This set of Mathematics Online Test for Engineering Entrance Exams focuses on “Calculus Application – Tangents and Normals – 2”.

1. What will be the equation of the normal to the parabola y^{2} = 5x that makes an angle 45° with the x axis?

a) 4(x – y) = 15

b) 4(x + y) = 15

c) 2(x – y) = 15

d) 2(x + y) = 15

View Answer

Explanation: The equation of the given parabola is, y

^{2}= 5x ……….(1)

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

2y = 5(dx/dy)

Or dx/dy = 2y/5

Take any point P((5/4)t

^{2}, (5/2)t). Then, the normal to the curve (1) at P is,

-[dx/dy]

_{P}= -(2*5t/2)/5 = -t

By the question, slope of the normal to the curve (1) at P is tan45°.

Thus, -t = 1

Or t = -1

So, the required equation of normal is,

y – 5t/2 = -t(x – 5t

^{2}/4)

Simplifying further we get,

4(x – y) = 15

2. What will be the co-ordinates of the foot of the normal to the parabola y^{2} = 5x that makes an angle 45° with the x axis?

a) (-5/4, 5/2)

b) (5/4, 5/2)

c) (5/4, -5/2)

d) (-5/4, -5/2)

View Answer

Explanation: The equation of the given parabola is, y

^{2}= 5x ……….(1)

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

2y = 5(dx/dy)

Or dx/dy = 2y/5

Take any point P((5/4)t

^{2}, (5/2)t). Then, the normal to the curve (1) at P is,

-[dx/dy]

_{P}= -(2*5t/2)/5 = -t

By the question, slope of the normal to the curve (1) at P is tan45°.

Thus, -t = 1

Or t = -1

So, the required equation of normal is,

y – 5t/2 = -t(x – 5t

^{2}/4)

Simplifying further we get,

4(x – y) = 15

The co-ordinates of the foot of the normal are, P((5/4)t

^{2}, (5/2)t).

As t = 1, so putting the value of t = 1, we get,

P = (5/4, 5/2).

3. What will be the equation of the normal to the parabola y^{2} = 3x which is perpendicular to the line y = 2x + 4?

a) 16x + 32y = 27

b) 16x – 32y = 27

c) 16x + 32y = -27

d) -16x + 32y = 27

View Answer

Explanation: Given, y

^{2}= 3x ……….(1) and y = 2x + 4 ……….(2)

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

2y = 3(dx/dy)

Or dx/dy = 2y/3

Let P (x

_{1}, y

_{1}) be any point on the parabola (1). Then the slope of the normal to the parabola (1) at point P is

-[dx/dy]

_{P}= -2y

_{1}/3

If the normal at the point P to the parabola (1) be perpendicular to the line (2) then we must have,

-2y

_{1}/3*2 = -1

Since the slope of the line (2) is 2

Or y

_{1}= 3/4

Since the point P(x

_{1}, y

_{1}) lies on (1) hence,

y

_{1}

^{2}= 3x

_{1}

As, y

_{1}= 3/4, so, x

_{1}= 3/16

Therefore, the required equation of the normal is

y – y

_{1}= -(2y

_{1})/3*(x – x

_{1})

Putting the value of x

_{1}and y

_{1}in the above equation we get,

16x + 32y = 27.

4. What will be the co-ordinates of the foot of the normal to the parabola y^{2} = 3x which is perpendicular to the line y = 2x + 4?

a) (-3/16, -3/4)

b) (-3/16, 3/4)

c) (3/16, -3/4)

d) (3/16, 3/4)

View Answer

Explanation: Given, y

^{2}= 3x ……….(1) and y = 2x + 4 ……….(2)

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

2y = 3(dx/dy)

Or dx/dy = 2y/3

Let P (x

_{1}, y

_{1}) be any point on the parabola (1). Then the slope of the normal to the parabola (1) at point P is

-[dx/dy]

_{P}= -2y

_{1}/3

If the normal at the point P to the parabola (1) be perpendicular to the line (2) then we must have,

-2y

_{1}/3*2 = -1

Since the slope of the line (2) is 2

Or y

_{1}= 3/4

Since the point P(x

_{1}, y

_{1}) lies on (1) hence,

y

_{1}

^{2}= 3x

_{1}

As, y

_{1}= 3/4, so, x

_{1}= 3/16

Therefore, the required equation of the normal is

y – y

_{1}= -(2y

_{1})/3*(x – x

_{1})

Putting the value of x

_{1}and y

_{1}in the above equation we get,

16x + 32y = 27

And the coordinates of the foot of the normal are (x

_{1}, y

_{1}) = (3/16, 3/4)

5. What will be the length of a tangent from the point (7, 2) to the circle 2x^{2} + 2y^{2} + 5 x + y = 15?

a) 10 units

b) 8 units

c) 6 units

d) 4 units

View Answer

Explanation: The given equation of the circle is,

2x

^{2}+ 2y

^{2}+ 5 x + y – 15 = 0

Or x

^{2}+ y

^{2}+ 5/2 x + y/2 – 15/2 = 0 ………..(1)

The required length of the tangent drawn from the point (7, 2) to the circle (1) is,

√(7

^{2}+ 2

^{2}+ 5/2 (7) + 1/2 – 15/2)

= √(49 + 4 + 35/2 + 1 – 15/2)

= √64

= 8 units.

6. If the normal to the ellipse x^{2} + 3y^{2} = 12 at the point be inclined at 60° to the major axis, then at what angle does the line joining the curve to the point is inclined to the same axis?

a) 90°

b) 45°

c) 60°

d) 30°

View Answer

Explanation: Given, x

^{2}+ 3y

^{2}= 12 Or x

^{2}/12 + y

^{2}/4 = 1

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

2x*(dx/dy) + 3*2y = 0

Or dx/dy = -3y/x

Suppose the normal to the ellipse (1) at the point P(√12cosθ, 2sinθ) makes an angle 60° with the major axis. Then, the slope of the normal at P is tan60°

Or -[dx/dy]

_{P}= tan60°

Or -(-(3*2sinθ)/√12cosθ) = √3

Or √3tanθ = √3

Or tanθ = 1

Now the centre of the ellipse (1) is C(0, 0)

Therefore, the slope of the line CP is,

(2sinθ – 0)/(√12cosθ – 0) = (1/√3)tanθ = 1/√3 [as, tanθ = 1]

Therefore, the line CP is inclined at 30° to the major axis.

7. What will be the equation of the circle which touches the line x + 2y + 5 = 0 and passes through the point of intersection of the circle x^{2} + y^{2} = 1 and x^{2} + y^{2} + 2x + 4y + 1 = 0?

a) x^{2} + y^{2} + 2x + y = 0

b) x^{2} + y^{2} + x + 2y = 1

c) x^{2} + y^{2} + x + 2y = 0

d) x^{2} + y^{2} + 2x + 2y = 1

View Answer

Explanation: The equation of any circle through the points of intersection of the given circle is,

x

^{2}+ y

^{2}+ 2x + 4y + 1 + k(x

^{2}+ y

^{2}– 1) = 0

x

^{2}+ y

^{2}+ 2x(1/(k + 1)) + 2*2y/(k + 1) + (1 – k)/(1 + k) = 0

Clearly, the co-ordinates of the center of the circle (1) are, (-1/(1 + k), -2/(1 + k)) and its radius,

= √[(1/(1 + k))

^{2}+ (2/(1 + k))

^{2}– ((1 – k)]/(1 + k))

= √(4 + k

^{2})/(1 + k)

Clearly, the line x + 2y + 5 = 0 is tangent to the circle (1), hence, the perpendicular distance of the line from the center of the circle = radius of the circle

± (-1/(1 + k))– 2(2/(1 + k)) + 5/ √(1

^{2}+ 2

^{2}) = √(4 + k

^{2})/(1 + k)

Or ±(5k/√5) = √(4 + k

^{2})

Or 5k

^{2}= 4 + k

^{2}

Or 4k

^{2}= 4

Or k = 1 [as, k ≠ -1]

Putting k = 1 in (1), equation of the given circle is,

x

^{2}+ y

^{2}+ x + 2y = 0

8. What will be the value of angle between the curves x^{2} – y^{2} = 2a^{2} and xv + y^{2} = 4a^{2}?

a) π/2

b) π/4

c) π/6

d) π/3

View Answer

Explanation: x

^{2}– y

^{2}= 2a

^{2}……….(1) and x

^{2}+ y

^{2}= 4a

^{2}……….(2)

Adding (1) and (2) we get, 2x

^{2}= 6a

^{2}

Again, (2) – (1) gives,

2y

^{2}= 2a

^{2}

Therefore, 2x

^{2}* 2y

^{2}= 6a

^{2}* 2a

^{2}

4x

^{2}y

^{2}= 12a

^{2}

Or x

^{2}y

^{2}= 3a

^{4}

Or 2xy = ±2√3

Differentiating both side of (1) and (2) with respect to x we get,

2x – 2y(dy/dx) = 0

Or dy/dx = x/y

And 2x + 2y(dy/dx) = 0

Ordy/dx = -x/y

Let (x, y) be the point of intersection of the curves(1) and (2) and m

_{1}and m

_{2}be the slopes of the tangents to the curves (1) and (2) respectively at the point (x, y); then,

m

_{1}= x/y and m

_{2}= -x/y

Now the angle between the curves (1) and (2) means the angle between the tangents to the curve at their point of intersection.

Therefore, if θ is the required angle between the curves (1) and (2), then

tanθ = |(m

_{1}– m

_{2})/(1 + m

_{1}m

_{2})|

Putting the value of m

_{1}, m

_{2}in the above equation we get,

tanθ = |2xy/(y

^{2}– x

^{2})|

As, 2xy = ±2√3a

^{2}and x

^{2}– y

^{2}= 2a

^{2}

tanθ = |±2√3a

^{2}/-2a

^{2}|

Or tanθ = √3

Thus, θ = π/3.

9. If the curves x^{2}/a + y^{2}/b = 1 and x^{2}/c + y^{2}/d = 1 intersect at right angles, then which one is the correct relation?

a) b – a = c – d

b) a + b = c + d

c) a – b = c – d

d) a – b = c + d

View Answer

Explanation: We have, x

^{2}/a + y

^{2}/b = 1 ……….(1) and x

^{2}/c + y

^{2}/d = 1 ……….(2)

Let, us assume curves (1) and (2) intersect at (x

_{1}, y

_{1}). Then

x

_{1}

^{2}/a + y

_{1}

^{2}/b = 1 ……….(3) and x

_{1}

^{2}/c + y

_{1}

^{2}/d = 1 ……….(4)

Differentiating both side of (1) and (2) with respect to x we get,

2x/a + 2y/b(dy/dx) = 0

Or dy/dx = -xb/ya

Let, m

_{1}and m

_{2}be the slopes of the tangents to the curves (1) and (2) respectively at the point (x

_{1}, y

_{1}); then,

m

_{1}= [dy/dx]

_{(x1, y1)}= -(bx

_{1}/ay

_{1}) and m

_{2}= [dy/dx]

_{(x1, y1)}= -(dx

_{1}/cy

_{1})

By question as the curves (1) and (2) intersects at right angle, so, m

_{1}m

_{2}= -1

Or -(bx

_{1}/ay

_{1})*-(dx

_{1}/cy

_{1}) = -1

Or bdx

_{1}

^{2}= -acy

_{1}

^{2}……….(5)

Now, (3) – (4) gives,

bdx

_{1}

^{2}(c – a) = acy

_{1}

^{2}(d – b) ……….(6)

Dividing (6) by (5) we get,

c – a = d – b

Or a – b = c – d.

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

To practice Mathematics Online Test for Engineering Entrance Exams, __ here is complete set of 1000+ Multiple Choice Questions and Answers__.

**Next Steps:**

- Get Free Certificate of Merit in Mathematics - Class 12
- Participate in Mathematics - Class 12 Certification Contest
- Become a Top Ranker in Mathematics - Class 12
- Take Mathematics - Class 12 Tests
- Chapterwise Practice Tests: Chapter 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
- Chapterwise Mock Tests: Chapter 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

**Related Posts:**

- Practice Class 11 - Mathematics MCQs
- Buy Class 12 - Mathematics Books
- Practice Class 12 - Chemistry MCQs
- Practice Class 12 - Physics MCQs
- Practice Class 12 - Biology MCQs