JEE Advanced 2008

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2}), y_{1} < 0, y_{2} < 0$ , be the end points of the latus rectum of the ellipse $x^{2} + 4 y^{2} = 4$ . The equations of parabolas with latus rectum $P Q$ are

Q: Why do we assume the parabola's axis is vertical?

Great question! Look at the latus rectum $PQ$. We found that it is a horizontal line segment. By the definition of a parabola, the latus rectum is always perpendicular to the axis of symmetry. Since the latus rectum is horizontal, the axis of symmetry must be vertical. This is why we immediately know the parabola opens either upwards or downwards.

Q: Why are there two possible parabolas?

Think about the vertex. The focus $M$ is the midpoint of the latus rectum. The vertex $V$ lies on the axis of symmetry at a distance $a'$ from the focus. Since the parabola can open upwards (vertex below the focus) or downwards (vertex above the focus), we have two distinct geometric possibilities. Both satisfy the condition of having the same latus rectum.

Q: Can we use the general equation of a parabola?

You could, but it is much more elegant to use the standard form $(x - h)^2 = \pm 4a'(y - k)$. Since we already know the focus $M$ and the parameter $a'$, we can easily find the vertex $V$ and plug these values directly into the standard form. It saves time and reduces the chance of algebraic errors.

Select Answer:

* Multiple Correct

Progressive Hints

Hint 1

Hint 1: Visualize the geometry. The latus rectum of an ellipse is a horizontal line segment. The problem asks for parabolas that share this segment as their own latus rectum. This means the parabola's focus must be the midpoint of this segment, and its length must be the same as the ellipse's latus rectum.

Hint 2

Hint 2: Recall the definition of a parabola's latus rectum: it is a chord passing through the focus, perpendicular to the axis, with length

4 a^{'}

. You need to find the length of

P Q

from the ellipse and equate it to

4 a^{'}

to find the focal length

a^{'}

Hint 3

Hint 3: Start by finding the foci of the ellipse

x^{2} + 4 y^{2} = 4

. The latus rectum is the vertical line through the focus. Substitute

x = \pm a e

into the ellipse equation to find the coordinates of

P

and

Q

. Once you have

P

and

Q

, the midpoint is your parabola's focus, and the distance

P Q

gives you

4 a^{'}

Visualized Solution

$S t an d a r d F or m o f E l l i p se$

Given Ellipse: $x^{2} + 4 y^{2} = 4$
Divide by $4$ : $\frac{x ^{2}}{4} + \frac{y ^{2}}{1} = 1$
Comparing with standard form, $a = 2$ and $b = 1$ .

$E cce n t r i c i t y$

Eccentricity formula: $e = 1 - \frac{b ^{2}}{a ^{2}}$
Substitute values: $e = 1 - \frac{1}{4}$
$e = \frac{3}{2}$

$F oc i o f t h e E l l i p se$

Foci coordinates: $(\pm a e, 0)$
Substitute $a$ and $e$ : $(\pm 2 \cdot \frac{3}{2}, 0)$
Foci are at $(\pm 3, 0)$

$E n d p o in t so f L a t u s R ec t u m$

Latus rectum passes through $x = \pm 3$
Substitute into ellipse: $3 + 4 y^{2} = 4 \Rightarrow 4 y^{2} = 1 \Rightarrow y = \pm \frac{1}{2}$
Given $y_{1}, y_{2} < 0$ , we select $y = - \frac{1}{2}$
Endpoints: $P (- 3, - \frac{1}{2})$ and $Q (3, - \frac{1}{2})$

$P a r ab o l a P a r am e t er s$

Length of Latus Rectum $P Q = 23$
For the required parabola, Latus Rectum $= 4 a^{'}$
$4 a^{'} = 23 \Rightarrow a^{'} = \frac{3}{2}$

$F oc u so f t h e P a r ab o l a$

Focus of parabola $M$ is the midpoint of $P Q$
$M = (\frac{- 3 + 3}{2}, \frac{- \frac{1}{2} - \frac{1}{2}}{2})$
$M = (0, - \frac{1}{2})$

$V er t i ceso f t h e P a r ab o l a s$

Axis of parabola is perpendicular to $P Q$ (i.e., the y-axis)
Vertex $V$ lies on the axis at distance $a^{'}$ from focus $M$
Two possibilities: Parabola opens upwards or downwards
$V = (0, - \frac{1}{2} \pm a^{'}) = (0, - \frac{1}{2} \pm \frac{3}{2})$

$C a se 1 : U pw a r d P a r ab o l a S e t u p$

Case 1: Upward opening parabola
Focus $M$ is above the vertex, so $V_{1} = (0, \frac{- 1 - 3}{2})$
Standard equation: $(x - h)^{2} = 4 a^{'} (y - k)$
Substitute: $x^{2} = 23 (y - (\frac{- 1 - 3}{2}))$

$C a se 1 : S im pl i f i c a t i o n$

Expand the equation: $x^{2} = 23 y - 23 (\frac{- 1 - 3}{2})$
Simplify the constant term: $- 3 (- 1 - 3) = 3 + 3$
Final Equation 1: $x^{2} - 23 y = 3 + 3$

$C a se 2 : D o w n w a r d P a r ab o l a S e t u p$

Case 2: Downward opening parabola
Focus $M$ is below the vertex, so $V_{2} = (0, \frac{- 1 + 3}{2})$
Standard equation: $(x - h)^{2} = - 4 a^{'} (y - k)$
Substitute: $x^{2} = - 23 (y - (\frac{- 1 + 3}{2}))$

$C a se 2 : S im pl i f i c a t i o n$

Expand the equation: $x^{2} = - 23 y + 23 (\frac{- 1 + 3}{2})$
Simplify the constant term: $3 (- 1 + 3) = - 3 + 3$
Final Equation 2: $x^{2} + 23 y = 3 - 3$

$C o n c l u s i o n$

The two possible parabolas are:
1. $x^{2} - 23 y = 3 + 3$
2. $x^{2} + 23 y = 3 - 3$
These match options 2 and 3.

00:00 / 00:00

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

Solution Diagram

Analyzing the Ellipse

We begin by deconstructing the given ellipse equation:

x^{2} + 4 y^{2} = 4

. To reveal its standard form, we divide the entire equation by

4

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

Here, the semi-major axis is

a = 2

and the semi-minor axis is

b = 1

. The eccentricity

e

is calculated as follows:

e = 1 - \frac{b ^{2}}{a ^{2}} = 1 - \frac{1}{4} = \frac{3}{2}

Locating the Latus Rectum

The latus rectum passes through the foci, which are located at

(\pm a e, 0)

. Substituting our values, we find the foci at

(\pm 3, 0)

The latus rectum is defined by the vertical lines

x = \pm 3

. Substituting

x = 3

into the ellipse equation:

3 + 4 y^{2} = 4 \Rightarrow 4 y^{2} = 1 \Rightarrow y = \pm \frac{1}{2}

Since the problem specifies

y < 0

, we identify the endpoints of the segment as

P (- 3, - \frac{1}{2})

and

Q (3, - \frac{1}{2})

. The length of this segment

P Q

23

Defining the Parabola

For a parabola, the length of the latus rectum is

4 a^{'}

. Setting

4 a^{'} = 23

, we find the focal length:

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

The focus of the parabola is the midpoint of

P Q

, which is

M (0, - \frac{1}{2})

. Since the latus rectum is horizontal, the axis of the parabola is the vertical

y

-axis.

Deriving the Equations

The vertex

V

lies on the

y

-axis at a distance

a^{'}

from the focus. This leads to two distinct cases based on the direction of the parabola.

Case 1: Upward Opening Parabola The vertex is

V_{1} = (0, - \frac{1}{2} - \frac{3}{2})

. Using the standard form

x^{2} = 4 a^{'} (y - k)

x^{2} = 23 (y - (\frac{- 1 - 3}{2}))

Expanding this, we obtain the first equation: $x^{2} - 23 y = 3 + 3$

Case 2: Downward Opening Parabola The vertex is

V_{2} = (0, - \frac{1}{2} + \frac{3}{2})

. Using the standard form

x^{2} = - 4 a^{'} (y - k)

x^{2} = - 23 (y - (\frac{- 1 + 3}{2}))

Simplifying this, we obtain the second equation: $x^{2} + 23 y = 3 - 3$

Common Traps & FAQs

Why do we assume the parabola's axis is vertical?

Great question! Look at the latus rectum

P Q

. We found that it is a horizontal line segment. By the definition of a parabola, the latus rectum is always perpendicular to the axis of symmetry. Since the latus rectum is horizontal, the axis of symmetry must be vertical. This is why we immediately know the parabola opens either upwards or downwards.

Why are there two possible parabolas?

Think about the vertex. The focus

M

is the midpoint of the latus rectum. The vertex

V

lies on the axis of symmetry at a distance

a^{'}

from the focus. Since the parabola can open upwards (vertex below the focus) or downwards (vertex above the focus), we have two distinct geometric possibilities. Both satisfy the condition of having the same latus rectum.

Can we use the general equation of a parabola?

You could, but it is much more elegant to use the standard form

(x - h)^{2} = \pm 4 a^{'} (y - k)

. Since we already know the focus

M

and the parameter

a^{'}

, we can easily find the vertex

V

and plug these values directly into the standard form. It saves time and reduces the chance of algebraic errors.

Animated Solution for Mathematics - Conic Sections: Let $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2}), y_{1} < 0, y_{2} < 0$ , be the end points of the latus rectum of the ellipse $x^{2} + 4 y^{2} = 4$ . The equations of parabolas with latus rectum $P Q$ are

Select Answer:

Visualized Solution

$S t an d a r d F or m o f E l l i p se$

$E cce n t r i c i t y$

$F oc i o f t h e E l l i p se$

$E n d p o in t so f L a t u s R ec t u m$

$P a r ab o l a P a r am e t er s$

$F oc u so f t h e P a r ab o l a$

$V er t i ceso f t h e P a r ab o l a s$

$C a se 1 : U pw a r d P a r ab o l a S e t u p$

$C a se 1 : S im pl i f i c a t i o n$

$C a se 2 : D o w n w a r d P a r ab o l a S e t u p$

$C a se 2 : S im pl i f i c a t i o n$

$C o n c l u s i o n$

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

Analyzing the Ellipse

Locating the Latus Rectum

Defining the Parabola

Deriving the Equations

Similar Questions

Let $P$ be a parabola with vertex $(2, 3)$ and directrix $2 x + y = 6$ . Let an ellipse $E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ of eccentricity $\frac{1}{2}$ pass through the focus of the parabola $P$ . Then the square of the length of the latus rectum of $E$ , is

The ellipse $x^{2} + 4 y^{2} = 4$ is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point $(4, 0)$ . Then the equation of the ellipse is:

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 :

Let a tangent to the curve $y^{2} = 24 x$ meet the curve $x y = 2$ at the points A and B. Then the mid points of such line segments AB lie on a parabola with the

For some $θ \in (0, \frac{π}{2})$ , if the eccentricity of the hyperbola, $x^{2} - y^{2} sec^{2} θ = 10$ is $5$ times the eccentricity of the ellipse, $x^{2} sec^{2} θ + y^{2} = 5$ , then the length of the latus rectum of the ellipse, is :

The length of the latus rectum of a parabola, whose vertex and focus are on the positive $x$ -axis at a distance $R$ and $S (> R)$ respectively from the origin, is:

Select Answer:

Visualized Solution

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

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

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case1:UpwardParabolaSetup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case1:Simplification

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case2:DownwardParabolaSetup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case2:Simplification

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

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

Analyzing the Ellipse

Locating the Latus Rectum

Defining the Parabola

Deriving the Equations

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a tangent to the curve y2=24x meet the curve xy=2 at the points A and B. Then the mid points of such line segments AB lie on a parabola with the

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The length of the latus rectum of a parabola, whose vertex and focus are on the positive x-axis at a distance R and S(>R) respectively from the origin, is:

$S t an d a r d F or m o f E l l i p se$

$E cce n t r i c i t y$

$F oc i o f t h e E l l i p se$

$E n d p o in t so f L a t u s R ec t u m$

$P a r ab o l a P a r am e t er s$

$F oc u so f t h e P a r ab o l a$

$V er t i ceso f t h e P a r ab o l a s$

$C a se 1 : U pw a r d P a r ab o l a S e t u p$

$C a se 1 : S im pl i f i c a t i o n$

$C a se 2 : D o w n w a r d P a r ab o l a S e t u p$

$C a se 2 : S im pl i f i c a t i o n$

$C o n c l u s i o n$

Let a tangent to the curve $y^{2} = 24 x$ meet the curve $x y = 2$ at the points A and B. Then the mid points of such line segments AB lie on a parabola with the

The length of the latus rectum of a parabola, whose vertex and focus are on the positive $x$ -axis at a distance $R$ and $S (> R)$ respectively from the origin, is: