Mathematics Multiple Choice Questions on “Calculus Application – Tangents and Normals – 1”.
1. What is the equation of the tangent at a specific point of y2 = 4ax at (0, 0)?
a) x = 0
b) x = 1
c) x = 2
d) x = 3
Answer: a
Clarification: Equation of the given parabola is y2 = 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 x3 – 3axy + y3 = 0 at (x, y)?
a) (x2 – ay)X + (y2 – ax)Y = -2axy
b) (x2 – ay)X + (y2 – ax)Y = 2axy
c) (x2 – ay)X + (y2 – ax)Y = axy
d) (x2 – ay)X + (y2 – ax)Y = -axy
Answer: c
Clarification: Equation of the given curve is, x3 – 3axy + y3 = 0 ……….(1)
Differentiating both sides with respect to x we get,
3x2 – 3a(x(dy/dx) + y) + 3y2(dy/dx) = 0
Or dy/dx = (ay – x2)/(y2 – ax)
So, it is clear that this can be written as,
Y – y = (dy/dx)(X – x)
Or Y – y = [(ay – x2)/(y2 – ax)](X – x)
Simplifying the above equation by cross multiplication, we get,
(x2 – ay)X + (y2 – ax)Y = x3 – 3axy + y3 + axy
Using (1),
(x2 – ay)X + (y2 – 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
Answer: b
Clarification: 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)
Answer: a
Clarification: 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 x2 + y2 – 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
Answer: a
Clarification: 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 y2 = 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
Answer: c
Clarification: Equation of the given parabola is, y2 = 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 (x1, y1) = (2t2, 4t) is,
y – y1 = [dy/dx](x1, y1) (x – 2t2)
y – 4t = [dy/dx](2(t2), 4t) (x – 2t2)
Putting the value of y = 4t in the equation dy/dx = 4/y, we get,
y – 4t = 4/4t(x – 2t2) ……….(2)
If the tangent to the parabola y2 = 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 x2 + y2 – 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
Answer: c
Clarification: 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 3x2 – 4y2 = 12 at the point (x1, y1)?
a) 3x1y + 4y1x + 7x1y1 = 0
b) 3x1y + 4y1x – 7x1y1 = 0
c) 3x1y – 4y1x – 7x1y1 = 0
d) 3x1y – 4y1x + 7x1y1 = 0
Answer: b
Clarification: Equation of the given hyperbola is, 3x2 – 4y2 = 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 (x1, y1) on it is,
y – y1 = -[dx/dy](x1, y1) (x – x1) = -4y1/3x1(x – x1)
Or 3x1y + 4y1x – 7x1y1 = 0
9. What is the nature of the straight line x + y + 7 = 0 to the hyperbola 3x2 – 4y2 = 12 whose normal is at the point (x1, y1)?
a) Chord to hyperbola
b) Tangent to hyperbola
c) Normal to hyperbola
d) Segment to hyperbola