JEE Main 2020 (5 September Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: 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:

Q: Why is the distance from the origin to the line equal to $c$?

The circle $x^2 + y^2 = c^2$ is centered at the origin with radius $c$. By definition, a line is tangent to a circle if and only if the perpendicular distance from the center to the line equals the radius. It is the most elegant way to link the line to the circle!

Select Answer:

Visualized Solution

Visualizing the Parabolas

Parabola 1: $y^{2} = 4 x$ (Rightward opening)
Parabola 2: $x^{2} = 4 y$ (Upward opening)
Goal: Find the equation of the common tangent.

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

Standard Tangent to $y^{2} = 4 a x$ : $y = m x + \frac{a}{m}$
For $y^{2} = 4 x$ : $4 a = 4 \Rightarrow a = 1$
Equation 1: $y = m x + \frac{1}{m}$

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

Standard Tangent to $x^{2} = 4 a y$ : $y = m x - a m^{2}$
For $x^{2} = 4 y$ : $4 a = 4 \Rightarrow a = 1$
Equation 2: $y = m x - m^{2}$

Finding the Common Slope $m$

For a common tangent, both equations must represent the same line.
Equating the y-intercepts: $\frac{1}{m} = - m^{2}$

Solving for $m$

$\frac{1}{m} = - m^{2}$
$m^{3} = - 1$
Solving for real $m$ : $m = - 1$

Equation of the Common Tangent

Substitute $m = - 1$ into $y = m x + \frac{1}{m}$
$y = (- 1) x + \frac{1}{- 1}$
$y = - x - 1$
Common Tangent: $x + y + 1 = 0$

The Circle Constraint

Circle: $x^{2} + y^{2} = c^{2}$
Center: $(0, 0)$ , Radius: $c$
Condition: The common tangent also touches this circle.

Applying the Distance Formula

For a line to touch a circle, the perpendicular distance from the center to the line must equal the radius.
Distance $d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$
Center $(x_{1}, y_{1}) = (0, 0)$ , Line: $x + y + 1 = 0$

Calculating $c$

Substitute $(0, 0)$ and $x + y + 1 = 0$ :
$c = \frac{∣1 ( 0 ) + 1 ( 0 ) + 1∣}{1 ^{2} + 1 ^{2}}$
$c = \frac{1}{1 + 1}$
$c = \frac{1}{2}$

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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 two parabolas:

y^{2} = 4 x

and

x^{2} = 4 y

. These curves represent parabolas opening along the positive

x

-axis and positive

y

-axis, respectively.

The Tangent Equations

For the parabola

y^{2} = 4 a x

, the equation of a tangent with slope

m

is given by:

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

Given

y^{2} = 4 x

, we have

a = 1

. Substituting this, the tangent equation becomes:

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

For the parabola

x^{2} = 4 a y

, the equation of a tangent with slope

m

is:

y = m x - a m^{2}

Again, with

a = 1

, this simplifies to:

y = m x - m^{2}

Finding the Common Tangent

For a line to be common to both parabolas, it must share the same slope

m

and the same

y

-intercept. By equating the intercepts from our two tangent equations, we obtain:

\frac{1}{m} = - m^{2}

Solving for

m

, we find

m^{3} = - 1

, which yields the real solution $m = - 1$ .

Substituting

m = - 1

back into either tangent equation, we get

y = - x - 1

. Rearranging this, the equation of the common tangent is:

$x + y + 1 = 0$

Relating to the Circle

We now consider the circle

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

. A line is tangent to a circle if the perpendicular distance from the center

(0, 0)

to the line is equal to the radius

c

Using the perpendicular distance formula

d = \frac{∣ A x _{0} + B y _{0} + C ∣}{A ^{2} + B ^{2}}

, we calculate:

c = \frac{∣1 ( 0 ) + 1 ( 0 ) + 1∣}{1 ^{2} + 1 ^{2}}

This simplifies to the final result:

$c = \frac{1}{2}$

Common Traps & FAQs

Wait, why did we use different formulas for the tangents of the two parabolas?

Great observation! The parabolas have different orientations.

y^{2} = 4 a x

is symmetric about the x-axis, while

x^{2} = 4 a y

is symmetric about the y-axis. Their tangent slopes behave differently relative to the axes, hence the distinct standard forms. Always check the orientation before choosing your formula!

Why do we equate the y-intercepts to find the slope

m

Because a line is uniquely defined by its slope and intercept. If two lines have the same slope

m

and the same y-intercept, they are identical. By forcing the intercepts to be equal, we ensure the tangent to the first parabola is the exact same line as the tangent to the second.

Why is the distance from the origin to the line equal to

c

The circle

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

is centered at the origin with radius

c

. By definition, a line is tangent to a circle if and only if the perpendicular distance from the center to the line equals the radius. It is the most elegant way to link the line to the circle!

Animated Solution for Mathematics - Conic Sections: 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:

Select Answer:

Visualized Solution

Visualizing the Parabolas

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

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

Finding the Common Slope $m$

Solving for $m$

Equation of the Common Tangent

The Circle Constraint

Applying the Distance Formula

Calculating $c$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent Equations

Finding the Common Tangent

Relating to the Circle

Similar Questions

Equation of a common tangent to the circle, $x^{2} + y^{2} - 6 x = 0$ and the parabola, $y^{2} = 4 x$ , 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

The equation of the common tangent touching the circle $(x - 3)^{2} + y^{2} = 9$ and the parabola $y^{2} = 4 x$ above the x-axis is

The tangent to the parabola $y^{2} = 4 x$ at the point where it intersects the circle $x^{2} + y^{2} = 5$ in the first quadrant, passes through the point :

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 ________.

The common tangent to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} + 6 x + 8 y - 24 = 0$ also passes through the point :-

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?

The number of common tangents to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} - 6 x - 8 y = 24$ is

Two circles each of radius 5 units touch each other at the point $(1, 2)$ . If the equation of their common tangent is $4 x + 3 y = 10$ , and $C_{1} (α, β)$ and $C_{2} (γ, δ)$ , $C_{1} \neq = C_{2}$ are their centres, then $∣ (α + β) (γ + δ) ∣$ is equal to .

Let $T_{1}, T_{2}$ be two tangents drawn from $(- 2, 0)$ onto the circle $C : x^{2} + y^{2} = 1$ . Determine the circles touching $C$ and $T_{1}, T_{2}$ as their pair of tangents. Further, find the equations of all possible common tangents to these circles, when taken two at a time.

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 the common tangent to the parabolas, y2=4x and x2=4y also touches the circle, x2+y2=c2, then c is equal to:

Select Answer:

Visualized Solution

Visualizing the Parabolas

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

Tangent to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2=4ay

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

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

Equation of the Common Tangent

The Circle Constraint

Applying the Distance Formula

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

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent Equations

Finding the Common Tangent

Relating to the Circle

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of a common tangent to the circle, x2+y2−6x=0 and the parabola, y2=4x, is:

.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; } The equation of the common tangent touching the circle (x−3)2+y2=9 and the parabola y2=4x above the x-axis is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangent to the parabola y2=4x at the point where it intersects the circle x2+y2=5 in the first quadrant, passes through the point :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a common tangent to the curves y2=4x and (x−4)2+y2=16 touch the curves at the points P and Q. Then (PQ)2 is equal to ________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The common tangent to the circles x2+y2=4 and x2+y2+6x+8y−24=0 also passes through the point :-

.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; } The number of common tangents to the circles x2+y2=4 and x2+y2−6x−8y=24 is

Animated Solution for Mathematics - Conic Sections: 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:

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

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

Finding the Common Slope $m$

Solving for $m$

Calculating $c$

Equation of a common tangent to the circle, $x^{2} + y^{2} - 6 x = 0$ and the parabola, $y^{2} = 4 x$ , 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

The equation of the common tangent touching the circle $(x - 3)^{2} + y^{2} = 9$ and the parabola $y^{2} = 4 x$ above the x-axis is

The tangent to the parabola $y^{2} = 4 x$ at the point where it intersects the circle $x^{2} + y^{2} = 5$ in the first quadrant, passes through the point :

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 ________.

The common tangent to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} + 6 x + 8 y - 24 = 0$ also passes through the point :-

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?

The number of common tangents to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} - 6 x - 8 y = 24$ is