Tangents and Normals 2 - Class 12 Maths MCQ

This set of Class 12 Maths Chapter 14 Multiple Choice Questions & Answers (MCQs) focuses on “Calculus Application – Tangents and Normals – 2”.

1. What will be the equation of the normal to the parabola y² = 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

Answer: a
Explanation: The equation of the given parabola is, y² = 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)t², (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 – 5t²/4)
Simplifying further we get,
4(x – y) = 15

2. What will be the co-ordinates of the foot of the normal to the parabola y² = 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

Answer: b
Explanation: The equation of the given parabola is, y² = 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)t², (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 – 5t²/4)
Simplifying further we get,
4(x – y) = 15
The co-ordinates of the foot of the normal are, P((5/4)t², (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 y² = 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

Answer: a
Explanation: Given, y² = 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 (x₁, y₁) be any point on the parabola (1). Then the slope of the normal to the parabola (1) at point P is
-[dx/dy]_P = -2y₁/3
If the normal at the point P to the parabola (1) be perpendicular to the line (2) then we must have,
-2y₁/3*2 = -1
Since the slope of the line (2) is 2
Or y₁ = 3/4
Since the point P(x₁, y₁) lies on (1) hence,
y₁² = 3x₁
As, y₁ = 3/4, so, x₁ = 3/16
Therefore, the required equation of the normal is
y – y₁ = -(2y₁)/3*(x – x₁)
Putting the value of x₁ and y₁ 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 y² = 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

Answer: d
Explanation: Given, y² = 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 (x₁, y₁) be any point on the parabola (1). Then the slope of the normal to the parabola (1) at point P is
-[dx/dy]_P = -2y₁/3
If the normal at the point P to the parabola (1) be perpendicular to the line (2) then we must have,
-2y₁/3*2 = -1
Since the slope of the line (2) is 2
Or y₁ = 3/4
Since the point P(x₁, y₁) lies on (1) hence,
y₁² = 3x₁
As, y₁ = 3/4, so, x₁ = 3/16
Therefore, the required equation of the normal is
y – y₁ = -(2y₁)/3*(x – x₁)
Putting the value of x₁ and y₁ in the above equation we get,
16x + 32y = 27
And the coordinates of the foot of the normal are (x₁, y₁) = (3/16, 3/4)

5. What will be the length of a tangent from the point (7, 2) to the circle 2x² + 2y² + 5 x + y = 15?
a) 10 units
b) 8 units
c) 6 units
d) 4 units
View Answer

Answer: b
Explanation: The given equation of the circle is,
2x² + 2y² + 5 x + y – 15 = 0
Or x² + y² + 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,
√(7² + 2² + 5/2 (7) + 1/2 – 15/2)
= √(49 + 4 + 35/2 + 1 – 15/2)
= √64
= 8 units.

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6. If the normal to the ellipse x² + 3y² = 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

Answer: d
Explanation: Given, x² + 3y² = 12 Or x²/12 + y²/4 = 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 x² + y² = 1 and x² + y² + 2x + 4y + 1 = 0?
a) x² + y² + 2x + y = 0
b) x² + y² + x + 2y = 1
c) x² + y² + x + 2y = 0
d) x² + y² + 2x + 2y = 1
View Answer

Answer: c
Explanation: The equation of any circle through the points of intersection of the given circle is,
x² + y² + 2x + 4y + 1 + k(x² + y² – 1) = 0
x² + y² + 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 + k²)/(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/ √(1² + 2²) = √(4 + k²)/(1 + k)
Or ±(5k/√5) = √(4 + k²)
Or 5k² = 4 + k²
Or 4k² = 4
Or k = 1 [as, k ≠ -1]
Putting k = 1 in (1), equation of the given circle is,
x² + y² + x + 2y = 0

8. What will be the value of angle between the curves x² – y² = 2a² and xv + y² = 4a²?
a) π/2
b) π/4
c) π/6
d) π/3
View Answer

Answer: d
Explanation: x² – y² = 2a² ……….(1) and x² + y² = 4a² ……….(2)
Adding (1) and (2) we get, 2x² = 6a²
Again, (2) – (1) gives,
2y² = 2a²
Therefore, 2x² * 2y² = 6a² * 2a²
4x²y² = 12a²
Or x²y² = 3a⁴
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 m₁ and m₂ be the slopes of the tangents to the curves (1) and (2) respectively at the point (x, y); then,
m₁ = x/y and m₂ = -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 = |(m₁ – m₂)/(1 + m₁m₂)|
Putting the value of m₁, m₂ in the above equation we get,
tanθ = |2xy/(y² – x²)|
As, 2xy = ±2√3a² and x² – y² = 2a²
tanθ = |±2√3a²/-2a²|
Or tanθ = √3
Thus, θ = π/3.

9. If the curves x²/a + y²/b = 1 and x²/c + y²/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

Answer: c
Explanation: We have, x²/a + y²/b = 1 ……….(1) and x²/c + y²/d = 1 ……….(2)
Let, us assume curves (1) and (2) intersect at (x₁, y₁). Then
x₁²/a + y₁²/b = 1 ……….(3) and x₁²/c + y₁²/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, m₁ and m₂ be the slopes of the tangents to the curves (1) and (2) respectively at the point (x₁, y₁); then,
m₁ = [dy/dx]_{(x1, y1)} = -(bx₁/ay₁) and m₂ = [dy/dx]_{(x1, y1)} = -(dx₁/cy₁)
By question as the curves (1) and (2) intersects at right angle, so, m₁m₂ = -1
Or -(bx₁/ay₁)*-(dx₁/cy₁) = -1
Or bdx₁² = -acy₁² ……….(5)
Now, (3) – (4) gives,
bdx₁²(c – a) = acy₁²(d – b) ……….(6)
Dividing (6) by (5) we get,
c – a = d – b
Or a – b = c – d.

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