JEE Main 2021 (February) (26 Feb Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: Let the normals at all the points on a given curve pass through a fixed point $(a, b)$ . If the curve passes through $(3, - 3)$ and $(4, - 22)$ , and given that $a - 22 b = 3$ , then $(a^{2} + b^{2} + ab)$ is equal to

Q: Why did the constant $c$ disappear when we subtracted the equations?

That is the beauty of linear systems! Since the constant $c$ appeared in both equations derived from the circle's general form, subtracting one equation from the other eliminates it entirely. This is a powerful technique in JEE problems; often, you do not need to solve for every variable individually if you can isolate the relationship between the ones you actually need.

Enter Numerical Value:

Visualized Solution

Understanding the Geometric Property

Property: If all normals to a curve pass through a fixed point $(a, b)$ , the curve is a circle centered at $(a, b)$ .
The fixed point $(a, b)$ represents the center of the circle.
Equation of normal at any point $(x, y)$ with slope $m = \frac{d y}{d x}$ is:
$Y - y = - \frac{1}{m} (X - x)$

Forming the Differential Equation

Normal passes through $(a, b)$ , so substitute $X = a$ and $Y = b$ :
$b - y = - \frac{1}{\frac{d y}{d x}} (a - x)$
Rearranging to separate variables:
$(b - y) d y = (x - a) d x$

Integrating the Equation

Integrating both sides:
$\int (b - y) d y = \int (x - a) d x$
$b y - \frac{y ^{2}}{2} = \frac{x ^{2}}{2} - a x + c$
Multiplying by $2$ and rearranging:
$x^{2} + y^{2} - 2 a x - 2 b y + 2 c = 0$

Substituting the First Point $(3, - 3)$

Curve passes through $(3, - 3)$ :
$b (- 3) - \frac{( - 3 ) ^{2}}{2} = \frac{3 ^{2}}{2} - a (3) + c$
$- 3 b - \frac{9}{2} = \frac{9}{2} - 3 a + c$
Multiplying by $2$ : $- 6 b - 9 = 9 - 6 a + 2 c$
$6 a - 6 b - 2 c = 18 ⟹ 3 a - 3 b - c = 9$ ... (i)

Substituting the Second Point $(4, - 22)$

Curve passes through $(4, - 22)$ :
$b (- 22) - \frac{( - 2 2 ) ^{2}}{2} = \frac{4 ^{2}}{2} - a (4) + c$
$- 22 b - 4 = 8 - 4 a + c$
$4 a - 22 b - c = 12$ ... (ii)

Eliminating the Constant $c$

Subtract equation (i) from (ii):
$(4 a - 22 b - c) - (3 a - 3 b - c) = 12 - 9$
$a - 22 b + 3 b = 3$
$a + (3 - 22) b = 3$ ... (iv)

Using the Given Constraint

Given constraint in the question:
$a - 22 b = 3$ ... (iii)
We now have a system of two equations:
1) $a + (3 - 22) b = 3$
2) $a - 22 b = 3$

Solving for $b$

Substitute $a = 3 + 22 b$ into equation (iv):
$(3 + 22 b) + (3 - 22) b = 3$
$3 + 22 b + 3 b - 22 b = 3$
$3 + 3 b = 3 ⟹ 3 b = 0 ⟹ b = 0$

Solving for $a$

Substitute $b = 0$ into equation (iii):
$a - 22 (0) = 3$
$a - 0 = 3$
$a = 3$
The fixed point $(a, b)$ is $(3, 0)$ .

Final Calculation

Calculate the required expression:
$a^{2} + b^{2} + ab$
Substitute $a = 3$ and $b = 0$ :
$3^{2} + 0^{2} + 3 (0) = 9 + 0 + 0 = 9$
Final Answer: $9$

00:00 / 00:00

The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Setup

Imagine you are standing on a curve, any curve. You draw a line perpendicular to the tangent at your feet—the normal.

If you extend that normal line and it always pierces through one specific, fixed point in space,

(a, b)

, the curve is perfectly symmetric with respect to that point. This geometric signature confirms that the curve is a circle.

The Calculus Bridge

We formalize this intuition by noting that the normal at any point

(x, y)

must pass through the center

(a, b)

. Let the slope of the tangent be

\frac{d y}{d x}

The slope of the normal is therefore

- \frac{1}{\frac{d y}{d x}}

. The equation of the normal line is given by:

Y - y = - \frac{1}{\frac{d y}{d x}} (X - x)

Since this line passes through

(a, b)

, we substitute

X = a

and

Y = b

to obtain:

b - y = - \frac{1}{\frac{d y}{d x}} (a - x)

Rearranging this expression leads to the fundamental differential equation:

(b - y) d y = (x - a) d x

The Integration

We integrate both sides of the differential equation:

\int (b - y) d y = \int (x - a) d x

Performing the integration yields:

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

Multiplying by

2

and rearranging terms, we arrive at the general equation of a circle:

x^{2} + y^{2} - 2 a x - 2 b y + 2 c = 0

The Algebraic Dance

We are given two points on the curve:

(3, - 3)

and

(4, - 22)

. Substituting

(3, - 3)

into the circle equation:

3^{2} + (- 3)^{2} - 2 a (3) - 2 b (- 3) + 2 c = 0

This simplifies to:

18 - 6 a + 6 b + 2 c = 0 \Rightarrow 3 a - 3 b - c = 9

Next, we substitute

(4, - 22)

into the equation:

4^{2} + (- 22)^{2} - 2 a (4) - 2 b (- 22) + 2 c = 0

Since

(- 22)^{2} = 8

, this simplifies to:

24 - 8 a + 42 b + 2 c = 0 \Rightarrow 4 a - 22 b - c = 12

Final Calculation

We now have a system of two equations with three unknowns, supplemented by the constraint

a - 22 b = 3

. By subtracting the two derived equations, we eliminate

c

(4 a - 22 b - c) - (3 a - 3 b - c) = 12 - 9

a + (3 - 22) b = 3

Solving this system alongside the constraint

a = 3 + 22 b

reveals that

b = 0

and

a = 3

. Thus, the center of the circle is

(3, 0)

The final value requested is

a^{2} + b^{2} + ab

3^{2} + 0^{2} + 3 (0) = * * 9 * *

Common Traps & FAQs

Why are we assuming it is a circle? Can it not be some other curve?

That is a brilliant question! In geometry, the normal to a curve at any point is the line perpendicular to the tangent. For a circle, the normal always passes through the center. If *all* normals pass through a fixed point, the distance from that point to any point on the curve must be constant (the radius). Thus, by definition, it must be a circle. No other curve possesses this property of radial symmetry where every normal converges to a single point.

Why did the constant

c

disappear when we subtracted the equations?

That is the beauty of linear systems! Since the constant

c

appeared in both equations derived from the circle's general form, subtracting one equation from the other eliminates it entirely. This is a powerful technique in JEE problems; often, you do not need to solve for every variable individually if you can isolate the relationship between the ones you actually need.

I got stuck on the algebra with the square roots. Is there a trick?

The trick is to not fear the radicals! When you substitute the point

(4, - 22)

, squaring it yields

8

. Keep your terms organized, and you will see that the radical terms often align or cancel out when you set up your system of equations. Always simplify the arithmetic before rushing into complex algebraic manipulation.

Animated Solution for Mathematics - Differential Equations: Let the normals at all the points on a given curve pass through a fixed point $(a, b)$ . If the curve passes through $(3, - 3)$ and $(4, - 22)$ , and given that $a - 22 b = 3$ , then $(a^{2} + b^{2} + ab)$ is equal to

Enter Numerical Value:

Visualized Solution

Understanding the Geometric Property

Forming the Differential Equation

Integrating the Equation

Substituting the First Point $(3, - 3)$

Substituting the Second Point $(4, - 22)$

Eliminating the Constant $c$

Using the Given Constraint

Solving for $b$

Solving for $a$

Final Calculation

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Calculus Bridge

The Integration

The Algebraic Dance

Final Calculation

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Enter Numerical Value:

Visualized Solution

Understanding the Geometric Property

Forming the Differential Equation

Integrating the Equation

Substituting the First Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (3,−3)

Substituting the Second Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (4,−22​)

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

Using the Given Constraint

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

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

Final Calculation

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Calculus Bridge

The Integration

The Algebraic Dance

Final Calculation

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