JEE Main 2022 (25 July Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Let the equation of two diameters of a circle $x^{2} + y^{2} - 2 x + 2 f y + 1 = 0$ be $2 p x - y = 1$ and $2 x + p y = 4 p$ . Then the slope $m \in (0, \infty)$ of the tangent to the hyperbola $3 x^{2} - y^{2} = 3$ passing through the centre of the circle is equal to ______.

Enter Numerical Value:

Visualized Solution

Identify the Circle's Centre

Circle equation: $x^{2} + y^{2} - 2 x + 2 f y + 1 = 0$
Compare with general form: $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
$2 g = - 2 \Rightarrow g = - 1$
Centre $(h, k) = (- g, - f) = (1, - f)$

Diameters and the Centre

Diameters: $2 p x - y = 1$ and $2 x + p y = 4 p$
Diameters always intersect at the centre $(1, - f)$ .
Substitute $(1, - f)$ in the first diameter: $2 p (1) - (- f) = 1$
$2 p + f = 1 \Rightarrow f = 1 - 2 p$

Forming the Quadratic in $p$

Substitute $(1, - f)$ in the second diameter: $2 (1) + p (- f) = 4 p$
Substitute $f = 1 - 2 p$ : $2 - p (1 - 2 p) = 4 p$
Expand: $2 - p + 2 p^{2} = 4 p$
Rearrange: $2 p^{2} - 5 p + 2 = 0$

Solving for $p$

Factorize $2 p^{2} - 5 p + 2 = 0$
Split the middle term: $2 p^{2} - 4 p - p + 2 = 0$
$2 p (p - 2) - 1 (p - 2) = 0$
$(2 p - 1) (p - 2) = 0$
Roots: $p = \frac{1}{2}$ or $p = 2$

Finding the Possible Centres

Case 1: If $p = \frac{1}{2}$ , $f = 1 - 2 (\frac{1}{2}) = 0$
Centre 1: $(1, 0)$
Case 2: If $p = 2$ , $f = 1 - 2 (2) = - 3$
Centre 2: $(1, 3)$

The Hyperbola and its Tangent

Hyperbola: $3 x^{2} - y^{2} = 3$
Standard form: $\frac{x ^{2}}{1} - \frac{y ^{2}}{3} = 1$
Here, $a^{2} = 1$ and $b^{2} = 3$
Equation of tangent with slope $m$ : $y = m x \pm a^{2} m^{2} - b^{2}$
$y = m x \pm m^{2} - 3$

Testing the First Centre $(1, 0)$

The tangent must pass through the circle's centre.
Test $(1, 0)$ : Substitute $x = 1, y = 0$ into tangent equation.
$0 = m (1) \pm m^{2} - 3$
$- m = \pm m^{2} - 3$
Square both sides: $m^{2} = m^{2} - 3 \Rightarrow 0 = - 3$
This is a contradiction. The centre cannot be $(1, 0)$ .

Testing the Second Centre $(1, 3)$

Test $(1, 3)$ : Substitute $x = 1, y = 3$ into tangent equation.
$3 = m (1) \pm m^{2} - 3$
$3 - m = \pm m^{2} - 3$
Square both sides: $(3 - m)^{2} = m^{2} - 3$
Expand: $9 - 6 m + m^{2} = m^{2} - 3$

Final Calculation for Slope $m$

Equation: $9 - 6 m + m^{2} = m^{2} - 3$
Cancel $m^{2}$ from both sides: $9 - 6 m = - 3$
Rearrange: $12 = 6 m$
Solve for $m$ : $m = 2$
Check condition: $m \in (0, \infty)$ . Since $2 > 0$ , it is valid.
Final Answer: $m = 2$

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The circle is defined by the equation

x^{2} + y^{2} - 2 x + 2 f y + 1 = 0

. By comparing this to the general form

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

, we identify the center of the circle.

Since

2 g = - 2

, we find

g = - 1

. The center of the circle is given by

(- g, - f)

, which simplifies to the point

(1, - f)

The Meeting Point of Diameters

In geometry, any diameter of a circle must pass through its center. Therefore, the center

(1, - f)

must satisfy the equations of the two given diameters:

2 p x - y = 1

and

2 x + p y = 4 p

Substituting the center into the first equation:

2 p (1) - (- f) = 1 ⟹ 2 p + f = 1 ⟹ f = 1 - 2 p

Substituting the center into the second equation:

2 (1) + p (- f) = 4 p ⟹ 2 - p f = 4 p

Substituting

f = 1 - 2 p

into the second equation:

2 - p (1 - 2 p) = 4 p

2 - p + 2 p^{2} = 4 p

2 p^{2} - 5 p + 2 = 0

Solving the quadratic equation

2 p^{2} - 5 p + 2 = 0

using the factorization

(2 p - 1) (p - 2) = 0

, we find two possible values for

p

p = \frac{1}{2}

and

p = 2

p = \frac{1}{2}

, then

f = 1 - 2 (\frac{1}{2}) = 0

, giving the center

(1, 0)

. If

p = 2

, then

f = 1 - 2 (2) = - 3

, giving the center

(1, 3)

The Hyperbola's Tangent Dance

The hyperbola is given by

3 x^{2} - y^{2} = 3

. Dividing by

3

, we obtain the standard form:

\frac{x ^{2}}{1} - \frac{y ^{2}}{3} = 1

Here,

a^{2} = 1

and

b^{2} = 3

. The equation of a tangent to this hyperbola with slope

m

is:

y = m x \pm a^{2} m^{2} - b^{2} ⟹ y = m x \pm m^{2} - 3

We test the candidate center

(1, 0)

by substituting it into the tangent equation:

0 = m (1) \pm m^{2} - 3 ⟹ - m = \pm m^{2} - 3

Squaring both sides yields

m^{2} = m^{2} - 3

, which simplifies to

0 = - 3

. This is a contradiction, meaning the center cannot be

(1, 0)

The Final Victory

We now test the candidate center

(1, 3)

by substituting it into the tangent equation:

3 = m (1) \pm m^{2} - 3 ⟹ 3 - m = \pm m^{2} - 3

Squaring both sides:

(3 - m)^{2} = m^{2} - 3

9 - 6 m + m^{2} = m^{2} - 3

The

m^{2}

terms cancel out, leaving:

9 - 6 m = - 3

6 m = 12 ⟹ m = 2

The slope is

2

, which satisfies the condition

m \in (0, \infty)

. The final result is $m = 2$ .

Common Traps & FAQs

Wait, why do we reject the first center

(1, 0)

? Is it just a guess?

It is not a guess! When we test a candidate center, we substitute its coordinates into the equation of the tangent. If that substitution leads to a mathematical contradiction (like

0 = - 3

), it proves that no such tangent with a real slope

m

can pass through that point. It is a rigorous elimination process.

Why can't we just use energy conservation or some other physics principle here?

This is a pure coordinate geometry problem, not a physics problem. In JEE Advanced, you must identify the 'domain' of the question. Here, the tools are algebraic manipulation, properties of conic sections, and the intersection of loci. Stick to the geometry toolkit!

Why did we square both sides when solving for

m

? Doesn't that introduce fake solutions?

You are absolutely right to be cautious! Squaring both sides can indeed introduce extraneous roots. That is why, after finding

m = 2

, we must verify it against the original equation

3 - m = \pm m^{2} - 3

. Since

3 - 2 = 1

and

2^{2} - 3 = 1

, the solution is valid.

Enter Numerical Value:

Visualized Solution

Identify the Circle's Centre

Diameters and the Centre

Forming the Quadratic in $p$

Solving for $p$

Finding the Possible Centres

The Hyperbola and its Tangent

Testing the First Centre $(1, 0)$

Testing the Second Centre $(1, 3)$

Final Calculation for Slope $m$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Meeting Point of Diameters

The Hyperbola's Tangent Dance

The Final Victory

Similar Questions

A hyperbola passes through the point $P (2, 3)$ and has foci at $(\pm 2, 0)$ . Then the tangent to this hyperbola at P also passes through the point:

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

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 .

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 equations of the tangents drawn from the origin to the circle $x^{2} + y^{2} - 2 r x - 2 h y + h^{2} = 0$ , are

Comprehension Passage

A possible equation of $L$ is

A common tangent of the two circles is

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:

Enter Numerical Value:

Visualized Solution

Identify the Circle's Centre

Diameters and the Centre

Forming the Quadratic in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

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

Finding the Possible Centres

The Hyperbola and its Tangent

Testing the First Centre .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,0)

Testing the Second Centre .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,3)

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

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Meeting Point of Diameters

The Hyperbola's Tangent Dance

The Final Victory

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A hyperbola passes through the point P(2​,3​) and has foci at (±2,0). Then the tangent to this hyperbola at P also passes through the point:

.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 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 equations of the tangents drawn from the origin to the circle x2+y2−2rx−2hy+h2=0, are

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A possible equation of L is

A common tangent of the two circles is

.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:

Forming the Quadratic in $p$

Solving for $p$

Testing the First Centre $(1, 0)$

Testing the Second Centre $(1, 3)$

Final Calculation for Slope $m$

A hyperbola passes through the point $P (2, 3)$ and has foci at $(\pm 2, 0)$ . Then the tangent to this hyperbola at P also passes through the point:

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 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 equations of the tangents drawn from the origin to the circle $x^{2} + y^{2} - 2 r x - 2 h y + h^{2} = 0$ , are

A possible equation of $L$ is

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: