JEE Main 2022 (26 June Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Let the common tangents to the curves $4 (x^{2} + y^{2}) = 9$ and $y^{2} = 4 x$ intersect at the point $Q$ . Let an ellipse, centered at the origin $O$ , has lengths of semi-minor and semi-major axes equal to $O Q$ and $6$ , respectively. If $e$ and $l$ respectively denote the eccentricity and the length of the latus rectum of this ellipse, then $\frac{l}{e ^{2}}$ is equal to

Q: Why do we reject the negative root when solving for $m^2$?

Great observation! When we solve the quadratic equation $9t^2 + 9t - 4 = 0$ (where $t = m^2$), we get two roots: $t = 1/3$ and $t = -4/3$. Since $m$ is the slope of a real line, $m^2$ must be a non-negative real number. A negative value for $m^2$ would imply an imaginary slope, which doesn't exist in the real coordinate plane. Thus, we discard $t = -4/3$.

Q: How do we know the intersection point $Q$ lies on the x-axis?

Look at the symmetry of the curves. The parabola $y^2 = 4x$ is symmetric about the x-axis, and the circle $x^2 + y^2 = 9/4$ is symmetric about both axes. Because the setup is perfectly symmetric, the two common tangents must also be symmetric with respect to the x-axis. Any two non-parallel lines that are symmetric about the x-axis must intersect on the x-axis itself.

Q: Can we use the discriminant method instead of the distance formula for the circle?

Absolutely! You can substitute $y = mx + 1/m$ into the circle equation $x^2 + y^2 = 9/4$, which results in a quadratic in $x$. For the line to be tangent, the discriminant ($D = B^2 - 4AC$) of this quadratic must be zero. Both methods are mathematically sound and will lead you to the exact same result. The distance formula is often faster, but the discriminant method is a robust fallback.

Enter Numerical Value:

Progressive Hints

Hint 1

The Spark: Visualize the geometry. You have a circle

x^{2} + y^{2} = (3/2)^{2}

and a parabola

y^{2} = 4 x

. A common tangent is a line that touches both curves simultaneously. The key is to express the tangent line in terms of its slope

m

and then use the condition that its perpendicular distance from the circle's center

(0, 0)

must equal the circle's radius.

Hint 2

The Arsenal: Recall the standard equation of a tangent to the parabola

y^{2} = 4 a x

with slope

m

, which is

y = m x + a / m

. For the circle

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

, the perpendicular distance from the origin to the line

m x - y + c = 0

is given by

d = ∣ c ∣/ m^{2} + 1

. Set this distance equal to the radius

r

Hint 3

The Execution: Substitute

a = 1

and

r = 3/2

into your equations. You will get the condition

∣1/ m ∣/ m^{2} + 1 = 3/2

. Square both sides to eliminate the square root and absolute value, leading to a biquadratic equation in

m

9 m^{4} + 9 m^{2} - 4 = 0

. Solve this for

m^{2}

to find the slopes.

Visualized Solution

Identify the Curves

Circle: $4 (x^{2} + y^{2}) = 9 \Rightarrow x^{2} + y^{2} = (\frac{3}{2})^{2}$
Radius of circle $r = \frac{3}{2}$
Parabola: $y^{2} = 4 x \Rightarrow a = 1$

General Tangent to Parabola

Equation of tangent to $y^{2} = 4 a x$ is $y = m x + \frac{a}{m}$
For our parabola ( $a = 1$ ), the tangent is $y = m x + \frac{1}{m}$
Rearranging: $m x - y + \frac{1}{m} = 0$

Tangency Condition for Circle

For the line to be tangent to the circle, perpendicular distance from center $(0, 0)$ must equal radius $r$ .
Distance formula: $d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$
$\frac{∣ m ( 0 ) - 0 + \frac{1}{m} ∣}{m ^{2} + ( - 1 ) ^{2}} = \frac{3}{2}$

Solving for Slope $m$ (Part 1)

Simplify the distance equation: $\frac{1}{∣ m ∣ m ^{2} + 1} = \frac{3}{2}$
Squaring both sides: $\frac{1}{m ^{2} ( m ^{2} + 1 )} = \frac{9}{4}$
Cross-multiplying: $9 m^{4} + 9 m^{2} = 4$
$9 m^{4} + 9 m^{2} - 4 = 0$

Solving for Slope $m$ (Part 2)

Let $t = m^{2}$ . The equation is $9 t^{2} + 9 t - 4 = 0$
Factorizing: $9 t^{2} + 12 t - 3 t - 4 = 0$
$3 t (3 t + 4) - 1 (3 t + 4) = 0 \Rightarrow (3 t - 1) (3 t + 4) = 0$
Since $t = m^{2} \geq 0$ , we reject $t = - \frac{4}{3}$
Therefore, $m^{2} = \frac{1}{3}$

Finding the Common Tangents

$m^{2} = \frac{1}{3} \Rightarrow m = \pm \frac{1}{3}$
Substitute $m$ back into $y = m x + \frac{1}{m}$
For $m = \frac{1}{3}$ : $y = \frac{1}{3} x + 3$
For $m = - \frac{1}{3}$ : $y = - \frac{1}{3} x - 3$

Intersection Point $Q$

To find intersection point $Q$ , solve the two tangent equations.
By symmetry, they intersect on the x-axis, so $y = 0$ .
$0 = \frac{1}{3} x + 3 \Rightarrow \frac{x}{3} = - 3 \Rightarrow x = - 3$
Point $Q$ is $(- 3, 0)$ and distance $O Q = 3$

Defining the Ellipse

An ellipse is centered at origin $O (0, 0)$ .
Semi-minor axis $b = O Q = 3$
Semi-major axis $a = 6$
Equation form: $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$

Calculating Eccentricity Squared $e^{2}$

Formula for eccentricity: $e^{2} = 1 - \frac{b ^{2}}{a ^{2}}$
Substitute $a = 6$ and $b = 3$ :
$e^{2} = 1 - \frac{3 ^{2}}{6 ^{2}} = 1 - \frac{9}{36}$
$e^{2} = 1 - \frac{1}{4} = \frac{3}{4}$

Length of Latus Rectum $l$

Formula for length of latus rectum: $l = \frac{2 b ^{2}}{a}$
Substitute $a = 6$ and $b = 3$ :
$l = \frac{2 ( 3 ^{2} )}{6} = \frac{2 \times 9}{6}$
$l = \frac{18}{6} = 3$

Final Calculation: $l / e^{2}$

We need to find the value of $\frac{l}{e ^{2}}$
Substitute $l = 3$ and $e^{2} = \frac{3}{4}$
$\frac{l}{e ^{2}} = \frac{3}{\frac{3}{4}}$
$\frac{l}{e ^{2}} = 3 \times \frac{4}{3} = 4$

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

Solution Diagram

Analyzing the Setup

We begin with our two geometric protagonists. The circle is defined by

4 (x^{2} + y^{2}) = 9

, which simplifies to

x^{2} + y^{2} = (3/2)^{2}

. Its radius is

r = 3/2

The parabola is

y^{2} = 4 x

, which matches the standard form

y^{2} = 4 a x

with

a = 1

. To find a common tangent, we start with the general equation of a tangent to the parabola with slope

m

, which is

y = m x + 1/ m

For this line to also be tangent to the circle, the perpendicular distance from the circle's center

(0, 0)

to the line

m x - y + 1/ m = 0

must be exactly equal to the radius

r = 3/2

. Using the distance formula, we establish the following relationship:

\frac{∣ m ( 0 ) - 0 + 1/ m ∣}{m ^{2} + ( - 1 ) ^{2}} = \frac{3}{2}

The Algebraic Unfolding

We simplify the distance equation to solve for the slope

m

\frac{1}{∣ m ∣ m ^{2} + 1} = \frac{3}{2}

Squaring both sides yields:

\frac{1}{m ^{2} ( m ^{2} + 1 )} = \frac{9}{4}

Cross-multiplying leads to the quartic equation

9 m^{2} (m^{2} + 1) = 4

, which rearranges to

9 m^{4} + 9 m^{2} - 4 = 0

. By substituting

t = m^{2}

, we obtain the quadratic equation

9 t^{2} + 9 t - 4 = 0

Factoring this quadratic gives

(3 t - 1) (3 t + 4) = 0

. Since

t = m^{2}

must be non-negative, we discard

t = - 4/3

and accept

m^{2} = 1/3

. This results in two possible slopes:

m = \pm 1/ 3

Final Calculation

Substituting the slopes back into the tangent equation

y = m x + 1/ m

, we find the two common tangents:

y = \frac{1}{3} x + 3 and y = - \frac{1}{3} x - 3

By symmetry, these lines intersect on the x-axis. Setting

y = 0

, we find

x = - 3

. Thus, the intersection point

Q

(- 3, 0)

, and the distance

O Q = 3

We construct the ellipse centered at the origin with semi-minor axis

b = O Q = 3

and semi-major axis

a = 6

. The eccentricity squared is:

e^{2} = 1 - \frac{b ^{2}}{a ^{2}} = 1 - \frac{9}{36} = \frac{3}{4}

The length of the latus rectum is:

l = \frac{2 b ^{2}}{a} = \frac{2 ( 9 )}{6} = 3

Finally, we calculate the required ratio:

\frac{l}{e ^{2}} = \frac{3}{3/4} = 4

Common Traps & FAQs

Why do we reject the negative root when solving for

m^{2}

Great observation! When we solve the quadratic equation

9 t^{2} + 9 t - 4 = 0

(where

t = m^{2}

), we get two roots:

t = 1/3

and

t = - 4/3

. Since

m

is the slope of a real line,

m^{2}

must be a non-negative real number. A negative value for

m^{2}

would imply an imaginary slope, which doesn't exist in the real coordinate plane. Thus, we discard

t = - 4/3

How do we know the intersection point

Q

lies on the x-axis?

Look at the symmetry of the curves. The parabola

y^{2} = 4 x

is symmetric about the x-axis, and the circle

x^{2} + y^{2} = 9/4

is symmetric about both axes. Because the setup is perfectly symmetric, the two common tangents must also be symmetric with respect to the x-axis. Any two non-parallel lines that are symmetric about the x-axis must intersect on the x-axis itself.

Can we use the discriminant method instead of the distance formula for the circle?

Absolutely! You can substitute

y = m x + 1/ m

into the circle equation

x^{2} + y^{2} = 9/4

, which results in a quadratic in

x

. For the line to be tangent, the discriminant (

D = B^{2} - 4 A C

) of this quadratic must be zero. Both methods are mathematically sound and will lead you to the exact same result. The distance formula is often faster, but the discriminant method is a robust fallback.

Enter Numerical Value:

Visualized Solution

Identify the Curves

General Tangent to Parabola

Tangency Condition for Circle

Solving for Slope $m$ (Part 1)

Solving for Slope $m$ (Part 2)

Finding the Common Tangents

Intersection Point $Q$

Defining the Ellipse

Calculating Eccentricity Squared $e^{2}$

Length of Latus Rectum $l$

Final Calculation: $l / e^{2}$

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

Analyzing the Setup

The Algebraic Unfolding

Final Calculation

Similar Questions

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

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

Let the eccentricity of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is reciprocal to that of the hyperbola $2 x^{2} - 2 y^{2} = 1$ . If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is _____.

Let the ellipse $E : \frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1$ and the hyperbola $H : \frac{x ^{2}}{16} - \frac{y ^{2}}{λ ^{2}} = - 1$ have the same foci. If $e$ and $L$ respectively denote the eccentricity and the length of the latus rectum of $H$ , then the value of $24 (e + L)$ is :

Enter Numerical Value:

Visualized Solution

Identify the Curves

General Tangent to Parabola

Tangency Condition for Circle

Solving for Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m (Part 1)

Solving for Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m (Part 2)

Finding the Common Tangents

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

Defining the Ellipse

Calculating Eccentricity Squared .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e2

Length of Latus Rectum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l/e2

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

Analyzing the Setup

The Algebraic Unfolding

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P(x1​,y1​) and Q(x2​,y2​),y1​<0,y2​<0, be the end points of the latus rectum of the ellipse x2+4y2=4. The equations of parabolas with latus rectum PQ are

Solving for Slope $m$ (Part 1)

Solving for Slope $m$ (Part 2)

Intersection Point $Q$

Calculating Eccentricity Squared $e^{2}$

Length of Latus Rectum $l$

Final Calculation: $l / e^{2}$

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