Let the tangent drawn to the parabola $y^2 = 24x$ at the point $(\alpha, \beta)$ is perpendicular to the line $2x + 2y = 5$. Then the normal to the hyperbola $\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$ at the point $(\alpha + 4, \beta + 4)$ does NOT pass through the point : Options: (A) $(25, 10)$ (B) $(20, 12)$ (C) $(30, 8)$ (D) $(15, 13)$

### Analyzing the Parabola and Tangent We begin with the parabola defined by the equation $y^2 = 24x$. We are given a point $(\alpha, \beta)$ on this curve such that the tangent at this point is perpendicular to the line $2x + 2y = 5$. To find the slope of the tangent, we differentiate the parabola equation with respect to $x$: $$2y \frac{dy}{dx} = 24 \implies \frac{dy}{dx} = \frac{12}{y}$$ At the point $(\alpha, \beta)$, the slope of the tangent is $m_1 = \frac{12}{\beta}$. The given line $2x + 2y = 5$ can be rewritten as $y = -x + \frac{5}{2}$, which has a slope $m_2 = -1$. ### Determining the Coordinates Since the tangent and the line are perpendicular, their slopes must satisfy the condition $m_1 \cdot m_2 = -1$. Substituting the known values: $$\left( \frac{12}{\beta} \right) \cdot (-1) = -1 \implies \beta = 12$$ Now, we substitute $\beta = 12$ into the parabola equation $y^2 = 24x$ to find $\alpha$: $$144 = 24\alpha \implies \alpha = 6$$ Thus, the coordinates of the point are $(6, 12)$. ### The Hyperbola and the Normal With $\alpha = 6$ and $\beta = 12$, the hyperbola is defined by: $$\frac{x^2}{36} - \frac{y^2}{144} = 1$$ We are tasked with finding the equation of the normal at the point $(\alpha + 4, \beta + 4)$, which corresponds to $(10, 16)$. We use the standard normal equation for a hyperbola $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$. Substituting $a^2 = 36$, $b^2 = 144$, $x_1 = 10$, and $y_1 = 16$: $$\frac{36x}{10} + \frac{144y}{16} = 36 + 144$$ ### Final Calculation and Verification Simplifying the equation above: $$3.6x + 9y = 180$$ Multiplying by $\frac{10}{18}$, we arrive at the linear equation: $$2x + 5y = 100$$ We test the provided points to identify which one does not satisfy this equation: * For $(25, 10)$: $2(25) + 5(10) = 50 + 50 = 100$ (Satisfies) * For $(20, 12)$: $2(20) + 5(12) = 40 + 60 = 100$ (Satisfies) * For $(30, 8)$: $2(30) + 5(8) = 60 + 40 = 100$ (Satisfies) * For $(15, 13)$: $2(15) + 5(13) = 30 + 65 = 95 \neq 100$ (Does not satisfy) The point that does not lie on the normal is **$(15, 13)$**.

JEE Main 2022 (26 July Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Let the tangent drawn to the parabola $y^{2} = 24 x$ at the point $(α, β)$ is perpendicular to the line $2 x + 2 y = 5$ . Then the normal to the hyperbola $\frac{x ^{2}}{α ^{2}} - \frac{y ^{2}}{β ^{2}} = 1$ at the point $(α + 4, β + 4)$ does NOT pass through the point :

Q: Wait, why do we differentiate the parabola? Can't we just use the slope-intercept form $y = mx + c$?

That is a great question! While you can use the condition $c = \frac{a}{m}$ for a tangent to a parabola, differentiating is often more robust and less prone to memory errors. By differentiating $y^2 = 24x$, we get $\frac{dy}{dx} = \frac{12}{y}$. This gives us the slope at any point $(\alpha, \beta)$ directly. It is a powerful, universal tool that works even if you forget the specific tangent formulas for conics.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the geometric connection. You are given a parabola and a line. The tangent to the parabola at

(α, β)

is perpendicular to the line

2 x + 2 y = 5

. Recall that the slope of the tangent is the derivative of the parabola's equation at that point. How does the perpendicularity condition relate the slope of the tangent to the slope of the given line?

Hint 2

Hint 2: Now that you have the slope, use the condition

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

to find

β

. Once you have

β

, substitute it back into the parabola's equation

y^{2} = 24 x

to unlock the value of

α

. This is your foundation for the next stage.

Hint 3

Hint 3: With

α

and

β

known, define the hyperbola

\frac{x ^{2}}{α ^{2}} - \frac{y ^{2}}{β ^{2}} = 1

. The problem asks for the normal at

(α + 4, β + 4)

. Use the standard formula for the normal to a hyperbola at a point

(x_{1}, y_{1})

, which is

\frac{a ^{2} x}{x _{1}} + \frac{b ^{2} y}{y _{1}} = a^{2} + b^{2}

. Substitute your values and check which option does not satisfy this linear equation.

Visualized Solution

Analyze the Parabola

Given Parabola: $y^{2} = 24 x$
Point on parabola: $(α, β)$

Slope of Tangent

Differentiating $y^{2} = 24 x$ with respect to $x$ :
$2 y \frac{d y}{d x} = 24 ⟹ \frac{d y}{d x} = \frac{12}{y}$
Slope of tangent $m_{1}$ at $(α, β)$ is: $m_{1} = \frac{12}{β}$

Slope of the Given Line

Given line: $2 x + 2 y = 5 ⟹ y = - x + \frac{5}{2}$
Slope of this line $m_{2} = - 1$

Condition for Perpendicularity

Condition: $m_{1} \cdot m_{2} = - 1$
Substituting values: $(\frac{12}{β}) \cdot (- 1) = - 1$
$⟹ \frac{12}{β} = 1 ⟹ β = 12$

Finding $α$

Point $(α, β)$ lies on $y^{2} = 24 x$
Substitute $β = 12$ : $1 2^{2} = 24 α$
$144 = 24 α ⟹ α = 6$
Point $(α, β) = (6, 12)$

Equation of the Hyperbola

Hyperbola: $\frac{x ^{2}}{α ^{2}} - \frac{y ^{2}}{β ^{2}} = 1$
Substitute $α = 6, β = 12$ :
$\frac{x ^{2}}{36} - \frac{y ^{2}}{144} = 1$
Here $a^{2} = 36$ and $b^{2} = 144$

Point for the Normal

Point for normal: $(α + 4, β + 4)$
Substitute $α = 6, β = 12$ :
Point $(x_{1}, y_{1}) = (6 + 4, 12 + 4) = (10, 16)$

Normal to Hyperbola

Equation of normal to $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ at $(x_{1}, y_{1})$ :
$\frac{a ^{2} x}{x _{1}} + \frac{b ^{2} y}{y _{1}} = a^{2} + b^{2}$

Setting up the Normal Equation

Substitute $a^{2} = 36, b^{2} = 144, x_{1} = 10, y_{1} = 16$ :
$\frac{36 x}{10} + \frac{144 y}{16} = 36 + 144$
$3.6 x + 9 y = 180$

Simplifying the Equation

Simplify $3.6 x + 9 y = 180$ :
Multiply by $10$ : $36 x + 90 y = 1800$
Divide by $18$ :
$2 x + 5 y = 100$

Checking the Options

Check points on $2 x + 5 y = 100$ :
(A) $(25, 10) : 2 (25) + 5 (10) = 50 + 50 = 100$
(B) $(20, 12) : 2 (20) + 5 (12) = 40 + 60 = 100$
(C) $(30, 8) : 2 (30) + 5 (8) = 60 + 40 = 100$
(D) $(15, 13) : 2 (15) + 5 (13) = 30 + 65 = 95 \neq = 100$

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Parabola and Tangent

We begin with the parabola defined by the equation

y^{2} = 24 x

. We are given a point

(α, β)

on this curve such that the tangent at this point is perpendicular to the line

2 x + 2 y = 5

To find the slope of the tangent, we differentiate the parabola equation with respect to

x

2 y \frac{d y}{d x} = 24 ⟹ \frac{d y}{d x} = \frac{12}{y}

At the point

(α, β)

, the slope of the tangent is

m_{1} = \frac{12}{β}

. The given line

2 x + 2 y = 5

can be rewritten as

y = - x + \frac{5}{2}

, which has a slope

m_{2} = - 1

Determining the Coordinates

Since the tangent and the line are perpendicular, their slopes must satisfy the condition

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

. Substituting the known values:

(\frac{12}{β}) \cdot (- 1) = - 1 ⟹ β = 12

Now, we substitute

β = 12

into the parabola equation

y^{2} = 24 x

to find

α

144 = 24 α ⟹ α = 6

Thus, the coordinates of the point are

(6, 12)

The Hyperbola and the Normal

With

α = 6

and

β = 12

, the hyperbola is defined by:

\frac{x ^{2}}{36} - \frac{y ^{2}}{144} = 1

We are tasked with finding the equation of the normal at the point

(α + 4, β + 4)

, which corresponds to

(10, 16)

. We use the standard normal equation for a hyperbola

\frac{a ^{2} x}{x _{1}} + \frac{b ^{2} y}{y _{1}} = a^{2} + b^{2}

Substituting

a^{2} = 36

b^{2} = 144

x_{1} = 10

, and

y_{1} = 16

\frac{36 x}{10} + \frac{144 y}{16} = 36 + 144

Final Calculation and Verification

Simplifying the equation above:

3.6 x + 9 y = 180

Multiplying by

\frac{10}{18}

, we arrive at the linear equation:

2 x + 5 y = 100

We test the provided points to identify which one does not satisfy this equation: For $(25, 10)$ : $2 (25) + 5 (10) = 50 + 50 = 100$ (Satisfies) For

(20, 12)

2 (20) + 5 (12) = 40 + 60 = 100

(Satisfies) For $(30, 8)$ : $2 (30) + 5 (8) = 60 + 40 = 100$ (Satisfies) For

(15, 13)

: $2(15) + 5(13) = 30 + 65 = 95 eq 100$ (Does not satisfy)

The point that does not lie on the normal is $(15, 13)$ .

Common Traps & FAQs

Wait, why do we differentiate the parabola? Can't we just use the slope-intercept form

y = m x + c

That is a great question! While you can use the condition

c = \frac{a}{m}

for a tangent to a parabola, differentiating is often more robust and less prone to memory errors. By differentiating

y^{2} = 24 x

, we get

\frac{d y}{d x} = \frac{12}{y}

. This gives us the slope at any point

(α, β)

directly. It is a powerful, universal tool that works even if you forget the specific tangent formulas for conics.

Is the normal equation

\frac{a ^{2} x}{x _{1}} + \frac{b ^{2} y}{y _{1}} = a^{2} + b^{2}

always valid for any point on the hyperbola?

Yes, absolutely! This is the standard equation of the normal to the hyperbola

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

at the point

(x_{1}, y_{1})

. It is derived from the fact that the normal is perpendicular to the tangent at that point. Memorizing this form is a huge time-saver for JEE Advanced, as it bypasses the need to derive the slope of the tangent and then find the negative reciprocal every single time.

I got the equation

2 x + 5 y = 100

. How do I know which point is the correct answer?

The question asks which point does NOT pass through the normal. This means you need to plug the coordinates of each option into the equation

2 x + 5 y = 100

. If the left side equals 100, the point lies on the line. If it does not equal 100, that is your answer! In this case,

(15, 13)

gives

30 + 65 = 95

, which is not 100, so that is the point that does not lie on the normal.

Select Answer:

Visualized Solution

Analyze the Parabola

Slope of Tangent

Slope of the Given Line

Condition for Perpendicularity

Finding $α$

Equation of the Hyperbola

Point for the Normal

Normal to Hyperbola

Setting up the Normal Equation

Simplifying the Equation

Checking the Options

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Parabola and Tangent

Determining the Coordinates

The Hyperbola and the Normal

Final Calculation and Verification

Similar Questions

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Select Answer:

Visualized Solution

Analyze the Parabola

Slope of Tangent

Slope of the Given Line

Condition for Perpendicularity

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

Equation of the Hyperbola

Point for the Normal

Normal to Hyperbola

Setting up the Normal Equation

Simplifying the Equation

Checking the Options

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Parabola and Tangent

Determining the Coordinates

The Hyperbola and the Normal

Final Calculation and Verification

Similar Questions

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