Question

Find the equation of the ellipse which passes through the point  $\left( { - 3,1} \right)$ and eccentricity $\dfrac{{\sqrt 2 }}{5}$ , with x-axis and its major axis and center at the origin.

Answer 1

Hint: We know the equation of ellipse passes through points  $(x,y)$ and center at origin is $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ . Given the points  $(x,y) = ( - 3,1)$ and eccentricity $e = \dfrac{{\sqrt 2 }}{5}$ . Using this we find the values of  ${a^2}$ and ${b^2}$ . Substituting this in the equation of the ellipse we get the required solution. (Remember the eccentricity of the ellipse).

Complete step-by-step answer:

Equation of ellipse with center at origin is  $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$         --- (1)

Given, the points $(x,y) = ( - 3,1)$ . Eccentricity $e = \dfrac{{\sqrt 2 }}{5}$ . 

We know the eccentricity of an ellipse is $ \Rightarrow e = \sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}} $ .

 $ \Rightarrow \sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}}  = \dfrac{{\sqrt 2 }}{5}$ 

Squaring on both side, 

 $ \Rightarrow {\left( {\sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}} } \right)^2} = {\left( {\dfrac{{\sqrt 2 }}{5}} \right)^2}$ 

 $ \Rightarrow \dfrac{{{a^2} - {b^2}}}{{{a^2}}} = \dfrac{2}{{25}}$ 

Cross multiplying we get,

 $ \Rightarrow 25({a^2} - {b^2}) = 2{a^2}$ 

 $ \Rightarrow 25{a^2} - 25{b^2} = 2{a^2}$ 

Separating a and b terms,

 $ \Rightarrow 25{a^2} - 2{a^2} = 25{b^2}$ 

 $ \Rightarrow 23{a^2} = 25{b^2}$ 

Equating for  ${a^2}$ , we get,

 $ \Rightarrow {a^2} = \dfrac{{25{b^2}}}{{23}}$ .       ---- (2)

We need to find the value of ${a^2}$ and ${b^2}$ , and substitute the value of ${a^2}$  in equation (1) and $(x,y) = ( - 3,1)$ .

We get

 $ \Rightarrow \dfrac{{{{( - 3)}^2}}}{{{a^2}}} + \dfrac{{{1^2}}}{{{b^2}}} = 1$ 

 $ \Rightarrow \dfrac{9}{{{a^2}}} + \dfrac{1}{{{b^2}}} = 1$ 

Taking L.C.M and simplifying 

 $ \Rightarrow \dfrac{{9{b^2} + {a^2}}}{{{a^2}{b^2}}} = 1$ 

Cross multiplication,

 $ \Rightarrow 9{b^2} + {a^2} = {a^2}{b^2}$ 

Substitute,  ${a^2} = \dfrac{{25{b^2}}}{{23}}$ . We get, 

 $ \Rightarrow 9{b^2} + \dfrac{{25{b^2}}}{{23}} = \dfrac{{25{b^2}}}{{23}}{b^2}$ 

Taking L.C.M in the left hand side,

 $ \Rightarrow \dfrac{{207{b^2} + 25{b^2}}}{{23}} = \dfrac{{25}}{{23}}{b^4}$ 

 $ \Rightarrow \dfrac{{232{b^2}}}{{23}} = \dfrac{{25{b^4}}}{{23}}$ 

Canceling 23 on both sides, 

 $ \Rightarrow 232{b^2} = 25{b^4}$ 

Divide by  ${b^2}$ on both side and rearranging, we get

 $ \Rightarrow {b^2} = \dfrac{{232}}{{25}}$ .

Now to find ${a^2}$ in substitute in equation (2) 

 $ \Rightarrow {a^2} = \dfrac{{25}}{{23}} \times \dfrac{{232}}{{25}}$ 

 $ \Rightarrow {a^2} = \dfrac{{232}}{{23}}$ 

Now to find the equation of ellipse substituting  ${a^2}$ and ${b^2}$  values in $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ . We get

 $\dfrac{{{x^2}}}{{\left( {\dfrac{{232}}{{23}}} \right)}} + \dfrac{{{y^2}}}{{\left( {\dfrac{{232}}{{25}}} \right)}} = 1$ 

Further simplification,

 $ \Rightarrow \dfrac{{23{x^2}}}{{232}} + \dfrac{{25{y^2}}}{{232}} = 1$ 

 $ \Rightarrow 23{x^2} + {25y^2} = 232$  Is the required equation.

So, the correct answer is “ $ 23{x^2} + {25y^2} = 232$ .

Note: In above all we did is using the eccentricity converting the ${a^2}$  term into  ${b^2}$ or we can also convert  ${b^2}$ into ${a^2}$ . So that we can find the value of one term easily and then the other. While finding the equation of ellipse do not substitute the value of x and y (points). Remember the formula of eccentricity of the ellipse.