JEE Main 2020 (7 January Shift 1)

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Animated Solution for Mathematics - Conic Sections: If $y = m x + 4$ is a tangent to both the parabolas, $y^{2} = 4 x$ and $x^{2} = 2 b y$ , then $b$ is equal to

Q: Wait, why can't we just use the formula $c = a/m$ for both parabolas?

That is a common trap! The formula $c = a/m$ is specifically derived for the parabola $y^2 = 4ax$. For a parabola like $x^2 = 4Ay$, the standard tangent equation is $y = mx - Am^2$. If you try to apply the same formula to both, you will get the wrong result. Always check the orientation of the parabola first!

Q: Why do we reject the solution $b = 0$?

Great observation! If $b = 0$, the equation $x^2 = 2by$ becomes $x^2 = 0$, which simplifies to $x = 0$. This is the equation of the y-axis, which is a straight line, not a parabola. Since the problem explicitly mentions two parabolas, $b = 0$ is a degenerate case that we must discard.

Q: Is there a faster way than using the discriminant?

Yes, you can use the condition of tangency for $x^2 = 4Ay$, which is $c = -Am^2$. Here, $4A = 2b$, so $A = b/2$. The tangent is $y = mx + 4$, so $c = 4$. Plugging these into the condition $c = -Am^2$ gives $4 = -(b/2)m^2$. Since we already know $m = 1/4$, you can solve for $b$ directly. Both methods are valid, but the discriminant method is a great fallback if you forget the specific formula!

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the two parabolas. One opens along the positive x-axis (

y^{2} = 4 x

), and the other opens along the positive y-axis (

x^{2} = 2 b y

). A common tangent is a line that satisfies the tangency condition for both curves simultaneously. Focus on the geometry of the y-intercept.

Hint 2

Hint 2 (The Arsenal): For a line

y = m x + c

to be tangent to

y^{2} = 4 a x

, the condition is

c = a / m

. For the second parabola

x^{2} = 4 A y

, the condition is

c = - A m^{2}

. Alternatively, you can substitute the line equation into the parabola equation and set the discriminant (

D

) of the resulting quadratic to zero.

Hint 3

Hint 3 (The Execution): Start by finding the slope

m

of the tangent using the first parabola

y^{2} = 4 x

. Once you have

m

, substitute the line

y = m x + 4

into the second parabola

x^{2} = 2 b y

to get a quadratic in

x

. Set the discriminant of this quadratic to zero to solve for

b

Visualized Solution

The Geometric Setup

We are given two parabolas: $y^{2} = 4 x$ and $x^{2} = 2 b y$ .
A common tangent line touches both curves.
The equation of this tangent is given as $y = m x + 4$ .
Our goal is to find the value of the constant $b$ .

Standard Tangent to $y^{2} = 4 a x$

Let's focus on the first parabola: $y^{2} = 4 x$ .
The standard equation of a tangent to $y^{2} = 4 a x$ in terms of slope $m$ is known.
Formula: $y = m x + \frac{a}{m}$ .

Finding Parameter $a$

Compare $y^{2} = 4 x$ with the standard form $y^{2} = 4 a x$ .
This gives $4 a = 4$ .
Therefore, $a = 1$ .
Substitute $a = 1$ into the tangent formula: $y = m x + \frac{1}{m}$ .

Determining the Slope $m$

We now have two equations for the same tangent line.
Given tangent: $y = m x + 4$ .
Derived tangent: $y = m x + \frac{1}{m}$ .
Comparing the y-intercepts: $\frac{1}{m} = 4$ .
Solving for $m$ : $m = \frac{1}{4}$ .

The Exact Tangent Equation

Substitute $m = \frac{1}{4}$ back into the tangent equation.
$y = \frac{1}{4} x + 4$ .
This line must also be a tangent to the second parabola, $x^{2} = 2 b y$ .

Intersection with Second Parabola

To find where the line touches the second parabola, we substitute $y$ into its equation.
Second parabola: $x^{2} = 2 b y$ .
Substitute $y = \frac{x}{4} + 4$ :
$x^{2} = 2 b (\frac{x}{4} + 4)$ .

Expanding the Equation

Let's expand the right side of the equation.
$x^{2} = 2 b (\frac{x}{4}) + 2 b (4)$ .
$x^{2} = \frac{b x}{2} + 8 b$ .

Forming the Quadratic Equation

Multiply the entire equation by 2 to remove the fraction.
$2 x^{2} = b x + 16 b$ .
Rearrange into standard quadratic form $A x^{2} + B x + C = 0$ :
$2 x^{2} - b x - 16 b = 0$ .

The Tangency Condition

The quadratic equation $2 x^{2} - b x - 16 b = 0$ represents the intersection points.
Since the line is a tangent, it touches the parabola at exactly one point.
Therefore, the quadratic equation must have equal roots.
Condition for equal roots: Discriminant $D = 0$ .

Applying $D = 0$

The formula for Discriminant is $D = B^{2} - 4 A C$ .
From $2 x^{2} - b x - 16 b = 0$ , we identify:
$A = 2$ , $B = - b$ , $C = - 16 b$ .
Substitute these into $D = 0$ :
$(- b)^{2} - 4 (2) (- 16 b) = 0$ .

Solving for $b$

Simplify the discriminant equation:
$b^{2} - 8 (- 16 b) = 0$ .
$b^{2} + 128 b = 0$ .
Factor out $b$ :
$b (b + 128) = 0$ .

Final Conclusion

The equation $b (b + 128) = 0$ gives two possible values: $b = 0$ or $b = - 128$ .
If $b = 0$ , the second parabola $x^{2} = 2 b y$ becomes $x^{2} = 0$ , which is just the y-axis, not a parabola.
Therefore, we reject $b = 0$ .
The final correct value is $b = - 128$ .

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two symmetric curves. One is a parabola opening to the right, defined by

y^{2} = 4 x

. The other is a parabola opening upwards, defined by

x^{2} = 2 b y

Our goal is to find the value of the constant

b

such that the line

y = m x + 4

is a common tangent to both curves.

The First Encounter

We begin with the first parabola,

y^{2} = 4 x

. In coordinate geometry, the line

y = m x + c

is tangent to

y^{2} = 4 a x

if and only if

c = \frac{a}{m}

Comparing

y^{2} = 4 x

with the standard form

y^{2} = 4 a x

, we identify

4 a = 4

, which implies

a = 1

. Given our tangent line

y = m x + 4

, the y-intercept is

c = 4

Applying the tangency condition:

4 = \frac{1}{m}

Solving this relation, we find the slope of the tangent line:

m = \frac{1}{4}

Thus, the equation of our common tangent is

y = \frac{1}{4} x + 4

The Bridge to the Second Curve

Next, we ensure this line touches the second parabola,

x^{2} = 2 b y

. We substitute the expression for

y

from our tangent line into the equation of the second parabola:

x^{2} = 2 b (\frac{x}{4} + 4)

Expanding this expression, we obtain:

x^{2} = \frac{b x}{2} + 8 b

Multiplying the entire equation by

2

to clear the fraction, we get

2 x^{2} = b x + 16 b

. Rearranging this into the standard quadratic form

A x^{2} + B x + C = 0

, we arrive at:

2 x^{2} - b x - 16 b = 0

The Moment of Truth

Since the line is a tangent to the second parabola, it must intersect at exactly one point. For a quadratic equation to have exactly one root, its discriminant

D

must be zero.

The discriminant is defined as

D = B^{2} - 4 A C

. Identifying the coefficients from

2 x^{2} - b x - 16 b = 0

, we have

A = 2

B = - b

, and

C = - 16 b

Substituting these into the discriminant formula:

(- b)^{2} - 4 (2) (- 16 b) = 0

Simplifying the expression yields:

b^{2} + 128 b = 0

Factoring out

b

, we get

b (b + 128) = 0

. This provides two potential values:

b = 0

b = - 128

Since

b = 0

would collapse the parabola into a line, we reject it. Therefore, the final value is:

$b = - 128$

Common Traps & FAQs

Wait, why can't we just use the formula

c = a / m

for both parabolas?

That is a common trap! The formula

c = a / m

is specifically derived for the parabola

y^{2} = 4 a x

. For a parabola like

x^{2} = 4 A y

, the standard tangent equation is

y = m x - A m^{2}

. If you try to apply the same formula to both, you will get the wrong result. Always check the orientation of the parabola first!

Why do we reject the solution

b = 0

Great observation! If

b = 0

, the equation

x^{2} = 2 b y

becomes

x^{2} = 0

, which simplifies to

x = 0

. This is the equation of the y-axis, which is a straight line, not a parabola. Since the problem explicitly mentions two parabolas,

b = 0

is a degenerate case that we must discard.

Is there a faster way than using the discriminant?

Yes, you can use the condition of tangency for

x^{2} = 4 A y

, which is

c = - A m^{2}

. Here,

4 A = 2 b

, so

A = b /2

. The tangent is

y = m x + 4

, so

c = 4

. Plugging these into the condition

c = - A m^{2}

gives

4 = - (b /2) m^{2}

. Since we already know

m = 1/4

, you can solve for

b

directly. Both methods are valid, but the discriminant method is a great fallback if you forget the specific formula!

Animated Solution for Mathematics - Conic Sections: If $y = m x + 4$ is a tangent to both the parabolas, $y^{2} = 4 x$ and $x^{2} = 2 b y$ , then $b$ is equal to

Select Answer:

Visualized Solution

The Geometric Setup

Standard Tangent to $y^{2} = 4 a x$

Finding Parameter $a$

Determining the Slope $m$

The Exact Tangent Equation

Intersection with Second Parabola

Expanding the Equation

Forming the Quadratic Equation

The Tangency Condition

Applying $D = 0$

Solving for $b$

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The First Encounter

The Bridge to the Second Curve

The Moment of Truth

Similar Questions

If $y = m x + 4$ is a tangent to both the parabolas $y^{2} = 4 x$ and $x^{2} = 2 b y$ . Then value of $b$ is

Let $L$ be a tangent line to the parabola $y^{2} = 4 x - 20$ at $(6, 2)$ . If $L$ is also a tangent to the ellipse $\frac{x ^{2}}{2} + \frac{y ^{2}}{b} = 1$ , then the value of $b$ is equal to :

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

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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:

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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

Let a line $y = m x (m > 0)$ intersect the parabola, $y^{2} = x$ at a point $P$ , other than the origin. Let the tangent to it at $P$ meet the $x$ -axis at the point $Q$ . If area $(Δ O P Q) = 4$ sq. units, then $m$ is equal to _________ .

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 y=mx+4 is a tangent to both the parabolas, y2=4x and x2=2by, then b is equal to

Select Answer:

Visualized Solution

The Geometric Setup

Standard Tangent to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2=4ax

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

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

The Exact Tangent Equation

Intersection with Second Parabola

Expanding the Equation

Forming the Quadratic Equation

The Tangency Condition

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D=0

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

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The First Encounter

The Bridge to the Second Curve

The Moment of Truth

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=mx+4 is a tangent to both the parabolas y2=4x and x2=2by. Then value of b is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let L be a tangent line to the parabola y2=4x−20 at (6,2). If L is also a tangent to the ellipse 2x2​+by2​=1, then the value of b is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=m1​x+c1​ and y=m2​x+c2​,m1​=m2​ are two common tangents of circle x2+y2=2 and parabola y2=x, then the value of 8∣m1​m2​∣ is equal to

.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 slope of the line touching both the parabolas y2=4x and x2=−32y is

.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

Animated Solution for Mathematics - Conic Sections: If $y = m x + 4$ is a tangent to both the parabolas, $y^{2} = 4 x$ and $x^{2} = 2 b y$ , then $b$ is equal to

Standard Tangent to $y^{2} = 4 a x$

Finding Parameter $a$

Determining the Slope $m$

Applying $D = 0$

Solving for $b$

If $y = m x + 4$ is a tangent to both the parabolas $y^{2} = 4 x$ and $x^{2} = 2 b y$ . Then value of $b$ is

Let $L$ be a tangent line to the parabola $y^{2} = 4 x - 20$ at $(6, 2)$ . If $L$ is also a tangent to the ellipse $\frac{x ^{2}}{2} + \frac{y ^{2}}{b} = 1$ , then the value of $b$ is equal to :

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

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 slope of the line touching both the parabolas $y^{2} = 4 x$ and $x^{2} = - 32 y$ 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