JEE Main 2023 (24 January Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: The equations of the sides AB and AC of a triangle ABC are $(λ + 1) x + λ y = 4$ and $λ x + (1 - λ) y + λ = 0$ respectively. Its vertex A is on the y-axis and its orthocentre is $(1, 2)$ . The length of the tangent from the point C to the part of the parabola $y^{2} = 6 x$ in the first quadrant is

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Since vertex

A

lies on the y-axis, its x-coordinate must be zero. Substitute

x = 0

into both given line equations to find the y-intercepts. Since both lines intersect at

A

, these y-intercepts must be identical, allowing you to solve for

λ

Hint 2

Hint 2: The Arsenal. The orthocentre

H (1, 2)

is the intersection of altitudes. Recall that the altitude from vertex

C

to side

A B

is perpendicular to

A B

. Use the slope of

A B

to find the slope of the altitude

C H

, and then use the point-slope form or slope formula with

H

and

C

to find the coordinates of

C

Hint 3

Hint 3: The Execution. Once you have point

C

, you need the tangent from

C

to the parabola

y^{2} = 6 x

. Use the condition that a line

y = m x + c

is tangent to

y^{2} = 4 a x

c = a / m

. Substitute the coordinates of

C

into this equation to solve for

m

, and remember to verify the point of contact lies in the first quadrant.

Visualized Solution

Identify Vertex A on the y-axis

Vertex $A$ lies on the y-axis $⟹ x = 0$
Equation of AB: $(λ + 1) x + λ y = 4$
Equation of AC: $λ x + (1 - λ) y + λ = 0$

Find y-intercepts of AB and AC

For AB at $x = 0$ : $λ y = 4 ⟹ y = \frac{4}{λ}$
For AC at $x = 0$ : $(1 - λ) y + λ = 0 ⟹ y = \frac{λ}{λ - 1}$

Solve for $λ$

Equating $y$ : $\frac{4}{λ} = \frac{λ}{λ - 1}$
$4 λ - 4 = λ^{2} ⟹ λ^{2} - 4 λ + 4 = 0$
$(λ - 2)^{2} = 0 ⟹ λ = 2$

Find Equations of AB and AC

Substitute $λ = 2$ into the equations:
Side AB: $3 x + 2 y = 4$
Side AC: $2 x - y + 2 = 0$
Vertex $A = (0, 2)$

Parametrize Vertex C

Point $C$ lies on $2 x - y + 2 = 0$
Let $x = α$ , then $y = 2 α + 2$
Coordinates of $C = (α, 2 α + 2)$

Use Orthocentre Property

Orthocentre $H = (1, 2)$
Altitude through $C$ is perpendicular to AB
Slope of AB ( $m_{A B}$ ) = $- \frac{3}{2}$
Slope of altitude CH ( $m_{C H}$ ) = $\frac{2}{3}$

Solve for Coordinates of C

$m_{C H} = \frac{( 2 α + 2 ) - 2}{α - 1} = \frac{2 α}{α - 1}$
$\frac{2 α}{α - 1} = \frac{2}{3} ⟹ 6 α = 2 α - 2$
$4 α = - 2 ⟹ α = - \frac{1}{2}$
Point $C = (- \frac{1}{2}, 1)$

Parabola and Tangent Setup

Parabola: $y^{2} = 6 x ⟹ 4 a = 6 ⟹ a = \frac{3}{2}$
Equation of tangent: $y = m x + \frac{a}{m}$
$y = m x + \frac{3}{2 m}$

Find Tangent Slope m

Tangent passes through $C (- \frac{1}{2}, 1)$
$1 = m (- \frac{1}{2}) + \frac{3}{2 m}$
$2 m = - m^{2} + 3 ⟹ m^{2} + 2 m - 3 = 0$
$(m + 3) (m - 1) = 0 ⟹ m = 1, - 3$

Identify Point of Contact T

Point of contact $T = (\frac{a}{m ^{2}}, \frac{2 a}{m})$
For $m = 1$ : $T = (\frac{3/2}{1 ^{2}}, \frac{2 ( 3/2 )}{1}) = (\frac{3}{2}, 3)$
Since $T$ is in the first quadrant, $m = 1$ is valid.

Calculate Length CT

Length $C T = (\frac{3}{2} - (- \frac{1}{2}))^{2} + (3 - 1)^{2}$
$C T = 2^{2} + 2^{2} = 4 + 4 = 8$
Final Answer: $22$

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

Vertex

A

lies on the y-axis, which implies its x-coordinate is

0

. Given the equations of sides

A B

and

A C

, we substitute

x = 0

to find the y-intercepts.

Since both lines pass through

A

, their y-intercepts must be equal:

\frac{4}{λ} = \frac{λ}{λ - 1}

Solving this leads to the quadratic equation:

λ^{2} - 4 λ + 4 = 0 \Rightarrow (λ - 2)^{2} = 0

Thus, we find

λ = 2

. Substituting this value back, the lines become

3 x + 2 y = 4

and

2 x - y + 2 = 0

, and vertex

A

is revealed as

(0, 2)

The Orthocentre

A Bridge to Point C

The orthocentre is given as

H (1, 2)

. The altitude from

C

A B

is perpendicular to

A B

The slope of

A B

- \frac{3}{2}

, so the slope of the altitude

C H

must be the negative reciprocal, which is

\frac{2}{3}

. Since

C

lies on the line

2 x - y + 2 = 0

, we parameterize it as

C (α, 2 α + 2)

Using the slope formula between

C (α, 2 α + 2)

and

H (1, 2)

\frac{( 2 α + 2 ) - 2}{α - 1} = \frac{2}{3}

This simplifies to

6 α = 2 α - 2

, which yields

α = - \frac{1}{2}

. Thus, point

C

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

The Parabolic Dance

For the parabola

y^{2} = 6 x

, we identify

4 a = 6

, so

a = \frac{3}{2}

. We seek the tangent from

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

to this parabola.

The condition for a line

y = m x + c

to be tangent to

y^{2} = 4 a x

c = \frac{a}{m}

. Substituting

C

into

y = m x + \frac{3}{2 m}

1 = m (- \frac{1}{2}) + \frac{3}{2 m}

Multiplying by

2 m

yields

2 m = - m^{2} + 3

, or

m^{2} + 2 m - 3 = 0

. Factoring gives

(m + 3) (m - 1) = 0

Final Calculation

Following the first-quadrant constraint, we choose

m = 1

. The point of contact

T

is given by:

T = (\frac{a}{m ^{2}}, \frac{2 a}{m}) = (\frac{3}{2}, 3)

The final distance

C T

is calculated as:

C T = (\frac{3}{2} - (- \frac{1}{2}))^{2} + (3 - 1)^{2} = 2^{2} + 2^{2} = 8

The final result is $22$ .

Common Traps & FAQs

Wait, why do we reject the negative root for the slope

m

Great observation! When we solve the quadratic equation for the slope

m

, we get two possible values. However, the problem explicitly states that the tangent touches the parabola in the first quadrant. The point of contact

T

is given by

(\frac{a}{m ^{2}}, \frac{2 a}{m})

. If you test

m = - 3

with

a = 3/2

, the y-coordinate becomes

2 (3/2) / (- 3) = - 1

, which is in the fourth quadrant. Thus, we must reject

m = - 3

and accept

m = 1

Why can't we just use the formula for the length of the tangent from a point to a parabola?

You absolutely can if you have the formula memorized! However, in JEE Advanced, deriving the point of contact is often safer because it forces you to verify the geometric constraints (like the quadrant condition). Calculating the distance between

C

and the point of contact

T

is a robust, foolproof method that minimizes the risk of misapplying a formula.

Is it necessary to find the coordinates of vertex B?

Not at all! The beauty of this problem lies in its efficiency. We only need the slope of

A B

to determine the slope of the altitude

C H

. Once we have the slope of

C H

and the coordinates of the orthocentre

H

, we can find

C

directly without ever needing to solve for

B

. Always look for the path of least resistance in coordinate geometry.

Select Answer:

Visualized Solution

Identify Vertex A on the y-axis

Find y-intercepts of AB and AC

Solve for $λ$

Find Equations of AB and AC

Parametrize Vertex C

Use Orthocentre Property

Solve for Coordinates of C

Parabola and Tangent Setup

Find Tangent Slope m

Identify Point of Contact T

Calculate Length CT

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Orthocentre

The Parabolic Dance

Final Calculation

Similar Questions

Let $A_{1}, B_{1}, C_{1}$ be three points in the $x y$ -plane. Suppose that the lines $A_{1} C_{1}$ and $B_{1} C_{1}$ are tangents to the curve $y^{2} = 8 x$ at $A_{1}$ and $B_{1}$ , respectively. If $O = (0, 0)$ and $C_{1} = (- 4, 0)$ , then which of the following statements is(are) TRUE?

If the tangents drawn at the points $P$ and $Q$ on the parabola $y^{2} = 2 x - 3$ intersect at the point $R (0, 1)$ , then the orthocentre of the triangle $P QR$ is :

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 :-

Let $O$ be the origin and $O P$ and $O Q$ be the tangents to the circle $x^{2} + y^{2} - 6 x + 4 y + 8 = 0$ at the points $P$ and $Q$ on it. If the circumcircle of the triangle $O P Q$ passes through the point $(α, \frac{1}{2})$ , then a value of $α$ is

Let the tangents at two points $A$ and $B$ on the circle $x^{2} + y^{2} - 4 x + 3 = 0$ meet at origin $O (0, 0)$ . Then the area of the triangle of $O A B$ is

If the tangents drawn at the point $O (0, 0)$ and $P (1 + 5, 2)$ on the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ intersect at the point $Q$ , then the area of the triangle $O P Q$ is equal to

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

Points $A, B$ and $C$ lie on the parabola $y^{2} = 4 a x$ . The tangents to the parabola at $A, B$ and $C$ , taken in pairs, intersect at points $P, Q$ and $R$ . Determine the ratio of the areas of the triangles $A B C$ and $P QR$ .

Select Answer:

Visualized Solution

Identify Vertex A on the y-axis

Find y-intercepts of AB and AC

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

Find Equations of AB and AC

Parametrize Vertex C

Use Orthocentre Property

Solve for Coordinates of C

Parabola and Tangent Setup

Find Tangent Slope m

Identify Point of Contact T

Calculate Length CT

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Orthocentre

The Parabolic Dance

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangents drawn at the points P and Q on the parabola y2=2x−3 intersect at the point R(0,1), then the orthocentre of the triangle PQR is :

.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; } Let the tangents at two points A and B on the circle x2+y2−4x+3=0 meet at origin O(0,0). Then the area of the triangle of OAB is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangents drawn at the point O(0,0) and P(1+5​,2) on the circle x2+y2−2x−4y=0 intersect at the point Q, then the area of the triangle OPQ is equal to

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

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

Solve for $λ$

If the tangents drawn at the points $P$ and $Q$ on the parabola $y^{2} = 2 x - 3$ intersect at the point $R (0, 1)$ , then the orthocentre of the triangle $P QR$ is :

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 :-

Let the tangents at two points $A$ and $B$ on the circle $x^{2} + y^{2} - 4 x + 3 = 0$ meet at origin $O (0, 0)$ . Then the area of the triangle of $O A B$ is

If the tangents drawn at the point $O (0, 0)$ and $P (1 + 5, 2)$ on the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ intersect at the point $Q$ , then the area of the triangle $O P Q$ is equal to

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