JEE(ADVANCED)-201

LEVELJEE Main

Animated Solution for Mathematics - Circles: For $a = 2$ , if a tangent is drawn to a suitable conic (Column 1) at the point of contact $(- 1, 1)$ , then which of the following options is the only CORRECT combination for obtaining its equation ?

Q: Why do we use $T=0$ instead of finding the slope using differentiation?

While differentiation is a powerful tool, the $T=0$ method is a specialized shortcut for conics. It is faster, less prone to calculation errors, and specifically designed for JEE-level efficiency. It directly gives you the equation of the tangent at a point on the curve.

Select Answer:

Visualized Solution

The Setup & Conic Assumption

Given parameter: $a = 2$
Point of contact: $P (- 1, 1)$
Testing combination: (I) (ii) (Q)
Assume Conic (I) is the standard circle: $x^{2} + y^{2} = a^{2}$

Forming the Circle Equation

Substitute $a = 2$ into $x^{2} + y^{2} = a^{2}$
$x^{2} + y^{2} = (2)^{2}$
$x^{2} + y^{2} = 2$

Verifying the Point $P (- 1, 1)$

Check if $P (- 1, 1)$ lies on $x^{2} + y^{2} = 2$
LHS: $(- 1)^{2} + (1)^{2}$
$= 1 + 1 = 2$
LHS = RHS, so the point lies on the circle.

Equation of Tangent using $T = 0$

Equation of tangent at $(x_{1}, y_{1})$ is given by $T = 0$
$x x_{1} + y y_{1} = a^{2}$
We will substitute $(x_{1}, y_{1}) = (- 1, 1)$ and $a^{2} = 2$

Deriving the Tangent Equation

$x (- 1) + y (1) = 2$
$- x + y = 2$
$y = x + 2$

Analyzing the Tangent

Compare $y = x + 2$ with $y = m x + c$
Slope $m = 1$
$y$ -intercept $c = 2$

The Standard Tangent Formula

Standard tangent in slope form for $x^{2} + y^{2} = a^{2}$ :
$y = m x + a 1 + m^{2}$
We need to verify if this matches our derived $y = x + 2$

Verifying the Tangent Equation

Substitute $m = 1$ and $a = 2$
$y = (1) x + (2) 1 + (1)^{2}$
$y = x + 22$
$y = x + 2$ (Matches exactly!)

Standard Point of Contact

Standard point of contact for $y = m x + a 1 + m^{2}$ :
$(\frac{- ma}{1 + m ^{2}}, \frac{a}{1 + m ^{2}})$

Verifying the Point of Contact

Substitute $m = 1$ and $a = 2$ :
$x$ -coordinate: $\frac{- ( 1 ) ( 2 )}{1 + ( 1 ) ^{2}} = \frac{- 2}{2} = - 1$
$y$ -coordinate: $\frac{2}{1 + ( 1 ) ^{2}} = \frac{2}{2} = 1$
Calculated Point: $(- 1, 1)$

Final Conclusion

Key Takeaway: The combination (I) (ii) (Q) is perfectly consistent for the circle $x^{2} + y^{2} = 2$ .
The derived tangent $y = x + 2$ matches the slope form.
The derived point of contact $(- 1, 1)$ matches the standard formula.

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

Imagine you are standing on the Cartesian plane, looking at a circle defined by the parameter

a = 2

. We begin by assuming the simplest case: Conic (I) is the standard circle

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

Substituting

a = 2

, we obtain the equation:

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

Before proceeding, we must verify if our point

P (- 1, 1)

actually lies on this circle. Substituting

x = - 1

and

y = 1

into the equation, we find:

(- 1)^{2} + (1)^{2} = 1 + 1 = 2

Since the left-hand side equals the right-hand side, the point

P

is confirmed to be on the circle.

The Master Equation

To find the tangent, we utilize the elegant

T = 0

method. For any conic, the equation of the tangent at point

(x_{1}, y_{1})

is given by

T = 0

For our circle, this formula simplifies to:

x x_{1} + y y_{1} = a^{2}

Substituting our point

P (- 1, 1)

and

a^{2} = 2

, we get:

x (- 1) + y (1) = 2

This simplifies to the linear equation:

- x + y = 2 \Rightarrow y = x + 2

Verification and Consistency

We must verify this result against the standard slope form of a tangent, defined as

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

. Comparing

y = x + 2

with

y = m x + c

, we identify the slope

m = 1

and the intercept

c = 2

Let us check if the standard formula yields the same intercept. With

m = 1

and

a = 2

, the intercept is:

a 1 + m^{2} = 2 1 + 1^{2} = 22 = 2

The values match perfectly. Finally, we verify the point of contact using the standard formula for the point of contact in slope form:

(\frac{- m a ^{2}}{c}, \frac{a ^{2}}{c}) or (\frac{- ma}{1 + m ^{2}}, \frac{a}{1 + m ^{2}})

Substituting

m = 1

and

a = 2

, we calculate:

(\frac{- ( 1 ) ( 2 )}{2}, \frac{2}{2}) = (- 1, 1)

Everything aligns. The combination (I)(ii)(Q) is a mathematically proven reality. This consistency confirms that the logic is sound and the geometric interpretation is correct.

Common Traps & FAQs

Why do we use

T = 0

instead of finding the slope using differentiation?

While differentiation is a powerful tool, the

T = 0

method is a specialized shortcut for conics. It is faster, less prone to calculation errors, and specifically designed for JEE-level efficiency. It directly gives you the equation of the tangent at a point on the curve.

How do I know if I should start with the circle or another conic?

In multiple-choice questions like this, the best strategy is to test the options. Start with the simplest conic (the circle) and verify if the given point and tangent conditions hold true. If they do, you have your answer! If not, move to the next option.

Is it possible for the point to lie on the conic but the tangent equation to be wrong?

Absolutely. A point can lie on a curve, but the tangent at that point is unique. If your derived tangent doesn't match the options, it means either your calculation is off or the conic assumption is incorrect. Always verify both the point of contact and the tangent equation.

Animated Solution for Mathematics - Circles: For $a = 2$ , if a tangent is drawn to a suitable conic (Column 1) at the point of contact $(- 1, 1)$ , then which of the following options is the only CORRECT combination for obtaining its equation ?

Select Answer:

Visualized Solution

The Setup & Conic Assumption

Forming the Circle Equation

Verifying the Point $P (- 1, 1)$

Equation of Tangent using $T = 0$

Deriving the Tangent Equation

Analyzing the Tangent

The Standard Tangent Formula

Verifying the Tangent Equation

Standard Point of Contact

Verifying the Point of Contact

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Master Equation

Verification and Consistency

Similar Questions

If the tangent to the parabola $y^{2} = x$ at a point $(α, β), (β > 0)$ is also a tangent to the ellipse, $x^{2} + 2 y^{2} = 1$ , then $α$ is equal to :

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:

Comprehension Passage

A possible equation of $L$ is

A common tangent of the two circles is

If $2 x - y + 1 = 0$ is a tangent to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{16} = 1$ , then which of the following CANNOT be sides of a right angled triangle ?

Comprehension Passage

The value of $r$ is

If $s t = 1$ , then the tangent at $P$ and the normal at $S$ to the parabola meet at a point whose ordinate is

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

If the tangents on the ellipse $4 x^{2} + y^{2} = 8$ at the points $(1, 2)$ and $(a, b)$ are perpendicular to each other, then $a^{2}$ is equal to :

The locus of a point $P (α, β)$ moving under the condition that the line $y = α x + β$ is a tangent to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ is

Select Answer:

Visualized Solution

The Setup & Conic Assumption

Forming the Circle Equation

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

Equation of Tangent using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=0

Deriving the Tangent Equation

Analyzing the Tangent

The Standard Tangent Formula

Verifying the Tangent Equation

Standard Point of Contact

Verifying the Point of Contact

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Master Equation

Verification and Consistency

Similar Questions

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.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:

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

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 2x−y+1=0 is a tangent to the hyperbola a2x2​−16y2​=1, then which of the following CANNOT be sides of a right angled triangle ?

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of r is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If st=1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a tangent to the circle x2+y2=1 intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangents on the ellipse 4x2+y2=8 at the points (1,2) and (a,b) are perpendicular to each other, then a2 is equal to :

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Verifying the Point $P (- 1, 1)$

Equation of Tangent using $T = 0$

If the tangent to the parabola $y^{2} = x$ at a point $(α, β), (β > 0)$ is also a tangent to the ellipse, $x^{2} + 2 y^{2} = 1$ , then $α$ is equal to :

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:

A possible equation of $L$ is

If $2 x - y + 1 = 0$ is a tangent to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{16} = 1$ , then which of the following CANNOT be sides of a right angled triangle ?

The value of $r$ is

If $s t = 1$ , then the tangent at $P$ and the normal at $S$ to the parabola meet at a point whose ordinate is

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

If the tangents on the ellipse $4 x^{2} + y^{2} = 8$ at the points $(1, 2)$ and $(a, b)$ are perpendicular to each other, then $a^{2}$ is equal to :

The locus of a point $P (α, β)$ moving under the condition that the line $y = α x + β$ is a tangent to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ is