Wait, why is the endpoint of the latus rectum $P(\lambda, 2\lambda)$? I thought it was $(\lambda, \lambda)$?

Great question! The latus rectum of the parabola $y^2 = 4\lambda x$ is the line $x = \lambda$. If you substitute $x = \lambda$ into the parabola's equation, you get $y^2 = 4\lambda^2$, which means $y = \pm 2\lambda$. Since the problem specifies the first quadrant, we take the positive root, giving us $P(\lambda, 2\lambda)$.

JEE Advanced 2020

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let $a, b$ and $λ$ be positive real numbers. Suppose $P$ is an end point of the latus rectum of the parabola $y^{2} = 4 λ x$ , and suppose the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ passes through the point $P$ . If the tangents to the parabola and the ellipse at the point $P$ are perpendicular to each other, then the eccentricity of the ellipse is

Q: How do I know if the ellipse is horizontal or vertical? Does it matter for the eccentricity formula?

It matters immensely! If $b^2 > a^2$, the major axis is vertical (along the $y$-axis), and the eccentricity is $e = \sqrt{1 - \frac{a^2}{b^2}}$. If $a^2 > b^2$, the major axis is horizontal (along the $x$-axis), and $e = \sqrt{1 - \frac{b^2}{a^2}}$. Always check the ratio of $a^2$ and $b^2$ first to determine the orientation.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the parabola

y^{2} = 4 λ x

. The latus rectum is the vertical line

x = λ

. The intersection point

P

is where this line meets the parabola in the first quadrant. Substitute

x = λ

into the parabola equation to find the coordinates of

P

Hint 2

Hint 2 (The Arsenal): The condition that the tangents are perpendicular means the product of their slopes must be

- 1

. Use implicit differentiation on both the parabola and the ellipse equations to find the general expression for the slope of the tangent at any point

(x, y)

Hint 3

Hint 3 (The Execution): Calculate the slope of the parabola tangent at

P

(

m_{1}

) and the slope of the ellipse tangent at

P

(

m_{2}

). Set

m_{1} \times m_{2} = - 1

to find the relationship between

a^{2}

and

b^{2}

, then use the eccentricity formula

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

(since

b^{2} > a^{2}

for a vertical ellipse).

Visualized Solution

Setup Parabola & Point $P$

Parabola: $y^{2} = 4 λ x$
Endpoint of latus rectum in 1st quadrant: $P (λ, 2 λ)$

Differentiating Parabola

Differentiating $y^{2} = 4 λ x$ with respect to $x$ :
$2 y \frac{d y}{d x} = 4 λ$
$\frac{d y}{d x} = \frac{2 λ}{y}$

Slope of Parabola Tangent

At $P (λ, 2 λ)$ :
$m_{1} = \frac{2 λ}{2 λ} = 1$

Setup Ellipse

Ellipse: $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
Passes through the same point $P (λ, 2 λ)$

Differentiating Ellipse

Differentiating ellipse equation:
$\frac{2 x}{a ^{2}} + \frac{2 y}{b ^{2}} \frac{d y}{d x} = 0$
$\frac{d y}{d x} = - \frac{b ^{2} x}{a ^{2} y}$

Slope of Ellipse Tangent

At $P (λ, 2 λ)$ :
$m_{2} = - \frac{b ^{2} λ}{a ^{2} ( 2 λ )} = - \frac{b ^{2}}{2 a ^{2}}$

Perpendicular Tangents

Tangents are perpendicular: $m_{1} \times m_{2} = - 1$

Finding Relation Between $a$ and $b$

$(1) \times (- \frac{b ^{2}}{2 a ^{2}}) = - 1$
$b^{2} = 2 a^{2} ⟹ \frac{a ^{2}}{b ^{2}} = \frac{1}{2}$
Since $b^{2} > a^{2}$ , the major axis is along the y-axis.

Eccentricity Formula

Eccentricity for vertical ellipse ( $b > a$ ):
$e = 1 - \frac{a ^{2}}{b ^{2}}$

Final Calculation

$e = 1 - \frac{1}{2}$
$e = \frac{1}{2}$

Conclusion

Key Takeaway: Perpendicular tangents at a common point link the derivatives of two curves.
Final Answer: $\frac{1}{2}$

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The parabola is defined by the equation

y^{2} = 4 λ x

, and the ellipse is defined by the equation:

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

The curves intersect at the endpoint of the latus rectum of the parabola in the first quadrant. For the parabola, the latus rectum is the vertical line

x = λ

Substituting

x = λ

into the parabola equation, we get

y^{2} = 4 λ^{2}

, which yields

y = 2 λ

. Thus, the point of intersection is

P (λ, 2 λ)

The Calculus of Tangents

To find the slope of the tangent to the parabola, we differentiate

y^{2} = 4 λ x

with respect to

x

2 y \frac{d y}{d x} = 4 λ \Rightarrow \frac{d y}{d x} = \frac{2 λ}{y}

At the point

P (λ, 2 λ)

, the slope

m_{1}

is calculated as:

m_{1} = \frac{2 λ}{2 λ} = 1

The Perpendicular Condition

The tangent to the ellipse at

P

is perpendicular to the tangent of the parabola. Since

m_{1} = 1

, the slope of the ellipse tangent

m_{2}

must satisfy

m_{1} \times m_{2} = - 1

, resulting in

m_{2} = - 1

Differentiating the ellipse equation

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

implicitly gives:

\frac{2 x}{a ^{2}} + \frac{2 y}{b ^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{b ^{2} x}{a ^{2} y}

Substituting the coordinates of

P (λ, 2 λ)

into this expression, we find:

m_{2} = - \frac{b ^{2} λ}{a ^{2} ( 2 λ )} = - \frac{b ^{2}}{2 a ^{2}}

The Final Synthesis

Equating the calculated slope

m_{2}

to the required value of

- 1

- \frac{b ^{2}}{2 a ^{2}} = - 1 \Rightarrow b^{2} = 2 a^{2} \Rightarrow \frac{a ^{2}}{b ^{2}} = \frac{1}{2}

The eccentricity

e

for an ellipse where

b^{2} > a^{2}

is given by the formula

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

. Substituting our ratio:

e = 1 - \frac{1}{2} = \frac{1}{2} = \frac{1}{2}

The final eccentricity of the ellipse is $\frac{1}{2}$ .

Common Traps & FAQs

Wait, why is the endpoint of the latus rectum

P (λ, 2 λ)

? I thought it was

(λ, λ)

Great question! The latus rectum of the parabola

y^{2} = 4 λ x

is the line

x = λ

. If you substitute

x = λ

into the parabola's equation, you get

y^{2} = 4 λ^{2}

, which means

y = \pm 2 λ

. Since the problem specifies the first quadrant, we take the positive root, giving us

P (λ, 2 λ)

Why did we differentiate implicitly? Can't we just use the tangent formula

y = m x + c

You absolutely could use the tangent formula, but implicit differentiation is a powerful, universal tool. It works for any curve, not just standard forms. It's the 'Swiss Army Knife' of calculus, and it helps us avoid messy algebraic substitutions early on.

How do I know if the ellipse is horizontal or vertical? Does it matter for the eccentricity formula?

It matters immensely! If

b^{2} > a^{2}

, the major axis is vertical (along the

y

-axis), and the eccentricity is

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

. If

a^{2} > b^{2}

, the major axis is horizontal (along the

x

-axis), and

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

. Always check the ratio of

a^{2}

and

b^{2}

first to determine the orientation.

Select Answer:

Visualized Solution

Setup Parabola & Point $P$

Differentiating Parabola

Slope of Parabola Tangent

Setup Ellipse

Differentiating Ellipse

Slope of Ellipse Tangent

Perpendicular Tangents

Finding Relation Between $a$ and $b$

Eccentricity Formula

Final Calculation

Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Calculus of Tangents

The Perpendicular Condition

The Final Synthesis

Similar Questions

If the maximum distance of normal to the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{b ^{2}} = 1, b < 2$ , from the origin is 1, then the eccentricity of the ellipse is:

Let $x = 4$ be a directrix to an ellipse whose centre is at the origin and its eccentricity is $\frac{1}{2}$ . If $P (1, β), β > 0$ is a point on this ellipse, then the equation of the normal to it at $P$ is

If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity $e$ of the ellipse satisfies :

If the tangent to the parabola $y^{2} = x$ at a point $(α, β), (β > 0)$ is also a tangent to the ellipse, $x^{2} + 2 y^{2} = 1$ , then $α$ is equal to :

Let $E$ be an ellipse whose axes are parallel to the co-ordinates axes, having its center at $(3, - 4)$ , one focus at $(4, - 4)$ and one vertex at $(5, - 4)$ . If $m x - y = 4, m > 0$ is a tangent to the ellipse $E$ , then the value of $5 m^{2}$ is equal to

If the tangents on the ellipse $4 x^{2} + y^{2} = 8$ at the points $(1, 2)$ and $(a, b)$ are perpendicular to each other, then $a^{2}$ is equal to :

The eccentricity of an ellipse whose centre is at the origin is $\frac{1}{2}$ . If one of its directrices is $x = - 4$ , then the equation of the normal to it at $(1, \frac{3}{2})$ is:

Select Answer:

Visualized Solution

Setup Parabola & Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Differentiating Parabola

Slope of Parabola Tangent

Setup Ellipse

Differentiating Ellipse

Slope of Ellipse Tangent

Perpendicular Tangents

Finding Relation Between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Eccentricity Formula

Final Calculation

Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Calculus of Tangents

The Perpendicular Condition

The Final Synthesis

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the maximum distance of normal to the ellipse 4x2​+b2y2​=1,b<2, from the origin is 1, then the eccentricity of the ellipse is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity e of the ellipse satisfies :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangent to the parabola y2=x at a point (α,β),(β>0) is also a tangent to the ellipse, x2+2y2=1, then α is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangents on the ellipse 4x2+y2=8 at the points (1,2) and (a,b) are perpendicular to each other, then a2 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The eccentricity of an ellipse whose centre is at the origin is 21​. If one of its directrices is x=−4, then the equation of the normal to it at (1,23​) is:

Setup Parabola & Point $P$

Finding Relation Between $a$ and $b$

If the maximum distance of normal to the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{b ^{2}} = 1, b < 2$ , from the origin is 1, then the eccentricity of the ellipse is:

If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity $e$ of the ellipse satisfies :

If the tangent to the parabola $y^{2} = x$ at a point $(α, β), (β > 0)$ is also a tangent to the ellipse, $x^{2} + 2 y^{2} = 1$ , then $α$ is equal to :

If the tangents on the ellipse $4 x^{2} + y^{2} = 8$ at the points $(1, 2)$ and $(a, b)$ are perpendicular to each other, then $a^{2}$ is equal to :

The eccentricity of an ellipse whose centre is at the origin is $\frac{1}{2}$ . If one of its directrices is $x = - 4$ , then the equation of the normal to it at $(1, \frac{3}{2})$ is: