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Animated Solution for Mathematics - Conic Sections: Equation of the ellipse whose axes are the axes of coordinates and which passes through the point $(- 3, 1)$ and has eccentricity $\frac{2}{5}$ is

Q: Wait, how do I know if $a$ is the semi-major axis or if the ellipse is vertical? Do I need to check the orientation first?

That is a brilliant question! In the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we define $a$ and $b$ as the semi-axes. If the math yields $a^2 > b^2$, it is horizontal; if $b^2 > a^2$, it is vertical. The beauty of the standard equation is that it is agnostic to the orientation. You do not need to assume anything; just solve for the parameters, and the equation will reveal the shape itself.

Q: Why do we use $b^2 = a^2(1-e^2)$? Is it always $a^2$ on the right side?

Yes, in the standard form where $a$ is the semi-major axis, this is the definition. If the ellipse were vertical, we would swap the roles, but the relationship $minor^2 = major^2(1-e^2)$ remains the universal truth. Stick to the standard form and let the variables define the shape. The eccentricity $e$ is always less than 1 for an ellipse, ensuring $1-e^2$ is positive, which keeps our $b^2$ value physically meaningful.

Select Answer:

Progressive Hints

Hint 1

Visualize the ellipse centered at the origin. The eccentricity

e

is the 'DNA' of the ellipse—it dictates how much it deviates from a circle. Before you touch the algebra, ask yourself: does this ellipse stretch along the x-axis or the y-axis? The standard form

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

is your canvas.

Hint 2

You have two unknowns:

a^{2}

and

b^{2}

. You need two independent equations. One comes from the point

(- 3, 1)

lying on the curve, and the other comes from the definition of eccentricity:

b^{2} = a^{2} (1 - e^{2})

. This is your golden key.

Hint 3

Start by substituting

e^{2} = \frac{2}{5}

into the eccentricity relation to find the ratio between

b^{2}

and

a^{2}

. Then, plug the point

(- 3, 1)

into the standard equation. You will have a system of two equations in two variables. Solve for

a^{2}

and

b^{2}

—the algebra will simplify beautifully!

Visualized Solution

Standard Equation of the Ellipse

Let the equation of the ellipse be:
$\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
The axes of the ellipse coincide with the coordinate axes, so the center is at $(0, 0)$ .

The Eccentricity Relation

Given eccentricity $e = \frac{2}{5}$
The relationship between $a$ , $b$ , and $e$ is:
$b^{2} = a^{2} (1 - e^{2})$

Relating $a^{2}$ and $b^{2}$

Substitute $e^{2} = \frac{2}{5}$ into the relation:
$b^{2} = a^{2} (1 - \frac{2}{5})$
$b^{2} = \frac{3 a ^{2}}{5}$ or $5 b^{2} = 3 a^{2}$

The Passing Point $(- 3, 1)$

The ellipse is specified to pass through the point $P (- 3, 1)$ .
This point must satisfy the standard equation of the ellipse.

Substituting $(- 3, 1)$

Substitute $x = - 3$ and $y = 1$ into $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ :
$\frac{( - 3 ) ^{2}}{a ^{2}} + \frac{1 ^{2}}{b ^{2}} = 1 ⟹ \frac{9}{a ^{2}} + \frac{1}{b ^{2}} = 1$

Eliminating $b^{2}$

We have: $\frac{9}{a ^{2}} + \frac{1}{b ^{2}} = 1$ and $b^{2} = \frac{3 a ^{2}}{5}$ .
Substitute $b^{2}$ into the point equation:
$\frac{9}{a ^{2}} + \frac{1}{\frac{3 a ^{2}}{5}} = 1 ⟹ \frac{9}{a ^{2}} + \frac{5}{3 a ^{2}} = 1$

Finding the Value of $a^{2}$

Find a common denominator for $\frac{9}{a ^{2}} + \frac{5}{3 a ^{2}} = 1$ :
$\frac{27}{3 a ^{2}} + \frac{5}{3 a ^{2}} = 1 ⟹ \frac{32}{3 a ^{2}} = 1$
$3 a^{2} = 32 ⟹ a^{2} = \frac{32}{3}$

Finding the Value of $b^{2}$

Substitute $a^{2} = \frac{32}{3}$ back into $b^{2} = \frac{3 a ^{2}}{5}$ :
$b^{2} = \frac{3}{5} \times \frac{32}{3}$
$b^{2} = \frac{32}{5}$

The Final Equation of the Ellipse

Substitute $a^{2} = \frac{32}{3}$ and $b^{2} = \frac{32}{5}$ into the standard form:
$\frac{x ^{2}}{\frac{32}{3}} + \frac{y ^{2}}{\frac{32}{5}} = 1$
$\frac{3 x ^{2}}{32} + \frac{5 y ^{2}}{32} = 1 ⟹ 3 x^{2} + 5 y^{2} = 32$

Conclusion and Key Takeaway

The correct option is Option 4: $3 x^{2} + 5 y^{2} - 32 = 0$ .
Key Takeaway: For any ellipse centered at the origin, the standard form $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ combined with the eccentricity relation $b^{2} = a^{2} (1 - e^{2})$ is a powerful starting point.

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The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Solution Diagram

Analyzing the Setup

Since the axes of the ellipse coincide with the coordinate axes and the center is at the origin

(0, 0)

, we invoke the standard equation of an ellipse:

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

This equation serves as the bedrock of our geometric construction, ensuring that every point

(x, y)

on the curve maintains the required balance.

The DNA of the Ellipse

Eccentricity

Eccentricity, denoted by

e

, defines how much our ellipse deviates from a circle. We are given

e = \frac{2}{5}

, which implies

e^{2} = \frac{2}{5}

The fundamental relationship connecting the semi-major axis

a

, the semi-minor axis

b

, and the eccentricity

e

is:

b^{2} = a^{2} (1 - e^{2})

Substituting our value for

e^{2}

, we find:

b^{2} = a^{2} (1 - \frac{2}{5}) = a^{2} (\frac{3}{5})

This yields the critical constraint:

5 b^{2} = 3 a^{2}

The Intersection of Constraints

The point

(- 3, 1)

lies on the ellipse, meaning it must satisfy the standard equation. Substituting

x = - 3

and

y = 1

into the equation, we get:

\frac{( - 3 ) ^{2}}{a ^{2}} + \frac{1 ^{2}}{b ^{2}} = 1 \Rightarrow \frac{9}{a ^{2}} + \frac{1}{b ^{2}} = 1

Now, we substitute

b^{2} = \frac{3 a ^{2}}{5}

into this point constraint:

\frac{9}{a ^{2}} + \frac{1}{\frac{3 a ^{2}}{5}} = 1 \Rightarrow \frac{9}{a ^{2}} + \frac{5}{3 a ^{2}} = 1

To solve for

a^{2}

, we find a common denominator:

\frac{27 + 5}{3 a ^{2}} = 1 \Rightarrow \frac{32}{3 a ^{2}} = 1

This leads us to the value

3 a^{2} = 32

, or

a^{2} = \frac{32}{3}

Final Calculation

With

a^{2} = \frac{32}{3}

, we determine

b^{2}

using our previous relation:

b^{2} = \frac{3}{5} \times (\frac{32}{3}) = \frac{32}{5}

Substituting these values back into the standard form

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

, we obtain:

\frac{x ^{2}}{\frac{32}{3}} + \frac{y ^{2}}{\frac{32}{5}} = 1

Simplifying this expression, we arrive at the final equation of the ellipse:

\frac{3 x ^{2}}{32} + \frac{5 y ^{2}}{32} = 1

Multiplying by

32

, we reach the final result:

$3 x^{2} + 5 y^{2} = 32$

Common Traps & FAQs

Wait, how do I know if

a

is the semi-major axis or if the ellipse is vertical? Do I need to check the orientation first?

That is a brilliant question! In the standard form

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

, we define

a

and

b

as the semi-axes. If the math yields

a^{2} > b^{2}

, it is horizontal; if

b^{2} > a^{2}

, it is vertical. The beauty of the standard equation is that it is agnostic to the orientation. You do not need to assume anything; just solve for the parameters, and the equation will reveal the shape itself.

Why do we use

b^{2} = a^{2} (1 - e^{2})

? Is it always

a^{2}

on the right side?

Yes, in the standard form where

a

is the semi-major axis, this is the definition. If the ellipse were vertical, we would swap the roles, but the relationship

min o r^{2} = maj o r^{2} (1 - e^{2})

remains the universal truth. Stick to the standard form and let the variables define the shape. The eccentricity

e

is always less than 1 for an ellipse, ensuring

1 - e^{2}

is positive, which keeps our

b^{2}

value physically meaningful.

Is there a faster way than solving the system of equations? This feels like a lot of algebra.

While it might feel like a lot, this is the most robust method. You are essentially finding the unique ellipse that satisfies two geometric constraints: its 'flatness' (eccentricity) and its 'position' (the point). There is no shortcut that bypasses these two fundamental constraints. Embrace the algebra—it is the language of the geometry!

Animated Solution for Mathematics - Conic Sections: Equation of the ellipse whose axes are the axes of coordinates and which passes through the point $(- 3, 1)$ and has eccentricity $\frac{2}{5}$ is

Select Answer:

Visualized Solution

Standard Equation of the Ellipse

The Eccentricity Relation

Relating $a^{2}$ and $b^{2}$

The Passing Point $(- 3, 1)$

Substituting $(- 3, 1)$

Eliminating $b^{2}$

Finding the Value of $a^{2}$

Finding the Value of $b^{2}$

The Final Equation of the Ellipse

Conclusion and Key Takeaway

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The DNA of the Ellipse

The Intersection of Constraints

Final Calculation

Similar Questions

An ellipse has eccentricity $\frac{1}{2}$ and one focus at the point $P (\frac{1}{2}, 1)$ . Its one directrix is the common tangent, nearer to the point $P$ , to the circle $x^{2} + y^{2} = 1$ and the hyperbola $x^{2} - y^{2} = 1$ . The equation of the ellipse, in the standard form, is.........

The eccentricity of an ellipse, with its centre at the origin, is $1/2$ . If one of the directrices is $x = 4$ , then the equation of the ellipse is:

An ellipse is drawn by taking a diameter of the circle $(x - 1)^{2} + y^{2} = 1$ as its semi-minor axis and a diameter of the circle $x^{2} + (y - 2)^{2} = 4$ as semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is:

Let a hyperbola passes through the focus of the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ . The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse, also the product of eccentricities of given ellipse and hyperbola is 1, then

Let the eccentricity of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ , be $\frac{1}{4}$ . If this ellipse passes through the point $(- 4 \frac{2}{5}, 3)$ , then $a^{2} + b^{2}$ is equal to :

The locus of mid-points of the line segments joining $(- 3, - 5)$ and the points on the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1$ is :

Let the maximum area of the triangle that can be inscribed in the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{4} = 1, a > 2$ , having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be $63$ . Then the eccentricity of the ellipse is :

The ellipse $E_{1} : \frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_{2}$ passing through the point $(0, 4)$ circumscribes the rectangle $R$ . The eccentricity of the ellipse $E_{2}$ is

Select Answer:

Visualized Solution

Standard Equation of the Ellipse

The Eccentricity Relation

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

The Passing Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (−3,1)

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (−3,1)

Eliminating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

The Final Equation of the Ellipse

Conclusion and Key Takeaway

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The DNA of the Ellipse

The Intersection of Constraints

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The eccentricity of an ellipse, with its centre at the origin, is 1/2. If one of the directrices is x=4, then the equation of the ellipse is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the eccentricity of an ellipse a2x2​+b2y2​=1,a>b, be 41​. If this ellipse passes through the point (−452​​,3), then a2+b2 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of mid-points of the line segments joining (−3,−5) and the points on the ellipse 4x2​+9y2​=1 is :

Relating $a^{2}$ and $b^{2}$

The Passing Point $(- 3, 1)$

Substituting $(- 3, 1)$

Eliminating $b^{2}$

Finding the Value of $a^{2}$

Finding the Value of $b^{2}$

The eccentricity of an ellipse, with its centre at the origin, is $1/2$ . If one of the directrices is $x = 4$ , then the equation of the ellipse is:

Let the eccentricity of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ , be $\frac{1}{4}$ . If this ellipse passes through the point $(- 4 \frac{2}{5}, 3)$ , then $a^{2} + b^{2}$ is equal to :

The locus of mid-points of the line segments joining $(- 3, - 5)$ and the points on the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1$ is :