JEE Main 2024 (06 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let $L_{1}, L_{2}$ be the lines passing through the point $P (0, 1)$ and touching the parabola $9 x^{2} + 12 x + 18 y - 14 = 0$ . Let $Q$ and $R$ be the points on the lines $L_{1}$ and $L_{2}$ such that the $△ P QR$ is an isosceles triangle with base $QR$ . If the slopes of the lines $QR$ are $m_{1}$ and $m_{2}$ , then $16 (m_{1}^{2} + m_{2}^{2})$ is equal to _______

Q: Why does the isosceles condition $PQ=PR$ mean the base $QR$ makes equal angles with the tangents?

In an isosceles triangle $PQR$ with base $QR$, the angles at the base are equal. Since $P$ is the vertex, the line $QR$ must be symmetric with respect to the two sides $PQ$ and $PR$. This symmetry means the line $QR$ must make the same angle with both tangents.

Enter Numerical Value:

Progressive Hints

Hint 1

The Spark: Visualize the parabola. It is not centered at the origin. The point

P (0, 1)

is the vertex of the angle formed by the two tangents. Before doing any algebra, recognize that the symmetry of the parabola dictates the behavior of these tangents.

Hint 2

The Arsenal: The condition that

△ P QR

is isosceles with base

QR

is the key. In any triangle, if the base is

QR

, the sides

P Q

and

P R

must be equal. This implies that the line

QR

must be equally inclined to the two tangents

L_{1}

and

L_{2}

. Recall the formula for the angle between two lines with slopes

m_{1}

and

m_{2}

Hint 3

The Execution: Let the slope of the line

QR

m

. The angle between

QR

and

L_{1}

(slope

m_{L 1}

) must equal the angle between

QR

and

L_{2}

(slope

m_{L 2}

). Use the formula

tan θ = ∣ \frac{m - m _{L 1}}{1 + m \cdot m _{L 1}} ∣ = ∣ \frac{m - m _{L 2}}{1 + m \cdot m _{L 2}} ∣

to set up your equation for

m

Visualized Solution

The Geometric Setup

Given Parabola: $9 x^{2} + 12 x + 18 y - 14 = 0$
Point $P (0, 1)$ lies outside the parabola.
Two tangents $L_{1}$ and $L_{2}$ are drawn from $P$ .

Completing the Square

Group $x$ terms: $9 (x^{2} + \frac{4}{3} x) = - 18 y + 14$
Add $(\frac{2}{3})^{2}$ inside the bracket:
$9 (x^{2} + \frac{4}{3} x + \frac{4}{9}) = - 18 y + 14 + 9 (\frac{4}{9})$

Standard Form of Parabola

Simplify the right side: $- 18 y + 18 = - 18 (y - 1)$
Standard Form: $9 (x + \frac{2}{3})^{2} = - 18 (y - 1)$
$(x + \frac{2}{3})^{2} = - 2 (y - 1)$

Coordinate Transformation

Shift origin to $(- \frac{2}{3}, 1)$
Let $X = x + \frac{2}{3}$ and $Y = y - 1$
New Parabola: $X^{2} = - 2 Y$
New Point $P$ : $X = 0 + \frac{2}{3} = \frac{2}{3}$ , $Y = 1 - 1 = 0 ⟹ P (\frac{2}{3}, 0)$

Tangent to $X^{2} = 4 aY$

Standard tangent equation: $Y = m X - a m^{2}$
Compare $X^{2} = - 2 Y$ with $X^{2} = 4 aY$
$4 a = - 2 ⟹ a = - \frac{1}{2}$
Tangent: $Y = m X - (- \frac{1}{2}) m^{2} = m X + \frac{1}{2} m^{2}$

Finding Tangent Slopes

Tangents pass through $P (\frac{2}{3}, 0)$
Substitute $X = \frac{2}{3}, Y = 0$ :
$0 = m (\frac{2}{3}) + \frac{1}{2} m^{2}$
Multiply by $6$ : $0 = 4 m + 3 m^{2}$

Slopes of $L_{1}$ and $L_{2}$

Factorize: $m (3 m + 4) = 0$
Roots: $m = 0$ and $m = - \frac{4}{3}$
Slope of $L_{1}$ , $m_{L 1} = 0$
Slope of $L_{2}$ , $m_{L 2} = - \frac{4}{3}$

The Isosceles Triangle Condition

$△ P QR$ is isosceles with base $QR$ .
This means $P Q = P R$ .
Therefore, the base $QR$ makes equal angles with $L_{1}$ and $L_{2}$ .
Let the slope of $QR$ be $m$ .

Angle Between Lines Formula

Angle $θ$ between lines with slopes $m_{a}$ and $m_{b}$ :
$tan θ = ∣ \frac{m _{a} - m _{b}}{1 + m _{a} m _{b}} ∣$
Equating angles: $∣ \frac{m - m _{L 1}}{1 + m \cdot m _{L 1}} ∣ = ∣ \frac{m - m _{L 2}}{1 + m \cdot m _{L 2}} ∣$

Substituting the Slopes

Substitute $m_{L 1} = 0$ and $m_{L 2} = - \frac{4}{3}$ :
$∣ \frac{m - 0}{1 + m ( 0 )} ∣ = ∣ \frac{m - ( - \frac{4}{3} )}{1 + m ( - \frac{4}{3} )} ∣$
$∣ m ∣ = ∣ \frac{m + \frac{4}{3}}{1 - \frac{4 m}{3}} ∣$

Simplifying the Modulus Equation

Multiply numerator and denominator by $3$ :
$∣ m ∣ = ∣ \frac{3 m + 4}{3 - 4 m} ∣$
This gives two cases:
Case 1: $m = \frac{3 m + 4}{3 - 4 m}$
Case 2: $m = - \frac{3 m + 4}{3 - 4 m}$

Solving Case 1

$m (3 - 4 m) = 3 m + 4$
$3 m - 4 m^{2} = 3 m + 4$
$- 4 m^{2} = 4 ⟹ m^{2} = - 1$
No real roots for $m$ .

Solving Case 2

$m (3 - 4 m) = - (3 m + 4)$
$3 m - 4 m^{2} = - 3 m - 4$
$4 m^{2} - 6 m - 4 = 0$
Divide by 2: $2 m^{2} - 3 m - 2 = 0$

Finding $m_{1}$ and $m_{2}$

Factorize: $2 m^{2} - 4 m + m - 2 = 0$
$2 m (m - 2) + 1 (m - 2) = 0$
$(2 m + 1) (m - 2) = 0$
Slopes are $m_{1} = 2$ and $m_{2} = - \frac{1}{2}$

Final Computation

We need to find: $16 (m_{1}^{2} + m_{2}^{2})$
Substitute $m_{1} = 2$ and $m_{2} = - \frac{1}{2}$ :
$16 (2^{2} + (- \frac{1}{2})^{2}) = 16 (4 + \frac{1}{4})$
$16 (\frac{17}{4}) = 4 \times 17 = 68$
Final Answer: 68

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The given parabola is defined by the equation

9 x^{2} + 12 x + 18 y - 14 = 0

. To simplify this, we complete the square for the

x

terms:

9 (x^{2} + \frac{4}{3} x + \frac{4}{9}) + 18 y - 14 - 4 = 0

This simplifies to

9 (x + \frac{2}{3})^{2} = - 18 (y - 1)

, which reduces to the standard form:

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

We define the shifted coordinates

X = x + \frac{2}{3}

and

Y = y - 1

. The parabola becomes

X^{2} = - 2 Y

, and the point

P (0, 1)

transforms to

P (\frac{2}{3}, 0)

in the

X Y

-plane.

The Tangent Hunt

Finding

L_{1}

and

L_{2}

For a parabola

X^{2} = 4 aY

, the equation of a tangent with slope

m

Y = m X - a m^{2}

. Comparing

X^{2} = - 2 Y

X^{2} = 4 aY

, we identify

4 a = - 2

, or

a = - \frac{1}{2}

The tangent equation is

Y = m X + \frac{1}{2} m^{2}

. Since the tangents pass through

P (\frac{2}{3}, 0)

, we substitute these coordinates:

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

Multiplying by

6

yields

3 m^{2} + 4 m = 0

, which factors as

m (3 m + 4) = 0

. Thus, the slopes of the two tangents are

m_{L 1} = 0

and

m_{L 2} = - \frac{4}{3}

The Geometry of Isosceles Triangles

Given that

△ P QR

is isosceles with base

QR

and

P Q = P R

, the line

QR

must be equally inclined to the two tangents

L_{1}

and

L_{2}

. This implies that

QR

is the angle bisector of the tangents.

Using the angle between two lines formula

tan θ = ∣ \frac{m _{a} - m _{b}}{1 + m _{a} m _{b}} ∣

, we set the angle between

QR

(slope

m

) and

L_{1}

equal to the angle between

QR

and

L_{2}

\frac{m - 0}{1 + m ( 0 )} = \frac{m - ( - 4/3 )}{1 + m ( - 4/3 )}

This simplifies to the following relation:

∣ m ∣ = \frac{3 m + 4}{3 - 4 m}

The Final Calculation

We solve for

m

by considering the two possible cases for the absolute value equality:

Case 1:

m = \frac{3 m + 4}{3 - 4 m} \Rightarrow 3 m - 4 m^{2} = 3 m + 4 \Rightarrow - 4 m^{2} = 4

. This yields

m^{2} = - 1

, which provides no real solutions.

Case 2:

m = - \frac{3 m + 4}{3 - 4 m} \Rightarrow 3 m - 4 m^{2} = - 3 m - 4

. Rearranging gives the quadratic equation:

4 m^{2} - 6 m - 4 = 0 \Rightarrow 2 m^{2} - 3 m - 2 = 0

Factoring the quadratic, we get

(2 m + 1) (m - 2) = 0

, resulting in slopes

m_{1} = 2

and

m_{2} = - \frac{1}{2}

. We are asked to calculate

16 (m_{1}^{2} + m_{2}^{2})

16 (2^{2} + (- \frac{1}{2})^{2}) = 16 (4 + \frac{1}{4}) = 16 (\frac{17}{4}) = 68

The final result is 68.

Common Traps & FAQs

Why do we shift the origin? Can't we just use the tangent formula for the original equation?

You absolutely could, but shifting the origin to the vertex of the parabola simplifies the equation to

X^{2} = 4 aY

, which makes the tangent formula

Y = m X - a m^{2}

much easier to apply. It is a strategic move to reduce algebraic errors, not a necessity.

Why does the isosceles condition

P Q = P R

mean the base

QR

makes equal angles with the tangents?

In an isosceles triangle

P QR

with base

QR

, the angles at the base are equal. Since

P

is the vertex, the line

QR

must be symmetric with respect to the two sides

P Q

and

P R

. This symmetry means the line

QR

must make the same angle with both tangents.

Wait, why did one of the cases for

m

result in

m^{2} = - 1

? Did I make a mistake?

You did not make a mistake! That is the beauty of the math. The equation

∣ m ∣ = ∣ \frac{m + 4/3}{1 - 4 m /3} ∣

leads to two cases. One case represents the line that is parallel to the axis of symmetry, which is impossible for a base of a triangle in this configuration. The math naturally filters out the impossible case.

Enter Numerical Value:

Visualized Solution

The Geometric Setup

Completing the Square

Standard Form of Parabola

Coordinate Transformation

Tangent to $X^{2} = 4 aY$

Finding Tangent Slopes

Slopes of $L_{1}$ and $L_{2}$

The Isosceles Triangle Condition

Angle Between Lines Formula

Substituting the Slopes

Simplifying the Modulus Equation

Solving Case 1

Solving Case 2

Finding $m_{1}$ and $m_{2}$

Final Computation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent Hunt

The Geometry of Isosceles Triangles

The Final Calculation

Similar Questions

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 __________

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 .............

If the tangent at a point P on the parabola $y^{2} = 3 x$ is parallel to the line $x + 2 y = 1$ and the tangents at the points Q and R on the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{1} = 1$ are perpendicular to the line $x - y = 2$ , then the area of the triangle PQR is:

Two tangent lines $l_{1}$ and $l_{2}$ are drawn from the point $(2, 0)$ to the parabola $2 y^{2} = - x$ . If the lines $l_{1}$ and $l_{2}$ are also tangent to the circle $(x - 5)^{2} + y^{2} = r$ , then $17 r$ is equal to

Enter Numerical Value:

Visualized Solution

The Geometric Setup

Completing the Square

Standard Form of Parabola

Coordinate Transformation

Tangent to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X2=4aY

Finding Tangent Slopes

Slopes of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

The Isosceles Triangle Condition

Angle Between Lines Formula

Substituting the Slopes

Simplifying the Modulus Equation

Solving Case 1

Solving Case 2

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m2​

Final Computation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent Hunt

The Geometry of Isosceles Triangles

The Final Calculation

Similar Questions

Tangent to $X^{2} = 4 aY$

Slopes of $L_{1}$ and $L_{2}$

Finding $m_{1}$ and $m_{2}$