Answer
Verified
478.2k+ 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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE