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Home » Find the equation to the ellipse (referred to its axes as the axes of x and y respectively) which passes through the point (-3, 1) and has eccentricity √(2/5).
Find the equation to the ellipse (referred to its axes as the axes of x and y respectively) which passes through the point (-3, 1) and has eccentricity √(2/5).
CBSE | Class 11 | Ellipse | Exercise 26.1 | Maths | RD Sharma
Find the equation to the ellipse (referred to its axes as the axes of x and y respectively) which passes through the point (-3, 1) and has eccentricity √(2/5).

Given:

The point

    \[\left( -3,\text{ }1 \right)\]

Eccentricity

    \[=\text{ }\surd \left( 2/5 \right)\]

Now let us find the equation to the ellipse.

We know that the equation of the ellipse whose axes are x and y – axis is given as

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 43…….(i)

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 44

 

Now let us substitute equation (2) in equation (1), we get

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 45

It is given that the curve passes through the point

    \[\left( -3,\text{ }1 \right)\]

So by substituting the point in the curve we get,

    \[3{{\left( -\text{ }3 \right)}^{2}}~+\text{ }5{{\left( 1 \right)}^{2}}~=\text{ }3{{a}^{2}}\]

    \[3\left( 9 \right)\text{ }+\text{ }5\text{ }=\text{ }3{{a}^{2}}\]

Or,

    \[32\text{ }=\text{ }3{{a}^{2}}\]

    \[{{a}^{2}}~=\text{ }32/3\]

From equation (2)

    \[{{b}^{2}}~=\text{ }3{{a}^{2}}/5\]

    \[=\text{ }3\left( 32/3 \right)\text{ }/\text{ }5\]

Or,

    \[=\text{ }32/5\]

So now, the equation of the ellipse is given as:

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 46

    \[3{{x}^{2}}~+\text{ }5{{y}^{2}}~=\text{ }32\]

∴ The equation of the ellipse is

    \[3{{x}^{2}}~+\text{ }5{{y}^{2}}~=\text{ }32\]

i

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