TANGENTS & NORMALS MCQs for JEE MATHEMATICS — Medium, Page 4

91 Medium

91 of 97

Application +4 | -1

The length of subnormal of parabola

{y^2} = 4ax

at any point is equal to

View answer and explanation

Correct option: D

2a

Option D (

\left( {\frac{3}{2},\; - \frac{{17}}{2}} \right)

) is correct because it is the point where the derivative of the function is zero, indicating a horizontal tangent.

Solution approach

Correct Answer + Core Reason: $\left( {\frac{3}{2},\; - \frac{{17}}{2}} \right)$ . The point is found by setting the derivative equal to zero.

Step-by-Step Solution:

Step 1: Given equation: $y = 2{x^2} - 6x - 4$ .

Step 2: Find the derivative: $\frac{{dy}}{{dx}} = 4x - 6$ .

Step 3: Set the derivative equal to zero: $4x - 6 = 0 \Rightarrow x = \frac{6}{4} = \frac{3}{2}$ .

Step 4: Substitute $x = \frac{3}{2}$ into the original equation: $y = 2{\left( {\frac{3}{2}} \right)^2} - 6\left( {\frac{3}{2}} \right) - 4 = 2\left( {\frac{9}{4}} \right) - 9 - 4 = \frac{9}{2} - 13 = \frac{{9 - 26}}{2} = - \frac{{17}}{2}$ .

Step 5: Final Answer: The point is $\left( {\frac{3}{2},\; - \frac{{17}}{2}} \right)$ .

Why other options are wrong:
- **Option A ( $\left( {\frac{3}{2},\frac{{13}}{2}} \right)$ ): The y-coordinate is calculated incorrectly.
- Option B ( $\left( { - \frac{5}{2},\; - \frac{{17}}{2}} \right)$ ): The x-coordinate is calculated incorrectly.
- Option C ( $\left( {\frac{3}{2},\frac{{17}}{2}} \right)$ ): The y-coordinate is calculated incorrectly.

Key Concepts:
- Finding the derivative of a polynomial function.
- Finding the point where the tangent is horizontal.

Memory Aids:
- Remember that the derivative is zero at a horizontal tangent.

Quick Tips:**
- Always double-check your calculations.

92 Medium

92 of 97

Application +4 | -1

The slope of the tangent to the curve

x = 3{t^2} + 1,\;y = {t^3} - 1

, at

x = 1

is

View answer and explanation

Correct option: A

0

Option A (0) is correct because the slope of the tangent is found by differentiating the parametric equations and evaluating at the given x value.

Solution approach

Correct Answer + Core Reason: 0. The slope of the tangent is found by differentiating the parametric equations and evaluating at the given x value.

Step-by-Step Solution:

Step 1: Given the parametric equations: $x = 3{t^2} + 1$ and $y = {t^3} - 1$ .

Step 2: Find the value of $t$ when $x = 1$ :
$1 = 3{t^2} + 1 \Rightarrow 3{t^2} = 0 \Rightarrow t = 0$

Step 3: Find $\frac{{dx}}{{dt}}$ and $\frac{{dy}}{{dt}}$ :
$\frac{{dx}}{{dt}} = 6t$ and $\frac{{dy}}{{dt}} = 3{t^2}$

Step 4: Find $\frac{{dy}}{{dx}}$ using the chain rule:
$\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{3{t^2}}}{{6t}} = \frac{t}{2}$

Step 5: Evaluate $\frac{{dy}}{{dx}}$ at $t = 0$ :
$\left( \frac{{dy}}{{dx}} \right)_{t=0} = \frac{0}{2} = 0$

Why other options are wrong:
- Option B: $\frac{1}{2}$ is incorrect. This might be obtained if the value of t is incorrectly calculated or if the expression for $\frac{{dy}}{{dx}}$ is not evaluated at the correct t value.
- Option C: $\infty$ is incorrect. This would imply that $\frac{{dx}}{{dt}} = 0$ while $\frac{{dy}}{{dt}} \ne 0$ , which is not the case here.
- Option D: $-2$ is incorrect. This value is not related to the correct derivative calculation.

Key Concepts:
- Parametric differentiation
- Slope of a tangent

Memory Aids:
- Remember the chain rule for parametric differentiation: $\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}$

Quick Tips:
- Always find the correct value of the parameter $t$ before evaluating the derivative.

93 Medium

93 of 97

Analyzing +4 | -1

If the curves

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\,and\,\frac{{{x^2}}}{{{l^2}}} - \frac{{{y^2}}}{{{m^2}}} = 1

cut each other orthogonally, then

View answer and explanation

Correct option: C

{a^2} - {b^2} = {l^2} + {m^2}

Option C (

{a^2} - {b^2} = {l^2} + {m^2}

) is correct because it is derived from the condition for orthogonal intersection of the two given curves.

Solution approach

Correct Answer + Core Reason: ${a^2} - {b^2} = {l^2} + {m^2}$ . This condition arises from the orthogonality condition of the curves.

Step-by-Step Solution:

Step 1: Write down the equations of the two curves.
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (1)
$\frac{x^2}{l^2} - \frac{y^2}{m^2} = 1$ (2)

Step 2: Differentiate both equations with respect to x.\nFor (1): $\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \Rightarrow (\frac{dy}{dx})_{C_1} = -\frac{b^2}{a^2} \frac{x}{y}$ \nFor (2): $\frac{2x}{l^2} - \frac{2y}{m^2} \frac{dy}{dx} = 0 \Rightarrow (\frac{dy}{dx})_{C_2} = \frac{m^2}{l^2} \frac{x}{y}$

Step 3: Apply the condition for orthogonal intersection: $( \frac{dy}{dx})_{C_1} (\frac{dy}{dx})_{C_2} = -1$ .
$(- \frac{b^2}{a^2} \frac{x}{y}) (\frac{m^2}{l^2} \frac{x}{y}) = -1 \Rightarrow \frac{b^2m^2x^2}{a^2l^2y^2} = 1 \Rightarrow b^2m^2x^2 = a^2l^2y^2$ (3)

Step 4: Subtract equation (2) from equation (1).
$x^2(\frac{1}{a^2} - \frac{1}{l^2}) + y^2(\frac{1}{b^2} + \frac{1}{m^2}) = 0$ (4)

Step 5: Substitute $y^2 = \frac{b^2m^2x^2}{a^2l^2}$ from (3) into (4).
$x^2(\frac{1}{a^2} - \frac{1}{l^2}) + \frac{b^2m^2x^2}{a^2l^2}(\frac{1}{b^2} + \frac{1}{m^2}) = 0 \Rightarrow (\frac{1}{a^2} - \frac{1}{l^2}) + \frac{b^2m^2}{a^2l^2}(\frac{1}{b^2} + \frac{1}{m^2}) = 0 \Rightarrow \frac{1}{a^2} - \frac{1}{l^2} + \frac{m^2}{a^2l^2} + \frac{b^2}{a^2l^2} = 0 \Rightarrow l^2 - a^2 + m^2 + b^2 = 0 \Rightarrow a^2 - b^2 = l^2 + m^2$

Why other options are wrong:
- Option A: ${a^2} + {b^2} = {l^2} + {m^2}$ is incorrect because it does not satisfy the orthogonality condition.
- Option B: ${a^2} - {b^2} = {l^2} - {m^2}$ is incorrect because it does not satisfy the orthogonality condition.
- Option D: ${a^2} + {b^2} = {l^2} - {m^2}$ is incorrect because it does not satisfy the orthogonality condition.

Key Concepts:
- Orthogonal intersection of curves.
- Differentiation of implicit functions.

Memory Aids:
- Two curves are orthogonal if the product of their slopes at the point of intersection is -1.

Quick Tips:
- When dealing with orthogonal curves, remember to differentiate implicitly and use the condition $m_1m_2 = -1$ .

94 Medium

94 of 97

Analyzing +4 | -1

The two curves

{x^3} - 3x{y^2} + 2 = 0

and

3{x^2}y - {y^3} - 2 = 0

View answer and explanation

Correct option: A

Cut at right angel

Option A (Cut at right angel) is correct because the product of the slopes of the tangents to the curves at the point of intersection is -1.

Solution approach

Correct Answer + Core Reason: Cut at right angel. The product of the slopes of the tangents is -1.

Step-by-Step Solution:

Step 1: Given curves: $x^3 - 3xy^2 + 2 = 0$ and $3x^2y - y^3 - 2 = 0$ .

Step 2: Differentiate the first equation with respect to $x$ : $3x^2 - 3y^2 - 6xy \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{x^2 - y^2}{2xy} = m_1$ .

Step 3: Differentiate the second equation with respect to $x$ : $6xy + 3x^2 \frac{dy}{dx} - 3y^2 \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{-2xy}{x^2 - y^2} = m_2$ .

Step 4: Find the product of the slopes: $m_1 \times m_2 = \frac{x^2 - y^2}{2xy} \times \frac{-2xy}{x^2 - y^2} = -1$ .

Step 5: Since the product of the slopes is -1, the curves intersect at right angles.

Step 6: Final Answer: The curves cut at right angles.

Why other options are wrong:
- Option B: Touch each other is incorrect because the product of the slopes is -1, not equal.
- Option C: Cut at an angle $\frac{\pi}{3}$ is incorrect because the product of the slopes is -1, implying a right angle.
- Option D: Cot at an angle $\frac{\pi}{4}$ is incorrect because the product of the slopes is -1, implying a right angle.

Key Concepts:
- Differentiation of implicit functions
- Condition for orthogonal intersection of curves

Memory Aids:
- Two curves are orthogonal if the product of their slopes at the point of intersection is -1.

Quick Tips:
- Remember to differentiate implicitly when $y$ is not explicitly defined as a function of $x$ .

95 Medium

95 of 97

Application +4 | -1

The tangent to the curve

y = 2{x^2} - x + 1

is parallel to the line

y = 3x + 9

at the point

View answer and explanation

Correct option: D

\left( {1,2} \right)

Option D ((1,2)) is correct because at this point, the derivative of the curve equals the slope of the line, indicating a parallel tangent.

Solution approach

Correct Answer + Core Reason: (1,2). The tangent is parallel when the derivative of the curve equals the slope of the line.

Step-by-Step Solution:

Step 1: Find the derivative of the curve $y = 2x^2 - x + 1$ : $\frac{{dy}}{{dx}} = 4x - 1$ .

Step 2: Find the slope of the line $y = 3x + 9$ . The slope is 3.

Step 3: Set the derivative equal to the slope to find the x-coordinate of the point where the tangent is parallel: $4x - 1 = 3 \Rightarrow 4x = 4 \Rightarrow x = 1$ .

Step 4: Substitute the x-coordinate into the equation of the curve to find the corresponding y-coordinate: $y = 2(1)^2 - 1 + 1 = 2 - 1 + 1 = 2$ .

Step 5: The point is (1, 2).

Why other options are wrong:
- Option A (3,9): This point does not lie on the curve $y = 2x^2 - x + 1$ . Substituting x=3 gives y = 2(9) - 3 + 1 = 16, not 9.
- Option B ((2,-1)): This point does not satisfy the condition that the derivative of the curve at that point equals the slope of the line. Also, the point (2, -1) does not lie on the curve.
- Option C ((2,1)): Substituting x=2 into the curve gives y = 2(4) - 2 + 1 = 7, not 1. Also, the derivative at x=2 is 4(2) - 1 = 7, which is not equal to 3.

Key Concepts:
- Derivatives as slopes of tangents
- Parallel lines have equal slopes

Memory Aids:
- The derivative of a function at a point gives the slope of the tangent at that point.

Quick Tips:
- Remember that parallel lines have the same slope.
- Always check if the point lies on the curve.

96 Medium

96 of 97

Application +4 | -1

The point

P

of the curve

{y^2} = 2{x^3}

such that the tangent at P is perpendicular to the line

4x - 3y + 2 = 0

is given by

View answer and explanation

Correct option: D

\left( {1/8,\; - 1/16} \right)

Option D ((1/8, -1/16)) is correct because the tangent at this point is perpendicular to the given line.

Solution approach

Correct Answer + Core Reason: (1/8, -1/16). The tangent is perpendicular when the product of the slopes is -1.

Step-by-Step Solution:

Step 1: Find the slope of the line $4x - 3y + 2 = 0$ . Rewrite the equation as $3y = 4x + 2 \Rightarrow y = \frac{4}{3}x + \frac{2}{3}$ . The slope is $\frac{4}{3}$ .

Step 2: Find the derivative of the curve $y^2 = 2x^3$ . Differentiating implicitly with respect to x gives $2y \frac{{dy}}{{dx}} = 6x^2 \Rightarrow \frac{{dy}}{{dx}} = \frac{{3x^2}}{y}$ .

Step 3: The tangent is perpendicular to the line, so the product of their slopes is -1: $\frac{{dy}}{{dx}} \times \frac{4}{3} = -1 \Rightarrow \frac{{dy}}{{dx}} = -\frac{3}{4}$ .

Step 4: Set the derivative equal to $-\frac{3}{4}$ : $\frac{{3x^2}}{y} = -\frac{3}{4} \Rightarrow y = -4x^2$ .

Step 5: Substitute $y = -4x^2$ into the equation of the curve $y^2 = 2x^3$ : $(-4x^2)^2 = 2x^3 \Rightarrow 16x^4 = 2x^3 \Rightarrow 8x^4 = x^3 \Rightarrow 8x = 1 \Rightarrow x = \frac{1}{8}$ .

Step 6: Find the corresponding y-coordinate: $y = -4(\frac{1}{8})^2 = -4(\frac{1}{64}) = -\frac{1}{16}$ .

Step 7: The point is (1/8, -1/16).

Why other options are wrong:
- Option A ((2,4)): This point does not satisfy the condition that the tangent at that point is perpendicular to the given line. Also, (2,4) does not lie on the curve.
- **Option B ((1, $\sqrt{2}$ )):** This point does not satisfy the condition that the tangent at that point is perpendicular to the given line. Also, $(\sqrt{2})^2 \neq 2(1)^3$ .
- Option C ((1/2, -1/2)): This point does not satisfy the condition that the tangent at that point is perpendicular to the given line. Also, (-1/2)² ≠ 2(1/2)³.

Key Concepts:
- Derivatives as slopes of tangents
- Perpendicular lines have slopes whose product is -1
- Implicit differentiation

Memory Aids:
- Remember that the product of slopes of perpendicular lines is -1.

Quick Tips:
- When dealing with implicit functions, use implicit differentiation.
- Always check if the point lies on the curve.

97 Medium

97 of 97

Application +4 | -1

The abscissa of the point on the curve

a{y^2} = {x^3}

, the normal at which cuts off equal intercepts from the coordinate axes is

View answer and explanation

Correct option: B

4\;a/9

Option B (

4\\;a/9

) is correct because it represents the x-coordinate where the normal to the curve cuts off equal intercepts, derived by equating the x and y intercepts of the normal's equation.

Solution approach

Correct Answer + Core Reason: $4\\;a/9$ . The abscissa is found by equating the intercepts of the normal to the curve.

Step-by-Step Solution:

Step 1: Find the derivative: $a{y^2} = {x^3} \Rightarrow 2ay\frac{{dy}}{{dx}} = 3{x^2} \Rightarrow \frac{{dy}}{{dx}} = \frac{{3{x^2}}}{{2ay}}$ .

Step 2: Let the point be $({{x_1},{y_1}})$ . Then $ay_1^2 = x_1^3$ .

Step 3: Slope of the normal at $({{x_1},{y_1}})$ is $- \frac{{2a{y_1}}}{{3x_1^2}}$ .

Step 4: Equation of the normal: $y - {y_1} = - \frac{{2a{y_1}}}{{3x_1^2}}(x - {x_1})$ .

Step 5: Find the x-intercept (y=0): $- {y_1} = - \frac{{2a{y_1}}}{{3x_1^2}}(x - {x_1}) \Rightarrow x = {x_1} + \frac{{3x_1^2}}{{2a}}$ .

Step 6: Find the y-intercept (x=0): $y - {y_1} = - \frac{{2a{y_1}}}{{3x_1^2}}( - {x_1}) \Rightarrow y = {y_1} + \frac{{2a{y_1}}}{{3{x_1}}}$ .

Step 7: Equate the intercepts: ${x_1} + \frac{{3x_1^2}}{{2a}} = {y_1} + \frac{{2a{y_1}}}{{3{x_1}}} \Rightarrow {x_1}\frac{{(2a + 3{x_1})}}{{2a}} = {y_1}\frac{{(3{x_1} + 2a)}}{{3{x_1}}} \Rightarrow 3x_1^2 = 2a{y_1}$ .

Step 8: Substitute $y_1 = \frac{{3x_1^2}}{{2a}}$ into $ay_1^2 = x_1^3$ : $a(\frac{{3x_1^2}}{{2a}})^2 = x_1^3 \Rightarrow a\frac{{9x_1^4}}{{4a^2}} = x_1^3 \Rightarrow 9x_1^4 = 4a x_1^3 \Rightarrow x_1 = \frac{{4a}}{9}$ .

Why other options are wrong:
- Option A (2\\;a/9): This is incorrect because it's a result of an error in equating intercepts or simplification.
- Option C (-4\\;a/9): This is incorrect because the abscissa cannot be negative in this context.
- Option D (-2\\;a/9): This is incorrect because the abscissa cannot be negative in this context.

Key Concepts:
- Derivatives
- Equation of a normal line
- Intercepts of a line

Memory Aids:
- Slope of normal = -1 / slope of tangent
- x-intercept: y=0, y-intercept: x=0

Quick Tips:
- Always find the derivative first.
- Remember to equate the intercepts correctly.
- Simplify the equation carefully.

Your Progress

Tangents & normals practice · Medium

Continue Your Practice

Your practice