JEE Main 2022 (26 July Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Differentiation: Let the function $f (x) = 2 x^{2} - lo g_{e} x, x > 0$ , be decreasing in $(0, a)$ and increasing in $(a, 4)$ . A tangent to the parabola $y^{2} = 4 a x$ at a point $P$ on it passes through the point $(8 a, 8 a - 1)$ but does not pass through the point $(- \frac{1}{a}, 0)$ . If the equation of the normal at $P$ is $\frac{x}{α} + \frac{y}{β} = 1$ , then $α + β$ is equal to ——————.

Q: How do I know if I should use the $y = mx + c$ form or the $yy_1 = 2a(x+x_1)$ form for the tangent?

Both are powerful tools! The $y = mx + c$ form is excellent when you know the slope, but here, we didn't know the slope initially—we knew the point of contact $P(x_1, y_1)$. The form $yy_1 = 2a(x+x_1)$ is specifically designed for when you are working with a point on the curve. It allowed us to directly relate the coordinates of $P$ to the external point $(4, 3)$, making the algebra much smoother.

Enter Numerical Value:

Progressive Hints

Hint 1

The function

f (x) = 2 x^{2} - lo g_{e} x

changes its behavior from decreasing to increasing at a specific point. Recall that for a differentiable function, this transition occurs at a local minimum where the derivative

f^{'} (x)

must be zero. Start by finding

f^{'} (x)

and solving

f^{'} (a) = 0

Hint 2

Once you have

a = \frac{1}{2}

, the parabola equation

y^{2} = 4 a x

becomes

y^{2} = 2 x

. The tangent to this parabola at point

P (x_{1}, y_{1})

is given by

y y_{1} = 2 a (x + x_{1})

. Use the fact that this tangent passes through the external point

(8 a, 8 a - 1)

to establish a linear relationship between

x_{1}

and

y_{1}

Hint 3

You will find two possible points of contact

P

by solving the system of equations formed by the tangent condition and the parabola equation. The problem provides a 'rejection condition': the tangent must not pass through

(- \frac{1}{a}, 0)

. Test both potential tangents against this point to identify the correct one, then find the normal's equation using the negative reciprocal of the tangent's slope.

Visualized Solution

Analyze the Function $f (x)$

Given function: $f (x) = 2 x^{2} - lo g_{e} x, x > 0$
Condition: Decreasing in $(0, a)$ and increasing in $(a, 4)$
This implies the derivative changes sign at $x = a$ , so $f^{'} (a) = 0$ .

Find the Value of $a$

$f^{'} (x) = 4 x - \frac{1}{x} = \frac{4 x ^{2} - 1}{x}$
Setting $f^{'} (a) = 0 ⟹ \frac{4 a ^{2} - 1}{a} = 0$
$4 a^{2} - 1 = 0 ⟹ a^{2} = \frac{1}{4}$
Since $a > 0$ , we get $a = \frac{1}{2}$ .

Identify the Parabola and External Point

Parabola equation: $y^{2} = 4 a x$
Substitute $a = \frac{1}{2} ⟹ y^{2} = 2 x$
External Point: $(8 a, 8 a - 1)$
Substitute $a = \frac{1}{2} ⟹ (8 (\frac{1}{2}), 8 (\frac{1}{2}) - 1) = (4, 3)$

Equation of the Tangent at $P (x_{1}, y_{1})$

Let the point of contact on the parabola be $P (x_{1}, y_{1})$ .
Since $P$ lies on $y^{2} = 2 x$ , we have $y_{1}^{2} = 2 x_{1}$ .
The equation of the tangent at $P$ is $y y_{1} = 1 (x + x_{1})$ .
This tangent must pass through the external point $(4, 3)$ .
Substituting $(4, 3)$ into the tangent equation: $3 y_{1} = 4 + x_{1}$ .

Solve for Point of Contact $P$

From the linear relation, express $x_{1}$ : $x_{1} = 3 y_{1} - 4$ .
Substitute this into the parabola equation $y_{1}^{2} = 2 x_{1}$ :
$y_{1}^{2} = 2 (3 y_{1} - 4) ⟹ y_{1}^{2} = 6 y_{1} - 8$
Rearranging gives a quadratic: $y_{1}^{2} - 6 y_{1} + 8 = 0$ .
Factoring: $(y_{1} - 2) (y_{1} - 4) = 0 ⟹ y_{1} = 2$ or $y_{1} = 4$ .

Find the Corresponding $x_{1}$ Coordinates

If $y_{1} = 2$ , then $x_{1} = 3 (2) - 4 = 2$ . Point is $(2, 2)$ .
If $y_{1} = 4$ , then $x_{1} = 3 (4) - 4 = 8$ . Point is $(8, 4)$ .
So, the two possible tangents touch the parabola at $(2, 2)$ and $(8, 4)$ .

Check the Rejection Condition

The problem states the tangent does NOT pass through $(- \frac{1}{a}, 0)$ .
Since $a = \frac{1}{2}$ , this point is $(- 2, 0)$ .
Let's check the tangent at $(2, 2)$ : $y (2) = 1 (x + 2) ⟹ 2 y = x + 2$ .
Substitute $(- 2, 0)$ into $2 y = x + 2$ : $2 (0) = - 2 + 2 ⟹ 0 = 0$ .
This tangent passes through $(- 2, 0)$ , so it must be rejected.

Identify the Correct Point $P$

Now check the tangent at $(8, 4)$ : $y (4) = 1 (x + 8) ⟹ 4 y = x + 8$ .
Substitute $(- 2, 0)$ into $4 y = x + 8$ : $4 (0) = - 2 + 8 ⟹ 0 \neq = 6$ .
This tangent does NOT pass through $(- 2, 0)$ .
Therefore, the correct point of contact is $P (8, 4)$ .

Find the Slope of the Normal at $P$

The accepted tangent is $4 y = x + 8 ⟹ y = \frac{1}{4} x + 2$ .
The slope of this tangent is $m_{T} = \frac{1}{4}$ .
The normal is perpendicular to the tangent.
Slope of the normal $m_{N} = - \frac{1}{m _{T}} = - 4$ .

Equation of the Normal

Point $P = (8, 4)$ and slope $m_{N} = - 4$ .
Using point-slope form: $y - y_{1} = m_{N} (x - x_{1})$
$y - 4 = - 4 (x - 8)$
$y - 4 = - 4 x + 32$
Rearranging: $4 x + y = 36$

Convert to Intercept Form

The problem asks to compare with $\frac{x}{α} + \frac{y}{β} = 1$ .
Divide the normal equation $4 x + y = 36$ by 36:
$\frac{4 x}{36} + \frac{y}{36} = 1$
Simplifying the fractions: $\frac{x}{9} + \frac{y}{36} = 1$
Comparing, we get $α = 9$ and $β = 36$ .

Final Calculation

We need to find the value of $α + β$ .
Substitute the values: $α + β = 9 + 36$
$α + β = 45$
Final Answer: 45

00:00 / 00:00

The Sigma Insight: Tangents, Normals and Rate Measure

Solution Diagram

Analyzing the Calculus Prelude

Unlocking the Constant

a

Consider the function

f (x) = 2 x^{2} - lo g_{e} x

. We are given that the function decreases on

(0, a)

and increases on

(a, 4)

This transition indicates a local minimum at

x = a

, where the derivative must vanish. We calculate the derivative as follows:

f^{'} (x) = \frac{d}{d x} (2 x^{2} - lo g_{e} x) = 4 x - \frac{1}{x}

Setting the derivative to zero at

x = a

4 a - \frac{1}{a} = 0 \Rightarrow 4 a^{2} = 1

Given the constraint

a > 0

, we find the value of the constant to be $a = \frac{1}{2}$ .

The Geometric Dance

Tangents and Parabolas

With

a = \frac{1}{2}

, the parabola

y^{2} = 4 a x

becomes

y^{2} = 2 x

. The external point

(8 a, 8 a - 1)

evaluates to

(4, 3)

Let

P (x_{1}, y_{1})

be a point on the parabola. The equation of the tangent at

P

is given by

y y_{1} = 2 a (x + x_{1})

, which simplifies to:

y y_{1} = x + x_{1}

Since this tangent passes through

(4, 3)

, we substitute the coordinates to obtain

3 y_{1} = 4 + x_{1}

, or

x_{1} = 3 y_{1} - 4

. Substituting this into the parabola equation

y_{1}^{2} = 2 x_{1}

y_{1}^{2} = 2 (3 y_{1} - 4) \Rightarrow y_{1}^{2} - 6 y_{1} + 8 = 0

Factoring the quadratic

(y_{1} - 2) (y_{1} - 4) = 0

reveals two potential points of contact:

(2, 2)

and

(8, 4)

The Filter

Choosing the Correct Path

The problem specifies that the tangent must not pass through

(- \frac{1}{a}, 0)

, which is the point

(- 2, 0)

. We test our candidates:

For

P (2, 2)

, the tangent equation is

2 y = x + 2

. Substituting

(- 2, 0)

gives

0 = - 2 + 2

, which is

0 = 0

. This point is rejected.

For

P (8, 4)

, the tangent equation is

4 y = x + 8

. Substituting

(- 2, 0)

gives

0 = - 2 + 8

, or

0 = 6

, which is false. Thus, the valid point of contact is $P (8, 4)$ .

The Final Stretch

The Normal and the Sum

The tangent equation is

4 y = x + 8

, or

y = \frac{1}{4} x + 2

. The slope of the tangent is

m_{T} = \frac{1}{4}

, implying the normal slope is

m_{N} = - 4

Using the point-slope form at

P (8, 4)

for the normal:

y - 4 = - 4 (x - 8) \Rightarrow 4 x + y = 36

To express this in intercept form

\frac{x}{α} + \frac{y}{β} = 1

, we divide by 36:

\frac{4 x}{36} + \frac{y}{36} = 1 \Rightarrow \frac{x}{9} + \frac{y}{36} = 1

We identify

α = 9

and

β = 36

. The final sum is:

α + β = 9 + 36 = * * 45 * *

Common Traps & FAQs

Wait, why do we reject one of the tangents? Can't a point have two tangents drawn to a parabola?

You are absolutely right! From an external point, you can generally draw two tangents to a parabola. However, the problem statement explicitly adds a constraint: the tangent must not pass through the point

(- \frac{1}{a}, 0)

. This is a filter designed to test your attention to detail. By checking both potential tangents against this specific coordinate, we identify which one violates the condition and must be discarded.

Why is the slope of the normal

- 4

Great question! We found the slope of the tangent line to be

m_{T} = \frac{1}{4}

by rewriting the tangent equation in the slope-intercept form

y = m x + c

. Since the normal line is, by definition, perpendicular to the tangent at the point of contact, its slope

m_{N}

must satisfy the condition

m_{N} \cdot m_{T} = - 1

. Therefore,

m_{N} = - \frac{1}{1/4} = - 4

. It is a fundamental geometric property that keeps our calculations elegant.

How do I know if I should use the

y = m x + c

form or the

y y_{1} = 2 a (x + x_{1})

form for the tangent?

Both are powerful tools! The

y = m x + c

form is excellent when you know the slope, but here, we didn't know the slope initially—we knew the point of contact

P (x_{1}, y_{1})

. The form

y y_{1} = 2 a (x + x_{1})

is specifically designed for when you are working with a point on the curve. It allowed us to directly relate the coordinates of

P

to the external point

(4, 3)

, making the algebra much smoother.

Enter Numerical Value:

Visualized Solution

Analyze the Function $f (x)$

Find the Value of $a$

Identify the Parabola and External Point

Equation of the Tangent at $P (x_{1}, y_{1})$

Solve for Point of Contact $P$

Find the Corresponding $x_{1}$ Coordinates

Check the Rejection Condition

Identify the Correct Point $P$

Find the Slope of the Normal at $P$

Equation of the Normal

Convert to Intercept Form

Final Calculation

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Calculus Prelude

The Geometric Dance

The Filter

The Final Stretch

Similar Questions

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Enter Numerical Value:

Visualized Solution

Analyze the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Find the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Identify the Parabola and External Point

Equation of the Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x1​,y1​)

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

Find the Corresponding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x1​ Coordinates

Check the Rejection Condition

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

Find the Slope of the Normal at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Equation of the Normal

Convert to Intercept Form

Final Calculation

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Calculus Prelude

The Geometric Dance

The Filter

The Final Stretch

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Solve for Point of Contact $P$

Find the Corresponding $x_{1}$ Coordinates

Identify the Correct Point $P$

Find the Slope of the Normal at $P$

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Consider: $f (x) = tan^{- 1} (\frac{1 + s i n x}{1 - s i n x}), x \in (0, \frac{π}{2})$ . A normal to $y = f (x)$ at $x = \frac{π}{6}$ also passes through the point

Let the normal at a point $P$ on the curve $y^{2} - 3 x^{2} + y + 10 = 0$ intersect the $y$ -axis at $(0, \frac{3}{2})$ . If $m$ is the slope of the tangent at $P$ to the curve, then $∣ m ∣$ is equal to

If $∣ f (x_{1}) - f (x_{2}) ∣ < (x_{1} - x_{2})^{2}$ , for all $x_{1}, x_{2} \in R$ . Find the equation of tangent to the curve $y = f (x)$ at the point $(1, 2)$ .

If the tangent to the curve, $y = f (x) = x lo g_{e} x$ , $(x > 0)$ at a point $(c, f (c))$ is parallel to the line-segment joining the points $(1, 0)$ and $(e, e)$ , then $c$ is equal to :

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