JEE Main 2019 (12 April)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: An ellipse, with foci at (0, 2) and (0, -2) and minor axis of length 4, passes through which of the following points ?

Q: Wait, why is the minor axis $2a = 4$? I thought the minor axis was always $2b$?

That is a classic trap! The labels '$a$' and '$b$' are just variables. In a vertical ellipse, the semi-minor axis is along the $x$-axis, so its length is $2a$. In a horizontal ellipse, it would be $2b$. Don't memorize the letter; memorize the geometry: the minor axis is the shorter one, and in our case, it lies on the $x$-axis.

Q: Why did we use $a^2 = b^2 - (be)^2$ instead of the standard formula?

Great observation! The formula $a^2 = b^2(1-e^2)$ is just a rearrangement of the fundamental geometric truth: $b^2 = a^2 + c^2$ (where $c$ is the focal distance). Using $c=2$ and $a=2$ directly is often faster and less prone to errors than calculating the eccentricity $e$ first.

Select Answer:

Progressive Hints

Hint 1

First, visualize the geometry. Since the foci are at

(0, 2)

and

(0, - 2)

, the major axis lies along the

y

-axis. This means our ellipse is 'tall' rather than 'wide'. What does this imply about the relationship between the denominators in the standard equation

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

Hint 2

Recall the fundamental relationship for an ellipse:

c^{2} = ∣ a^{2} - b^{2} ∣

, where

c

is the distance from the center to the focus. Here,

c = 2

. You are given the minor axis length as

4

. Since the ellipse is vertical, the minor axis corresponds to the

x

-direction, so

2 a = 4

. Can you now find

a

and then use the focal relation to find

b

Hint 3

Once you have

a^{2}

and

b^{2}

, write the full equation of the ellipse. To check which point lies on it, simply substitute the

(x, y)

coordinates of each option into the equation

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

. The point that yields a sum of

1

is your winner.

Visualized Solution

Visualizing the Foci

Foci are given at $(0, 2)$ and $(0, - 2)$ .
Since the $x$ -coordinates are zero, they lie on the $y$ -axis.

Orientation of the Ellipse

The major axis always passes through the foci.
Therefore, the ellipse is vertical (major axis along the $y$ -axis).
Center is the midpoint of the foci: $(0, 0)$ .

Minor Axis Length

Length of the minor axis is given as $4$ .
For a vertical ellipse, the minor axis lies along the $x$ -axis.
Formula for minor axis length: $2 a = 4$ .

Calculating Semi-Minor Axis $a$

$2 a = 4$
Dividing by $2$ : $a = 2$ .
The semi-minor axis is $2$ .

Focal Distance Relation

For a vertical ellipse, the foci are at $(0, \pm b e)$ .
We are given the foci at $(0, \pm 2)$ .
Therefore, $b e = 2$ .

Eccentricity Formula

The fundamental relation for a vertical ellipse ( $b > a$ ) is:
$a^{2} = b^{2} (1 - e^{2})$
Expanding the bracket: $a^{2} = b^{2} - b^{2} e^{2}$
Which can be written as: $a^{2} = b^{2} - (b e)^{2}$

Substituting Known Values

We know $a = 2$ and $b e = 2$ .
Substitute these into $a^{2} = b^{2} - (b e)^{2}$ :
$(2)^{2} = b^{2} - (2)^{2}$

Solving for $b^{2}$

Evaluate the squares: $4 = b^{2} - 4$
Add $4$ to both sides: $b^{2} = 4 + 4$
$b^{2} = 8$

Forming the Equation

The standard equation of an ellipse centered at origin is:
$\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
We have $a^{2} = 2^{2} = 4$ and $b^{2} = 8$ .

Final Ellipse Equation

Substituting the values:
$\frac{x ^{2}}{4} + \frac{y ^{2}}{8} = 1$
This is the mathematical model of our vertical ellipse.

Testing the Given Points

We need to find which point lies on the ellipse.
A point $(x, y)$ lies on the ellipse if it satisfies $\frac{x ^{2}}{4} + \frac{y ^{2}}{8} = 1$ .
Let's test Option 4: $(2, 2)$ .

Substituting $(2, 2)$

Substitute $x = 2$ and $y = 2$ into the Left Hand Side (L.H.S).
L.H.S $= \frac{( 2 ) ^{2}}{4} + \frac{( 2 ) ^{2}}{8}$

Evaluating the Expression

$(2)^{2} = 2$ and $2^{2} = 4$ .
L.H.S $= \frac{2}{4} + \frac{4}{8}$
Simplify fractions: $\frac{1}{2} + \frac{1}{2}$
L.H.S $= 1$

Final Conclusion

Since L.H.S $=$ R.H.S, the point $(2, 2)$ satisfies the equation.
Therefore, the ellipse passes through $(2, 2)$ .
Correct Option: 4

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two points:

(0, 2)

and

(0, - 2)

. These are your anchors, your foci.

In the world of conic sections, the ellipse is defined by the sum of distances from any point on its boundary to these two foci being constant.

Because these foci share an

x

-coordinate of

0

, they sit perfectly on the

y

-axis. This tells us immediately that our ellipse is not lying flat; it is standing tall. It is a vertical ellipse.

Unlocking the Dimensions

Now, we are told the minor axis has a length of

4

. In our vertical orientation, the minor axis is the horizontal stretch of the ellipse.

The total length is

2 a = 4

, which simplifies to

a = 2

. This is our semi-minor axis, meaning the ellipse extends from

- 2

2

along the

x

-axis.

We know the focal distance

c

2

. The relationship between these parameters for a vertical ellipse is governed by the equation:

b^{2} = a^{2} + c^{2}

Substituting our values, we get:

b^{2} = 2^{2} + 2^{2} = 4 + 4 = 8

We have our parameters:

a^{2} = 4

and

b^{2} = 8

The Equation of Elegance

With

a^{2}

and

b^{2}

in hand, we can construct the standard equation of our ellipse centered at the origin:

\frac{x ^{2}}{4} + \frac{y ^{2}}{8} = 1

This equation is the DNA of our curve. Any point

(x, y)

that lies on this ellipse must satisfy this equality.

The Final Verification

We are looking for a point

(x, y)

that makes the left-hand side equal to

1

. Let's test the point

(2, 2)

Substituting these values into the equation, we get:

\frac{( 2 ) ^{2}}{4} + \frac{2 ^{2}}{8}

Calculating the squares, we obtain:

\frac{2}{4} + \frac{4}{8} = \frac{1}{2} + \frac{1}{2} = 1

The math holds up perfectly. We have successfully navigated the geometry, derived the equation, and verified our result. The point $(2, 2)$ lies on the ellipse.

Common Traps & FAQs

Wait, why is the minor axis

2 a = 4

? I thought the minor axis was always

2 b

That is a classic trap! The labels '

a

' and '

b

' are just variables. In a vertical ellipse, the semi-minor axis is along the

x

-axis, so its length is

2 a

. In a horizontal ellipse, it would be

2 b

. Don't memorize the letter; memorize the geometry: the minor axis is the shorter one, and in our case, it lies on the

x

-axis.

Why did we use

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

instead of the standard formula?

Great observation! The formula

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

is just a rearrangement of the fundamental geometric truth:

b^{2} = a^{2} + c^{2}

(where

c

is the focal distance). Using

c = 2

and

a = 2

directly is often faster and less prone to errors than calculating the eccentricity

e

first.

Can I just check the options without finding the full equation?

You could, but that's risky! If you don't have the equation, you're just guessing. By deriving the equation

\frac{x ^{2}}{4} + \frac{y ^{2}}{8} = 1

, you gain a complete understanding of the curve. It's the difference between guessing a password and knowing the algorithm.

Animated Solution for Mathematics - Conic Sections: An ellipse, with foci at (0, 2) and (0, -2) and minor axis of length 4, passes through which of the following points ?

Select Answer:

Visualized Solution

Visualizing the Foci

Orientation of the Ellipse

Minor Axis Length

Calculating Semi-Minor Axis $a$

Focal Distance Relation

Eccentricity Formula

Substituting Known Values

Solving for $b^{2}$

Forming the Equation

Final Ellipse Equation

Testing the Given Points

Substituting $(2, 2)$

Evaluating the Expression

Final Conclusion

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

Analyzing the Setup

Unlocking the Dimensions

The Equation of Elegance

The Final Verification

Similar Questions

Let the length of the latus rectum of an ellipse with its major axis along x-axis and centre at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?

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:

A hyperbola having the transverse axis of length $2$ has the same foci as that of the ellipse of $3 x^{2} + 4 y^{2} = 12$ , then this hyperbola does not pass through which of the following points?

The locus of the mid point of the line segment joining the point $(4, 3)$ and the points on the ellipse $x^{2} + 2 y^{2} = 4$ is an ellipse with eccentricity :

In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at $(0, 53)$ , then the length of its latus rectum is:

Axis of a parabola lies along x-axis. If its vertex and focus are at distances 2 and 4 respectively from the origin on the positive x-axis then which of the following points does not lie on it ?

Let $S$ and $S^{'}$ be the foci of the ellipse and $B$ be any one of the extremities of its minor axis. If $S^{'} B S$ is a right angled triangle with right angle at $B$ and area $(S^{'} B S) = 8$ sq. units, then the length of a latus rectum of the ellipse is :

If the length of the latus rectum of an ellipse is 4 units and the distance between a focus and its nearest vertex on the major axis is $\frac{3}{2}$ units, then its eccentricity is :

An ellipse has its center at $(1, - 2)$ , one focus at $(3, - 2)$ and one vertex at $(5, - 2)$ . Then the length of its latus rectum is :

Animated Solution for Mathematics - Conic Sections: An ellipse, with foci at (0, 2) and (0, -2) and minor axis of length 4, passes through which of the following points ?

Select Answer:

Visualized Solution

Visualizing the Foci

Orientation of the Ellipse

Minor Axis Length

Calculating Semi-Minor Axis .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Focal Distance Relation

Eccentricity Formula

Substituting Known Values

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

Forming the Equation

Final Ellipse Equation

Testing the Given Points

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (2​,2)

Evaluating the Expression

Final Conclusion

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

Analyzing the Setup

Unlocking the Dimensions

The Equation of Elegance

The Final Verification

Similar Questions

Let the length of the latus rectum of an ellipse with its major axis along x-axis and centre at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the mid point of the line segment joining the point (4,3) and the points on the ellipse x2+2y2=4 is an ellipse with eccentricity :

Axis of a parabola lies along x-axis. If its vertex and focus are at distances 2 and 4 respectively from the origin on the positive x-axis then which of the following points does not lie on it ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } An ellipse has its center at (1,−2), one focus at (3,−2) and one vertex at (5,−2). Then the length of its latus rectum is :

Calculating Semi-Minor Axis $a$

Solving for $b^{2}$

Substituting $(2, 2)$

The locus of the mid point of the line segment joining the point $(4, 3)$ and the points on the ellipse $x^{2} + 2 y^{2} = 4$ is an ellipse with eccentricity :

An ellipse has its center at $(1, - 2)$ , one focus at $(3, - 2)$ and one vertex at $(5, - 2)$ . Then the length of its latus rectum is :