JEE Main 2023 (06 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let the tangent to the curve $x^{2} + 2 x - 4 y + 9 = 0$ at the point $P (1, 3)$ on it meet the $y$ -axis at $A$ . Let the line passing through $P$ and parallel to the line $x - 3 y = 6$ meet the parabola $y^{2} = 4 x$ at $B$ . If $B$ lies on the line $2 x - 3 y = 8$ , then $(A B)^{2}$ is equal to _______.

Q: Wait, why do we use the $T=0$ method instead of finding the derivative $dy/dx$?

Both methods are mathematically equivalent! The $T=0$ method is a powerful shortcut for conic sections that saves precious time in the exam hall. It avoids the potential for sign errors during differentiation and substitution. However, if you prefer calculus, you are absolutely free to use $dy/dx$—just ensure you evaluate the slope at $(1, 3)$ correctly.

Q: I keep getting a different slope for the parallel line. What am I doing wrong?

Check the standard form! For a line $ax + by + c = 0$, the slope is $-a/b$. For $x - 3y = 6$, rewrite it as $3y = x - 6$, so $y = (1/3)x - 2$. The slope is clearly $1/3$. If you are getting $-3$, you might be confusing the slope of the perpendicular line instead.

Q: Does the order of points in the distance formula $(AB)^2$ matter?

Not at all! Because we are squaring the differences $(x_2 - x_1)^2$ and $(y_2 - y_1)^2$, the sign of the difference disappears. Whether you calculate $A$ to $B$ or $B$ to $A$, the result will be identical. Symmetry is your friend in geometry!

Enter Numerical Value:

Visualized Solution

The Curve and Point $P$

Given curve: $x^{2} + 2 x - 4 y + 9 = 0$
Point on the curve: $P (1, 3)$

Tangent Equation Formula

To find the tangent at $(x_{1}, y_{1})$ , use $T = 0$
$x^{2} \to x x_{1}$ , $x \to \frac{x + x _{1}}{2}$ , $y \to \frac{y + y _{1}}{2}$
$x x_{1} + (x + x_{1}) - 2 (y + y_{1}) + 9 = 0$

Substituting Point $P$

Substitute $(x_{1}, y_{1}) = (1, 3)$ into the $T = 0$ equation
$x (1) + (x + 1) - 2 (y + 3) + 9 = 0$

Simplifying the Tangent

Expand: $x + x + 1 - 2 y - 6 + 9 = 0$
Combine terms: $2 x - 2 y + 4 = 0$
Divide by $2$ : $x - y + 2 = 0$

Finding Point $A$

The tangent meets the $y$ -axis at point $A$
On the $y$ -axis, $x = 0$
Substitute $x = 0$ : $0 - y + 2 = 0 \Rightarrow y = 2$
Coordinates of $A$ : $(0, 2)$

Slope of the Parallel Line

A new line passes through $P$ and is parallel to $x - 3 y = 6$
Slope of $x - 3 y = 6$ is $m = \frac{1}{3}$
Parallel lines have the same slope, so $m_{L} = \frac{1}{3}$

Equation of Line $L$

Use point-slope form: $y - y_{1} = m (x - x_{1})$
Point $P (1, 3)$ and slope $m = \frac{1}{3}$
Substitute: $y - 3 = \frac{1}{3} (x - 1)$

Simplifying Line $L$

Multiply by $3$ : $3 (y - 3) = x - 1$
Expand: $3 y - 9 = x - 1$
Rearrange: $x - 3 y + 8 = 0$

Intersection with Parabola

Line $L$ : $x = 3 y - 8$
Second curve is a parabola: $y^{2} = 4 x$
Substitute $x$ from line $L$ into the parabola's equation
$y^{2} = 4 (3 y - 8)$

Solving the Quadratic Equation

Expand: $y^{2} = 12 y - 32$
Rearrange: $y^{2} - 12 y + 32 = 0$
Factorize: $(y - 4) (y - 8) = 0$
Roots: $y = 4$ or $y = 8$

Potential Points for $B$

Use $x = 3 y - 8$ to find corresponding $x$ -coordinates
If $y = 4$ : $x = 3 (4) - 8 = 12 - 8 = 4 \Rightarrow (4, 4)$
If $y = 8$ : $x = 3 (8) - 8 = 24 - 8 = 16 \Rightarrow (16, 8)$
Two possible points for $B$ : $(4, 4)$ and $(16, 8)$

Identifying Point $B$

Constraint: $B$ must lie on the line $2 x - 3 y = 8$
Check $(4, 4)$ : $2 (4) - 3 (4) = 8 - 12 = - 4 \neq = 8$ (Rejected)
Check $(16, 8)$ : $2 (16) - 3 (8) = 32 - 24 = 8$ (Accepted)
Therefore, $B$ is $(16, 8)$

Setting up $(A B)^{2}$

We have $A (0, 2)$ and $B (16, 8)$
Distance formula squared: $(A B)^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Substitute coordinates: $(A B)^{2} = (16 - 0)^{2} + (8 - 2)^{2}$

Calculating the Final Answer

$(A B)^{2} = 1 6^{2} + 6^{2}$
$1 6^{2} = 256$
$6^{2} = 36$
$(A B)^{2} = 256 + 36 = 292$

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at a complex landscape defined by two distinct curves. One is a parabola, a graceful arc, and the other is a line cutting through the space.

Our goal is to navigate this landscape, find the hidden points of intersection, and calculate the distance between two specific landmarks, points

A

and

B

. This is a story of how different geometric entities interact.

The Tangent at Point $P$

We begin with the curve

x^{2} + 2 x - 4 y + 9 = 0

. We are interested in the behavior of this curve at the specific point

P (1, 3)

To find the tangent line at this point, we employ the elegant 'T=0' method. By replacing

x^{2}

with

x x_{1}

x

with

\frac{x + x _{1}}{2}

, and

y

with

\frac{y + y _{1}}{2}

, we transform the quadratic nature of the curve into the linear equation of the tangent.

Substituting

P (1, 3)

into this transformation, we get:

x (1) + (x + 1) - 2 (y + 3) + 9 = 0

Simplifying this, we arrive at

2 x - 2 y + 4 = 0

, or more simply,

x - y + 2 = 0

. This line represents the 'direction' of the curve at

P

When this line meets the

y

-axis, we set

x = 0

, yielding

y = 2

. Thus, our first landmark, point

A

, is found at

(0, 2)

The Parallel Path to $B$

Now, we shift our focus. We need a line passing through

P (1, 3)

that is parallel to

x - 3 y = 6

Parallel lines share the same slope. By rearranging

x - 3 y = 6

into

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

, we identify the slope

m = \frac{1}{3}

Using the point-slope form

y - 3 = \frac{1}{3} (x - 1)

, we derive the equation of our new line:

x - 3 y + 8 = 0

. This line is our path to the second landmark,

B

We are told that

B

lies on the intersection of this line and the parabola

y^{2} = 4 x

. By substituting

x = 3 y - 8

into the parabola's equation, we create a quadratic equation:

y^{2} = 4 (3 y - 8) \Rightarrow y^{2} - 12 y + 32 = 0

Factoring this, we find

y = 4

y = 8

. This gives us two potential candidates for

B

(4, 4)

and

(16, 8)

The Final Verification and Distance

The problem provides a final constraint:

B

must lie on the line

2 x - 3 y = 8

Testing our candidates, we find that

(4, 4)

fails the test, but

(16, 8)

satisfies it perfectly:

2 (16) - 3 (8) = 32 - 24 = 8

With

A (0, 2)

and

B (16, 8)

firmly in our grasp, the final step is to calculate the square of the distance between them. Using the distance formula:

(A B)^{2} = (16 - 0)^{2} + (8 - 2)^{2}

(A B)^{2} = 1 6^{2} + 6^{2} = 256 + 36 = 292

The journey is complete. We have navigated the curves, filtered the candidates, and arrived at the final, elegant result of 292.

Common Traps & FAQs

Wait, why do we use the

T = 0

method instead of finding the derivative

d y / d x

Both methods are mathematically equivalent! The

T = 0

method is a powerful shortcut for conic sections that saves precious time in the exam hall. It avoids the potential for sign errors during differentiation and substitution. However, if you prefer calculus, you are absolutely free to use

d y / d x

—just ensure you evaluate the slope at

(1, 3)

correctly.

Why did we get two points for

B

? How do I know which one to pick?

Great observation! A line intersecting a parabola will generally yield two points of intersection. The problem provides an extra constraint:

B

must lie on the line

2 x - 3 y = 8

. This is a classic JEE technique to filter out the 'extraneous' solution. Always check your candidates against all given conditions!

Is there a faster way to calculate

(A B)^{2}

without finding the coordinates of

A

and

B

While you could use vector notation or coordinate geometry properties, finding the coordinates is the most robust and error-proof path here. Since the coordinates are integers, the distance formula is very efficient. Don't look for shortcuts when the standard path is already a straight line!

I keep getting a different slope for the parallel line. What am I doing wrong?

Check the standard form! For a line

a x + b y + c = 0

, the slope is

- a / b

. For

x - 3 y = 6

, rewrite it as

3 y = x - 6

, so

y = (1/3) x - 2

. The slope is clearly

1/3

. If you are getting

- 3

, you might be confusing the slope of the perpendicular line instead.

Does the order of points in the distance formula

(A B)^{2}

matter?

Not at all! Because we are squaring the differences

(x_{2} - x_{1})^{2}

and

(y_{2} - y_{1})^{2}

, the sign of the difference disappears. Whether you calculate

A

B

B

A

, the result will be identical. Symmetry is your friend in geometry!

Enter Numerical Value:

Visualized Solution

The Curve and Point $P$

Tangent Equation Formula

Substituting Point $P$

Simplifying the Tangent

Finding Point $A$

Slope of the Parallel Line

Equation of Line $L$

Simplifying Line $L$

Intersection with Parabola

Solving the Quadratic Equation

Potential Points for $B$

Identifying Point $B$

Setting up $(A B)^{2}$

Calculating the Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent at Point $P$

The Parallel Path to $B$

The Final Verification and Distance

Similar Questions

Let the tangent to the parabola $y^{2} = 12 x$ at the point $(3, α)$ be perpendicular to the line $2 x + 2 y = 3$ . Then the square of distance of the point $(6, - 4)$ from the normal to the hyperbola $α^{2} x^{2} - 9 y^{2} = 9 α^{2}$ at its point $(α - 1, α + 2)$ is equal to .............

The tangents at the point $A (1, 3)$ and $B (1, - 1)$ on the parabola $y^{2} - 2 x - 2 y = 1$ meet at the point $P$ . Then the area (in unit $^{2}$ ) of the triangle $P A B$ is :-

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

Comprehension Passage

The coordinates of $A$ and $B$ are

The orthocenter of the triangle $P A B$ is

The equation of the locus of the point whose distances from the point $P$ and the line $A B$ are equal, is

Let $x^{2} + y^{2} + A x + B y + C = 0$ be a circle passing through $(0, 6)$ and touching the parabola $y = x^{2}$ at $(2, 4)$ . Then $A + C$ is equal to ____.

Let a line perpendicular to the line $2 x - y = 10$ touch the parabola $y^{2} = 4 (x - 9)$ at the point $P$ . The distance of the point $P$ from the centre of the circle $x^{2} + y^{2} - 14 x - 8 y + 56 = 0$ is __________

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

Comprehension Passage

A possible equation of $L$ is

A common tangent of the two circles is

Let a tangent to the Curve $9 x^{2} + 16 y^{2} = 144$ intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is _____.

Enter Numerical Value:

Visualized Solution

The Curve and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Tangent Equation Formula

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

Simplifying the Tangent

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

Slope of the Parallel Line

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

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

Intersection with Parabola

Solving the Quadratic Equation

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

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

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (AB)2

Calculating the Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent at Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

The Parallel Path to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

The Final Verification and Distance

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangents at the point A(1,3) and B(1,−1) on the parabola y2−2x−2y=1 meet at the point P. Then the area (in unit2) of the triangle PAB is :-

.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

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coordinates of A and B are

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the locus of the point whose distances from the point P and the line AB are equal, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let x2+y2+Ax+By+C=0 be a circle passing through (0,6) and touching the parabola y=x2 at (2,4). Then A+C is equal to ____.

.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

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a tangent to the Curve 9x2+16y2=144 intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is _____.

The Curve and Point $P$

Substituting Point $P$

Finding Point $A$

Equation of Line $L$

Simplifying Line $L$

Potential Points for $B$

Identifying Point $B$

Setting up $(A B)^{2}$

The Tangent at Point $P$

The Parallel Path to $B$

The tangents at the point $A (1, 3)$ and $B (1, - 1)$ on the parabola $y^{2} - 2 x - 2 y = 1$ meet at the point $P$ . Then the area (in unit $^{2}$ ) of the triangle $P A B$ is :-

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 coordinates of $A$ and $B$ are

The orthocenter of the triangle $P A B$ is

The equation of the locus of the point whose distances from the point $P$ and the line $A B$ are equal, is

Let $x^{2} + y^{2} + A x + B y + C = 0$ be a circle passing through $(0, 6)$ and touching the parabola $y = x^{2}$ at $(2, 4)$ . Then $A + C$ is equal to ____.

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

A possible equation of $L$ is

Let a tangent to the Curve $9 x^{2} + 16 y^{2} = 144$ intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is _____.