JEE Main 2022 (26 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: If $m$ is the slope of a common tangent to the curves $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ and $x^{2} + y^{2} = 12$ , then $12 m^{2}$ is equal to :

Q: Why can we just equate the square roots? What about the $\pm$ sign?

Excellent observation. When we equate the $y$-intercepts, we are essentially saying the distance from the origin to the line must be the same for both curves. Since we are squaring both sides of the equation $\sqrt{a^2m^2 + b^2} = r\sqrt{1 + m^2}$, the $\pm$ signs disappear, effectively capturing all possible common tangents at once.

Select Answer:

Visualized Solution

Visualizing the Curves

Given Ellipse: $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$
Given Circle: $x^{2} + y^{2} = 12$
Goal: Find $12 m^{2}$ where $m$ is the slope of their common tangent.

Tangent to an Ellipse

Standard equation of a tangent to an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ in slope form:
$y = m x \pm a^{2} m^{2} + b^{2}$

Substitution for Ellipse

For our ellipse: $a^{2} = 16$ , $b^{2} = 9$
Substitute into the tangent formula:
$y = m x \pm 16 m^{2} + 9$

Tangent to a Circle

Standard equation of a tangent to a circle $x^{2} + y^{2} = r^{2}$ in slope form:
$y = m x \pm r 1 + m^{2}$

Substitution for Circle

For our circle: $r^{2} = 12$
Substitute into the tangent formula:
$y = m x \pm 12 (1 + m^{2})$

The Condition for Common Tangent

For a common tangent, the lines must be identical.
Therefore, their $y$ -intercepts must be equal:
$16 m^{2} + 9 = 12 (1 + m^{2})$

Squaring Both Sides

To remove the square roots, square both sides:
$16 m^{2} + 9 = 12 (1 + m^{2})$

Expanding the Equation

Expand the right-hand side:
$16 m^{2} + 9 = 12 + 12 m^{2}$

Rearranging Terms

Group the $m^{2}$ terms on one side and constants on the other:
$16 m^{2} - 12 m^{2} = 12 - 9$

Solving for $m^{2}$

Simplify both sides:
$4 m^{2} = 3$

Finding $12 m^{2}$

The question asks for the value of $12 m^{2}$ .
Multiply both sides of $4 m^{2} = 3$ by $3$ :
$3 \cdot (4 m^{2}) = 3 \cdot 3$
$12 m^{2} = 9$

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We are tasked with finding the common tangent to the ellipse defined by

\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1

and the circle defined by

x^{2} + y^{2} = 12

A common tangent is a line that satisfies the tangency conditions for both curves simultaneously. We will utilize the slope-intercept form of tangent equations to bridge these two geometries.

The Tangent Toolkit

For an ellipse of the form

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

the equation of a tangent with slope

m

is given by

y = m x \pm a^{2} m^{2} + b^{2}

Given our ellipse parameters

a^{2} = 16

and

b^{2} = 9

, the family of tangents is:

y = m x \pm 16 m^{2} + 9

For a circle

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

, the tangent equation is

y = m x \pm r 1 + m^{2}

. With

r^{2} = 12

, the tangent equation for our circle becomes:

y = m x \pm 12 (1 + m^{2})

The Bridge of Equality

If a line is tangent to both curves, it must share the same slope

m

and the same

y

-intercept. Therefore, we equate the constant terms of the two tangent equations:

16 m^{2} + 9 = 12 (1 + m^{2})

To solve for

m

, we square both sides to eliminate the radicals:

16 m^{2} + 9 = 12 (1 + m^{2})

The Algebraic Dance

Expanding the right side of the equation, we obtain:

16 m^{2} + 9 = 12 + 12 m^{2}

Subtracting

12 m^{2}

from both sides yields:

4 m^{2} + 9 = 12

Subtracting

9

from both sides, we find:

4 m^{2} = 3

The problem asks for the value of

12 m^{2}

. By multiplying both sides of our result by

3

, we get:

3 \cdot (4 m^{2}) = 3 \cdot 3

12 m^{2} = 9

The final value is 9.

Common Traps & FAQs

Wait, why do we use the slope form

y = m x \pm c

instead of the point form?

Great question! The point form requires you to know the exact coordinate of contact

(x_{1}, y_{1})

, which we don't have. The slope form is a powerful shortcut because it directly introduces the slope

m

as a variable, allowing us to solve for it immediately using the tangency condition.

Why can we just equate the square roots? What about the

\pm

sign?

Excellent observation. When we equate the

y

-intercepts, we are essentially saying the distance from the origin to the line must be the same for both curves. Since we are squaring both sides of the equation

a^{2} m^{2} + b^{2} = r 1 + m^{2}

, the

\pm

signs disappear, effectively capturing all possible common tangents at once.

Is it possible for the common tangent to not exist? How would the math tell us?

Yes, absolutely! If the curves were positioned such that one was entirely inside the other without touching, the resulting equation for

m^{2}

would yield a negative value or an impossible condition. In such a case, the math would tell us that no real slope

m

exists, meaning no common tangent can be drawn.

Animated Solution for Mathematics - Conic Sections: If $m$ is the slope of a common tangent to the curves $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ and $x^{2} + y^{2} = 12$ , then $12 m^{2}$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Curves

Tangent to an Ellipse

Substitution for Ellipse

Tangent to a Circle

Substitution for Circle

The Condition for Common Tangent

Squaring Both Sides

Expanding the Equation

Rearranging Terms

Solving for $m^{2}$

Finding $12 m^{2}$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent Toolkit

The Bridge of Equality

The Algebraic Dance

Similar Questions

The distance of the point $(6, - 22)$ from the common tangent $y = m x + c, m > 0$ , of the curves $x = 2 y^{2}$ and $x = 1 + y^{2}$ is

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

If $y = m_{1} x + c_{1}$ and $y = m_{2} x + c_{2}, m_{1} \neq = m_{2}$ are two common tangents of circle $x^{2} + y^{2} = 2$ and parabola $y^{2} = x$ , then the value of $8∣ m_{1} m_{2} ∣$ is equal to

Let a common tangent to the curves $y^{2} = 4 x$ and $(x - 4)^{2} + y^{2} = 16$ touch the curves at the points $P$ and $Q$ . Then $(P Q)^{2}$ is equal to ________.

If the line $y = m x + c$ is a common tangent to the hyperbola $\frac{x ^{2}}{100} - \frac{y ^{2}}{64} = 1$ and the circle $x^{2} + y^{2} = 36$ , then which one of the following is true?

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 ____.

If the common tangent to the parabolas, $y^{2} = 4 x$ and $x^{2} = 4 y$ also touches the circle, $x^{2} + y^{2} = c^{2}$ , then $c$ is equal to:

Let A be a point on the x-axis. Common tangents are drawn from A to the curves $x^{2} + y^{2} = 8$ and $y^{2} = 16 x$ . If one of these tangents touches the two curves at Q and R, then $(QR)^{2}$ is equal to

The equation of a common tangent to the curves, $y^{2} = 16 x$ and $x y = - 4$ is :

Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If m is the slope of a common tangent to the curves 16x2​+9y2​=1 and x2+y2=12, then 12m2 is equal to :

Select Answer:

Visualized Solution

Visualizing the Curves

Tangent to an Ellipse

Substitution for Ellipse

Tangent to a Circle

Substitution for Circle

The Condition for Common Tangent

Squaring Both Sides

Expanding the Equation

Rearranging Terms

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m2

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 12m2

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent Toolkit

The Bridge of Equality

The Algebraic Dance

Similar Questions

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.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

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line y=mx+c is a common tangent to the hyperbola 100x2​−64y2​=1 and the circle x2+y2=36, then which one of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the common tangent to the parabolas, y2=4x and x2=4y also touches the circle, x2+y2=c2, then c is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a common tangent to the curves, y2=16x and xy=−4 is :

Animated Solution for Mathematics - Conic Sections: If $m$ is the slope of a common tangent to the curves $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ and $x^{2} + y^{2} = 12$ , then $12 m^{2}$ is equal to :

Solving for $m^{2}$

Finding $12 m^{2}$

The distance of the point $(6, - 22)$ from the common tangent $y = m x + c, m > 0$ , of the curves $x = 2 y^{2}$ and $x = 1 + y^{2}$ is

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

If $y = m_{1} x + c_{1}$ and $y = m_{2} x + c_{2}, m_{1} \neq = m_{2}$ are two common tangents of circle $x^{2} + y^{2} = 2$ and parabola $y^{2} = x$ , then the value of $8∣ m_{1} m_{2} ∣$ is equal to

Let a common tangent to the curves $y^{2} = 4 x$ and $(x - 4)^{2} + y^{2} = 16$ touch the curves at the points $P$ and $Q$ . Then $(P Q)^{2}$ is equal to ________.

If the line $y = m x + c$ is a common tangent to the hyperbola $\frac{x ^{2}}{100} - \frac{y ^{2}}{64} = 1$ and the circle $x^{2} + y^{2} = 36$ , then which one of the following is true?

If the common tangent to the parabolas, $y^{2} = 4 x$ and $x^{2} = 4 y$ also touches the circle, $x^{2} + y^{2} = c^{2}$ , then $c$ is equal to:

The equation of a common tangent to the curves, $y^{2} = 16 x$ and $x y = - 4$ is :