JEE Main 2022 (28 July Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: For the hyperbola $H : x^{2} - y^{2} = 1$ and the ellipse $E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b > 0$ , let the (1) eccentricity of $E$ be reciprocal of the eccentricity of $H$ , and (2) the line $y = \frac{5}{2} x + K$ be a common tangent of $E$ and $H$ . Then $4 (a^{2} + b^{2})$ is equal to ______.

Q: Can I just use the discriminant method instead of memorizing these tangency formulas?

Absolutely! You can always substitute $y = mx + K$ into the curve equation and set the discriminant $D = 0$. However, in JEE Advanced, time is your most precious resource. Memorizing these standard conditions allows you to jump straight to the algebraic manipulation, saving you valuable minutes for more complex problems.

Enter Numerical Value:

Visualized Solution

Analyze Hyperbola $H$

Hyperbola $H : x^{2} - y^{2} = 1$
This is a rectangular hyperbola where $a_{H} = 1$ and $b_{H} = 1$

Calculate Eccentricity $e_{H}$

Eccentricity of $H$ : $e_{H} = 1 + \frac{b _{H}^{2}}{a _{H}^{2}}$
Substituting $a_{H} = 1, b_{H} = 1$ : $e_{H} = 1 + 1 = 2$

Relate $e_{E}$ and $e_{H}$

Given: $e_{E} = \frac{1}{e _{H}}$
Substituting $e_{H} = 2$ : $e_{E} = \frac{1}{2}$

Link $a^{2}$ and $b^{2}$ for Ellipse

For Ellipse $E$ : $e_{E}^{2} = 1 - \frac{b ^{2}}{a ^{2}}$
Substituting $e_{E}^{2} = (\frac{1}{2})^{2} = \frac{1}{2}$ :
$\frac{1}{2} = 1 - \frac{b ^{2}}{a ^{2}}$

Establish $a^{2} = 2 b^{2}$

Rearranging: $\frac{b ^{2}}{a ^{2}} = 1 - \frac{1}{2} = \frac{1}{2}$
Therefore: $a^{2} = 2 b^{2}$

Identify Common Tangent

Common Tangent: $y = \frac{5}{2} x + K$
Comparing with $y = m x + c$ :
$m = \frac{5}{2} ⟹ m^{2} = \frac{5}{2}$

Tangency Condition for $H$

Condition for tangency to $H$ :
$c^{2} = a_{H}^{2} m^{2} - b_{H}^{2}$

Calculate $K^{2}$ from $H$

Substituting $a_{H} = 1, b_{H} = 1, m^{2} = \frac{5}{2}$ :
$K^{2} = 1 (\frac{5}{2}) - 1 = \frac{3}{2}$

Tangency Condition for $E$

Condition for tangency to $E$ :
$c^{2} = a^{2} m^{2} + b^{2}$

Substitute into $E$ Condition

Substitute $K^{2} = \frac{3}{2}, m^{2} = \frac{5}{2}$ and $a^{2} = 2 b^{2}$ :
$\frac{3}{2} = (2 b^{2}) (\frac{5}{2}) + b^{2}$

Solve for $b^{2}$

Simplifying: $\frac{3}{2} = 5 b^{2} + b^{2} = 6 b^{2}$
Solving for $b^{2}$ : $b^{2} = \frac{3}{2 \times 6} = \frac{3}{12} = \frac{1}{4}$

Solve for $a^{2}$

Using $a^{2} = 2 b^{2}$ :
$a^{2} = 2 (\frac{1}{4}) = \frac{1}{2}$

Final Calculation

Calculate $4 (a^{2} + b^{2})$ :
$4 (\frac{1}{2} + \frac{1}{4}) = 4 (\frac{3}{4}) = 3$

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Hyperbola's Secret

We begin with the hyperbola

H : x^{2} - y^{2} = 1

. This is a rectangular hyperbola, a special case where the branches are perfectly balanced.

By comparing this to the standard form

\frac{x ^{2}}{a _{H}^{2}} - \frac{y ^{2}}{b _{H}^{2}} = 1

, we identify that

a_{H} = 1

and

b_{H} = 1

The eccentricity of a hyperbola is defined as

e_{H} = 1 + \frac{b _{H}^{2}}{a _{H}^{2}}

. Substituting our values, we find:

e_{H} = 1 + 1 = 2

This is the fundamental constant of our hyperbola, the key that will unlock the ellipse.

The Ellipse's Constraint

The problem provides a relationship: the eccentricity of the ellipse

E

is the reciprocal of the hyperbola's eccentricity. Thus,

e_{E} = \frac{1}{e _{H}} = \frac{1}{2}

Now, we turn to the ellipse

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

. The eccentricity of an ellipse is governed by

e_{E}^{2} = 1 - \frac{b ^{2}}{a ^{2}}

Squaring our eccentricity, we get

e_{E}^{2} = \frac{1}{2}

. Substituting this into our equation:

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

This is the geometric constraint that binds the dimensions of our ellipse.

The Common Tangent

We are given the line

y = \frac{5}{2} x + K

. Comparing this to

y = m x + c

, we identify the slope

m = \frac{5}{2}

, so

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

For the hyperbola, the tangency condition is

c^{2} = a_{H}^{2} m^{2} - b_{H}^{2}

. Plugging in our values:

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

For the ellipse, the tangency condition is

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

. Substituting

K^{2} = \frac{3}{2}

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

, and our constraint

a^{2} = 2 b^{2}

, we get:

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

This simplifies to

\frac{3}{2} = 5 b^{2} + b^{2} = 6 b^{2}

. Solving for

b^{2}

, we find

b^{2} = \frac{3}{12} = \frac{1}{4}

Consequently,

a^{2} = 2 b^{2} = 2 (\frac{1}{4}) = \frac{1}{2}

The Final Calculation

We have arrived at the finish line. We need to calculate

4 (a^{2} + b^{2})

Substituting our values:

4 (\frac{1}{2} + \frac{1}{4}) = 4 (\frac{3}{4}) = 3

The elegance of the final cancellation is the reward for our persistence. The final result is 3.

Common Traps & FAQs

Wait, why do we use

c^{2} = a^{2} m^{2} - b^{2}

for the hyperbola but

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

for the ellipse?

That is a fantastic observation! The difference arises from the standard equations of the curves. For the hyperbola

x^{2} / a^{2} - y^{2} / b^{2} = 1

, the tangency condition is derived by setting the discriminant of the quadratic equation (formed by substituting the line into the hyperbola) to zero, which results in

c^{2} = a^{2} m^{2} - b^{2}

. For the ellipse

x^{2} / a^{2} + y^{2} / b^{2} = 1

, the same process yields

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

. They are distinct because the hyperbola has a negative sign in its standard equation, which flips the sign in the tangency condition.

Can I just use the discriminant method instead of memorizing these tangency formulas?

Absolutely! You can always substitute

y = m x + K

into the curve equation and set the discriminant

D = 0

. However, in JEE Advanced, time is your most precious resource. Memorizing these standard conditions allows you to jump straight to the algebraic manipulation, saving you valuable minutes for more complex problems.

Why is the hyperbola

x^{2} - y^{2} = 1

called a rectangular hyperbola?

A hyperbola is called 'rectangular' or 'equilateral' when its asymptotes are perpendicular to each other. This happens when

a = b

. In your equation

x^{2} - y^{2} = 1

, we have

a = 1

and

b = 1

, so

a = b

. This symmetry makes the eccentricity calculation much simpler, as you saw!

Enter Numerical Value:

Visualized Solution

Analyze Hyperbola $H$

Calculate Eccentricity $e_{H}$

Relate $e_{E}$ and $e_{H}$

Link $a^{2}$ and $b^{2}$ for Ellipse

Establish $a^{2} = 2 b^{2}$

Identify Common Tangent

Tangency Condition for $H$

Calculate $K^{2}$ from $H$

Tangency Condition for $E$

Substitute into $E$ Condition

Solve for $b^{2}$

Solve for $a^{2}$

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Hyperbola's Secret

The Ellipse's Constraint

The Common Tangent

The Final Calculation

Similar Questions

If two tangents drawn from a point $(α, β)$ lying on the ellipse $25 x^{2} + 4 y^{2} = 1$ to the parabola $y^{2} = 4 x$ are such that the slope of one tangent is four times the other, then the value of $(10 α + 5)^{2} + (16 β^{2} + 50)^{2}$ equals ____.

Let the eccentricity of the hyperbola $H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be $\frac{5}{2}$ and length of its latus rectum be $62$ . If $y = 2 x + c$ is a tangent to the hyperbola $H$ , then the value of $c^{2}$ is equal to:

Let the eccentricity of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be $\frac{5}{4}$ . If the equation of the normal at the point $(\frac{8}{5}, \frac{12}{5})$ on the hyperbola is $85 x + β y = λ$ , then $λ - β$ 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 :

If $a > 2 b > 0$ then the positive value of $m$ for which $y = m x - b 1 + m^{2}$ is a common tangent to $x^{2} + y^{2} = b^{2}$ and $(x - a)^{2} + y^{2} = b^{2}$ is

Enter Numerical Value:

Visualized Solution

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

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

Relate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } eE​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } eH​

Link .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 for Ellipse

Establish .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2=2b2

Identify Common Tangent

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

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K2 from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H

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

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

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

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

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Hyperbola's Secret

The Ellipse's Constraint

The Common Tangent

The Final Calculation

Similar Questions

.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; } If a>2b>0 then the positive value of m for which y=mx−b1+m2​ is a common tangent to x2+y2=b2 and (x−a)2+y2=b2 is

Analyze Hyperbola $H$

Calculate Eccentricity $e_{H}$

Relate $e_{E}$ and $e_{H}$

Link $a^{2}$ and $b^{2}$ for Ellipse

Establish $a^{2} = 2 b^{2}$

Tangency Condition for $H$

Calculate $K^{2}$ from $H$

Tangency Condition for $E$

Substitute into $E$ Condition

Solve for $b^{2}$

Solve for $a^{2}$

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 :

If $a > 2 b > 0$ then the positive value of $m$ for which $y = m x - b 1 + m^{2}$ is a common tangent to $x^{2} + y^{2} = b^{2}$ and $(x - a)^{2} + y^{2} = b^{2}$ is