{1}{5}\times \dfrac{4}{1}$
$\therefore {{e}^{2}}=1-\dfrac{4}{5}$
$\therefore {{e}^{2}}=\dfrac{5-4}{5}$
\[\therefore {{e}^{2}}=\dfrac{1}{5}\]
Taking square root on both sides we will get,
\[\therefore e=\dfrac{1}{\sqrt{5}}\] ……………………………………………… (2)
Foci:
As ellipse has two foci namely A and B we will write the formulae of both.
Foci of vertical ellipse is given by,
A (0, be) And B (0,-be)
If we put the values of equation (1) and equation (2) in above points we will get,
\[\therefore \] \[A\left( 0,\dfrac{1}{2}\times \dfrac{1}{\sqrt{5}} \right)\] And \[B\left( 0,-\dfrac{1}{2}\times \dfrac{1}{\sqrt{5}} \right)\]
\[\therefore A\left( 0,\dfrac{1}{2\sqrt{5}} \right)\] And \[B\left( 0,-\dfrac{1}{2\sqrt{5}} \right)\]
Therefore Foci of ellipse is \[\left( 0,\pm \dfrac{1}{2\sqrt{5}} \right)\]
Latus Rectum:
Formula for latus rectum of vertical ellipse is given by,
\[L.R.=\dfrac{2{{a}^{2}}}{b}\]
If we put the values of equation (1) in the above equation we will get,
\[\therefore L.R.=\dfrac{2{{\left( \dfrac{1}{\sqrt{5}} \right)}^{2}}}{\left( \dfrac{1}{2} \right)}\]
\[\therefore L.R.=\dfrac{2\times \dfrac{1}{5}}{\left( \dfrac{1}{2} \right)}\]
\[\therefore L.R.=2\times \dfrac{1}{5}\times \dfrac{2}{1}\]
\[\therefore L.R.=\dfrac{4}{5}\]
Therefore the value of Latus Rectum is \[\dfrac{4}{5}\].

Note: Students generally forgot to convert the equation in the standard form and consider the values of ‘a’ and ‘b’ directly as 5 and 4 respectively in this case i.e. $5{{x}^{2}}+4{{y}^{2}}=1$ and the whole answer becomes wrong therefore always convert the equation into the standard form.