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 “ $23{x^2} + {25y^2} = 232$ .

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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.