Mathematics Online Test for Engineering Entrance Exams 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
Answer: a
Clarification: 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)
Answer: b
Clarification: 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
Answer: a
Clarification: 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)
Answer: d
Clarification: 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
Answer: b
Clarification: 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.
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°
Answer: d
Clarification: 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
Answer: c
Clarification: 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
Answer: d
Clarification: 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
Answer: c
Clarification: 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.