JEE Main 2013

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Given : A circle, $2 x^{2} + 2 y^{2} = 5$ and a parabola, $y^{2} = 45 x$ . Statement-1 : An equation of a common tangent to these curves is $y = x + 5$ . Statement-2 : If the line, $y = m x + \frac{5}{m} (m \neq = 0)$ is their common tangent, then $m$ satisfies $m^{4} - 3 m^{2} + 2 = 0$ .

Q: Wait, Statement 2 is true, but it's not the explanation? How can an equation be true but not the explanation?

Think of it this way: $2+2=4$ is true, but it doesn't explain why the sky is blue. Statement 2 provides a polynomial that happens to share the correct roots ($m=\pm 1$), but it is not the polynomial derived from the tangency condition. In JEE, 'correct explanation' means the logical derivation must lead to that statement.

Select Answer:

Visualized Solution

Visualizing the Curves

Circle: $2 x^{2} + 2 y^{2} = 5$
Parabola: $y^{2} = 45 x$
A common tangent is a line that touches both curves at exactly one point each.

Standardizing the Equations

Standard form of the Circle: $x^{2} + y^{2} = \frac{5}{2}$
Radius of the circle: $r = \frac{5}{2}$
Standard form of the Parabola: $y^{2} = 4 a x$ , where $a = 5$

General Tangent to Parabola

The general equation of a tangent to the parabola $y^{2} = 4 a x$ is:
$y = m x + \frac{a}{m}$
Substituting $a = 5$ :
$y = m x + \frac{5}{m}$

Tangency Condition for Circle

Condition for a line to be tangent to a circle: Distance from center = Radius
Center of circle: $(0, 0)$ , Radius: $r = \frac{5}{2}$
Equation of the tangent line: $m x - y + \frac{5}{m} = 0$

Applying the Distance Formula

Distance formula: $d = \frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}}$
$\frac{∣ m ( 0 ) - 0 + \frac{5}{m} ∣}{m ^{2} + ( - 1 ) ^{2}} = \frac{5}{2}$
$\frac{\frac{5}{∣ m ∣}}{m ^{2} + 1} = \frac{5}{2}$

Squaring Both Sides

Squaring both sides to remove the square root and absolute value:
$\frac{5}{m ^{2} ( m ^{2} + 1 )} = \frac{5}{2}$
Canceling $5$ from both sides:
$\frac{1}{m ^{2} ( m ^{2} + 1 )} = \frac{1}{2}$

Forming the Polynomial

Cross-multiplying the equation:
$m^{2} (m^{2} + 1) = 2$
Expanding the brackets:
$m^{4} + m^{2} - 2 = 0$

Solving for $m$

Factoring the biquadratic equation:
$(m^{2} + 2) (m^{2} - 1) = 0$
Since $m^{2} \geq 0$ , we reject $m^{2} = - 2$ .
This leaves $m^{2} = 1$ , which means $m = \pm 1$ .

Equations of Common Tangents

For $m = 1$ , the tangent is: $y = (1) x + \frac{5}{1} ⟹ y = x + 5$
For $m = - 1$ , the tangent is: $y = (- 1) x + \frac{5}{- 1} ⟹ y = - x - 5$

Evaluating Statement-1

Statement-1: $y = x + 5$ is a common tangent.
This matches our derived equation for $m = 1$ .
Therefore, Statement-1 is True.

Evaluating Statement-2

Statement-2: If $y = m x + \frac{5}{m}$ is a common tangent, then $m^{4} - 3 m^{2} + 2 = 0$ .
Let's check our valid slopes $m = \pm 1$ in this equation:
$(\pm 1)^{4} - 3 (\pm 1)^{2} + 2 = 1 - 3 + 2 = 0$
Therefore, Statement-2 is also True.

Checking the Explanation

Both statements are true.
However, the actual tangency condition gave $m^{4} + m^{2} - 2 = 0$ , not $m^{4} - 3 m^{2} + 2 = 0$ .
Statement-2 just happens to share the roots $m = \pm 1$ .
Conclusion: Statement-2 is not a correct explanation for Statement-1.

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

Solution Diagram

Analyzing the Setup

We are given a circle

2 x^{2} + 2 y^{2} = 5

and a parabola

y^{2} = 45 x

. First, we simplify the circle equation by dividing by

2

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

This reveals the radius

r = \frac{5}{2}

. For the parabola

y^{2} = 45 x

, we compare it to the standard form

y^{2} = 4 a x

to identify the parameter

a = 5

The Bridge

Tangency Condition

The general equation of a tangent to the parabola

y^{2} = 4 a x

with slope

m

y = m x + \frac{a}{m}

. Substituting our value of

a

, the tangent line equation becomes:

y = m x + \frac{5}{m}

This line is tangent to the parabola by construction. To ensure it is also tangent to the circle, we must satisfy the geometric condition that the perpendicular distance from the center

(0, 0)

to the line equals the radius

r

The Collision

Distance Formula

We rewrite the tangent line as

m x - y + \frac{5}{m} = 0

. The perpendicular distance

d

from the origin

(0, 0)

to this line is:

d = \frac{∣ m ( 0 ) - 0 + \frac{5}{m} ∣}{m ^{2} + ( - 1 ) ^{2}} = \frac{\frac{5}{∣ m ∣}}{m ^{2} + 1}

Setting

d = r = \frac{5}{2}

, we obtain the following equation:

\frac{\frac{5}{∣ m ∣}}{m ^{2} + 1} = \frac{5}{2}

The Algebra

Solving for Slope

Squaring both sides to eliminate the radicals, we get:

\frac{5}{m ^{2} ( m ^{2} + 1 )} = \frac{5}{2}

Canceling the

5

and cross-multiplying yields

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

, which simplifies to the biquadratic equation:

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

Factoring this expression, we find

(m^{2} + 2) (m^{2} - 1) = 0

. Since

m^{2}

must be positive, we reject

m^{2} = - 2

, leaving

m^{2} = 1

, or

m = \pm 1

Final Evaluation

For

m = 1

, the tangent equation becomes

y = x + 5

. Thus, Statement 1 is true.

Regarding Statement 2, which proposes

m^{4} - 3 m^{2} + 2 = 0

: while

m^{2} = 1

is a root of this equation, it is not the correct condition derived from the geometry of the problem. Therefore, Statement 2 is a distractor and does not provide the correct explanation for Statement 1.

Common Traps & FAQs

Why can't we just substitute the line equation into the circle equation and set the discriminant to zero?

You absolutely can! That is a perfectly valid algebraic path. However, using the distance formula is often faster and less prone to calculation errors when dealing with circles centered at the origin. It keeps the geometry front and center.

Wait, Statement 2 is true, but it's not the explanation? How can an equation be true but not the explanation?

Think of it this way:

2 + 2 = 4

is true, but it doesn't explain why the sky is blue. Statement 2 provides a polynomial that happens to share the correct roots (

m = \pm 1

), but it is not the polynomial derived from the tangency condition. In JEE, 'correct explanation' means the logical derivation must lead to that statement.

How do I know if I should use the slope form or the point form?

Slope form is usually better for common tangents because the slope

m

is the common link between the two curves. It allows you to express the tangency condition for both curves in terms of the same variable

m

Select Answer:

Visualized Solution

Visualizing the Curves

Standardizing the Equations

General Tangent to Parabola

Tangency Condition for Circle

Applying the Distance Formula

Squaring Both Sides

Forming the Polynomial

Solving for $m$

Equations of Common Tangents

Evaluating Statement-1

Evaluating Statement-2

Checking the Explanation

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

Analyzing the Setup

The Bridge

The Collision

The Algebra

Final Evaluation

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Curves

Standardizing the Equations

General Tangent to Parabola

Tangency Condition for Circle

Applying the Distance Formula

Squaring Both Sides

Forming the Polynomial

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

Equations of Common Tangents

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Checking the Explanation

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

Analyzing the Setup

The Bridge

The Collision

The Algebra

Final Evaluation

Similar Questions

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