This set of Class 12 Maths Chapter 14 Multiple Choice Questions & Answers (MCQs) focuses on “Calculus Application – Tangents and Normals – 1”.

1. What is the equation of the tangent at a specific point of y^{2} = 4ax at (0, 0)?

a) x = 0

b) x = 1

c) x = 2

d) x = 3

View Answer

Explanation: Equation of the given parabola is y

^{2}= 4ax ……….(1)

Differentiating both side of (1) with respect to x we get,

2y(dy/dx) = 4a

Or dy/dx = 2a/y

Clearly dy/dx does not exist at (0, 0). Hence, the tangent to the parabola (1) at (0, 0) is parallel to y axis.

Again, the tangent passes through (0, 0). Therefore, the required tangent to the parabola (1) at (0, 0) is the y-axis and hence the required equation of the tangent is x = 0.

2. If X and Y are given as current co-ordinates, what is the equation of the tangent at a specific point of x^{3} – 3axy + y^{3} = 0 at (x, y)?

a) (x^{2} – ay)X + (y^{2} – ax)Y = -2axy

b) (x^{2} – ay)X + (y^{2} – ax)Y = 2axy

c) (x^{2} – ay)X + (y^{2} – ax)Y = axy

d) (x^{2} – ay)X + (y^{2} – ax)Y = -axy

View Answer

Explanation: Equation of the given curve is, x

^{3}– 3axy + y

^{3}= 0 ……….(1)

Differentiating both sides with respect to x we get,

3x

^{2}– 3a(x(dy/dx) + y) + 3y

^{2}(dy/dx) = 0

Or dy/dx = (ay – x

^{2})/(y

^{2}– ax)

So, it is clear that this can be written as,

Y – y = (dy/dx)(X – x)

Or Y – y = [(ay – x

^{2})/(y

^{2}– ax)](X – x)

Simplifying the above equation by cross multiplication, we get,

(x

^{2}– ay)X + (y

^{2}– ax)Y = x

^{3}– 3axy + y

^{3}+ axy

Using (1),

(x

^{2}– ay)X + (y

^{2}– ax)Y = axy

3. What will be the equation of the normal to the hyperbola xy = 4 at the point (2, 2)?

a) x + y = 0

b) x – y = 0

c) 2x – y = 0

d) x + 2y = 0

View Answer

Explanation: Equation of the given hyperbola is, xy = 4 ……….(1)

Differentiating both side of (1) with respect to y, we get,

y*(dx/dy) + x(1) = 0

Or dx/dy = -(x/y)

Thus, the required equation of the normal to the hyperbola at (2, 2) is,

y – 2 = -[dx/dy]

_{(2, 2)}(x – 2) = -(-2/2)(x – 2)

So, from here,

y – 2 = x – 2

Or x – y = 0

4. At which point does the normal to the hyperbola xy = 4 at (2, 2) intersects the hyperbola again?

a) (-2, -2)

b) (-2, 2)

c) (2, -2)

d) (0, 2)

View Answer

Explanation: Equation of the given hyperbola is, xy = 4 ……….(1)

Differentiating both side of (1) with respect to y, we get,

y*(dx/dy) + x(1) = 0

Or dx/dy = -(x/y)

Thus, the required equation of the normal to the hyperbola at (2, 2) is,

y – 2 = -[dx/dy]

_{(2, 2)}(x – 2) = -(-2/2)(x – 2)

So, from here,

y – 2 = x – 2

Or x – y = 0 ……….(2)

Solving the equation (1) and (2) we get,

x = 2 and y = 2 or x = -2 and y = -2

Thus, the line (2) intersects the hyperbola (1) at (2, 2) and (-2, -2).

Hence, the evident is that the normal at (2, 2) to the hyperbola (1) again intersects it at (-2, -2).

5. What will be the equation of the tangent to the circle x^{2} + y^{2} – 6x + 4y – 7 = 0, which are perpendicular to the straight line 2x – y + 3 = 0?

a) x + 2y – 9 = 0

b) x + 2y + 9 = 0

c) x + 2y – 10 = 0

d) x + 2y + 10 = 0

View Answer

Explanation: The equation of any straight line perpendicular to the line 2x – y + 3 = 0 is,

x + 2y + k = 0 ……….(1)

Now, the co-ordinate of the center of the circle (3, -2) and its radius is,

√(9 + 4 – (-7) = 2√5

If straight line (1) be tangent to the given circle then, the perpendicular distance of the point (3, -2) from the line (1) = radius of the circle

Thus, ±(3 + 2(-2) + k)/√(1 + 4)

Or k – 1 = 2√5 * √5

So, k = 1 ± 10

= 11 or -9

Putting the value of k in (1) we get,

x + 2y + 11 = 0 and x + 2y – 9 = 0

6. What is the equation of the tangent to the parabola y^{2} = 8x, which is inclined at an angle of 45° with the x axis?

a) x + y – 2 = 0

b) x + y + 2 = 0

c) x – y + 2 = 0

d) x – y – 2 = 0

View Answer

Explanation: Equation of the given parabola is, y

^{2}= 8x ……….(1)

Differentiating both sides with respect to x,

2y(dy/dx) = 8

Or dy/dx = 4/y

Thus, equation of the tangent to the parabola (1) at (x

_{1}, y

_{1}) = (2t

^{2}, 4t) is,

y – y

_{1}= [dy/dx]

_{(x1, y1)}(x – 2t

^{2})

y – 4t = [dy/dx]

_{(2(t2), 4t)}(x – 2t

^{2})

Putting the value of y = 4t in the equation dy/dx = 4/y, we get,

y – 4t = 4/4t(x – 2t

^{2}) ……….(2)

If the tangent to the parabola y

^{2}= 8x, which is inclined at an angle of 45° with the x axis,

Then, slope of tangent (2) = tan 45° = 1

Thus, 4/4t = 1

Or t = 1

Thus, required equation of the tangent is,

y– 4 = 1(x – 2)

Putting, t = 1 in (2),

So, x – y + 2 = 0

7. What will be the equation of the tangent to the circle x^{2} + y^{2} – 6x + 4y – 7 = 0, which are perpendicular to the straight line 2x – y + 3 = 0?

a) x – 2y + 11 = 0

b) x – 2y – 11 = 0

c) x + 2y + 11 = 0

d) x + 2y – 11 = 0

View Answer

Explanation: The equation of any straight line perpendicular to the line 2x – y + 3 = 0 is,

x + 2y + k = 0 ……….(1)

Now, the co-ordinate of the center of the circle (3, -2) and its radius is,

√(9 + 4 – (-7) = 2√5

If straight line (1) be tangent to the given circle then, the perpendicular distance of the point (3, -2) from the line (1) = radius of the circle

Thus, ±(3 + 2(-2) + k)/√(1 + 4)

Or k – 1 = 2√5 * √5

So, k = 1 ± 10

= 11 or -9

Putting the value of k in (1) we get,

x + 2y + 11 = 0 and x + 2y – 9 = 0

8. What will be the equation of normal to the hyperbola 3x^{2} – 4y^{2} = 12 at the point (x_{1}, y_{1})?

a) 3x_{1}y + 4y_{1}x + 7x_{1}y_{1} = 0

b) 3x_{1}y + 4y_{1}x – 7x_{1}y_{1} = 0

c) 3x_{1}y – 4y_{1}x – 7x_{1}y_{1} = 0

d) 3x_{1}y – 4y_{1}x + 7x_{1}y_{1} = 0

View Answer

Explanation: Equation of the given hyperbola is, 3x

^{2}– 4y

^{2}= 12 ……….(1)

Differentiating both sides of (1) with respect to y we get,

3*2x(dy/dx) – 4*(2y) = 0

Or dx/dy = 4y/3x

Therefore, the equation of the normal to the hyperbola (1) at the point (x

_{1}, y

_{1}) on it is,

y – y

_{1}= -[dx/dy]

_{(x1, y1)}(x – x

_{1}) = -4y

_{1}/3x

_{1}(x – x

_{1})

Or 3x

_{1}y + 4y

_{1}x – 7x

_{1}y

_{1}= 0

9. What is the nature of the straight line x + y + 7 = 0 to the hyperbola 3x^{2} – 4y^{2} = 12 whose normal is at the point (x_{1}, y_{1})?

a) Chord to hyperbola

b) Tangent to hyperbola

c) Normal to hyperbola

d) Segment to hyperbola

View Answer

Explanation: Equation of the given hyperbola is, 3x

^{2}– 4y

^{2}= 12 ……….(1)

Differentiating both sides of (1) with respect to y we get,

3*2x(dy/dx) – 4*(2y) = 0

Or dx/dy = 4y/3x

Therefore, the equation of the normal to the hyperbola (1) at the point (x

_{1}, y

_{1}) on it is,

y – y

_{1}= -[dx/dy]

_{(x1, y1)}(x – x

_{1}) = -4y

_{1}/3x

_{1}(x – x

_{1})

Or 3x

_{1}y + 4y

_{1}x – 7x

_{1}y

_{1}= 0

Now, if possible, let us assume that the straight line

x + y + 7 = 0 ………..(2)

This line is normal to the hyperbola (1) at the point (x

_{1}, y

_{1}). Then, the equation (2) and (3) must be identical. Hence, we have,

3x

_{1}/1 = 4y

_{1}/1 = -7x

_{1}y

_{1}/7

So, x

_{1}= -4 and y

_{1}= -3

Now, 3x

_{1}

^{2}– 4y

_{1}

^{2}= 3(-4)

^{2}– 4(-3)

^{2}= 12

This shows the point (-4, -3) lies on the hyperbola (1).

Thus, it is evident that the straight line (3) is normal to the hyperbola (1).

10. What is the foot of the normal if the straight line x + y + 7 = 0 is normal to the hyperbola 3x^{2} – 4y^{2} = 12 whose normal is at the point (x_{1}, y_{1})?

a) (4, 3)

b) (-4, 3)

c) (4, -3)

d) (-4, -3)

View Answer

Explanation: Equation of the given hyperbola is, 3x

^{2}– 4y

^{2}= 12 ……….(1)

Differentiating both sides of (1) with respect to y we get,

3*2x(dy/dx) – 4*(2y) = 0

Or dx/dy = 4y/3x

Therefore, the equation of the normal to the hyperbola (1) at the point (x

_{1}, y

_{1}) on it is,

y – y

_{1}= -[dx/dy]

_{(x1, y1)}(x – x

_{1}) = -4y

_{1}/3x

_{1}(x – x

_{1})

Or 3x

_{1}y + 4y

_{1}x – 7x

_{1}y

_{1}= 0

Now, if possible, let us assume that the straight line

x + y + 7 = 0 ………..(2)

This line is normal to the hyperbola (1) at the point (x

_{1}, y

_{1}). Then, the equation (2) and (3) must be identical. Hence, we have,

3x

_{1}/1 = 4y

_{1}/1 = -7x

_{1}y

_{1}/7

So, x

_{1}= -4 and y

_{1}= -3

Now, 3x

_{1}

^{2}– 4y

_{1}

^{2}= 3(-4)

^{2}– 4(-3)

^{2}= 12

This shows the point (-4, -3) lies on the hyperbola (1).

So, it’s the normal to the hyperbola.

Thus, it is evident that the straight line (3) is normal to the hyperbola (1); the co-ordinate foot is (-4, -3).

**Sanfoundry Global Education & Learning Series – Mathematics – Class 12**.

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