Conic Sections – Parabola, Ellipse, Hyperbola Quiz - ICSE Class 11 Mathematics
Free practice quiz: Conic Sections – Parabola, Ellipse, Hyperbola (ICSE Class 11 Mathematics). 41 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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Find the locus of a point which moves such that its distance from the point (2, 3) is always twice its distance from the line x + y = 1.
- x^2 - 5y^2 - 6xy - 10x + 14y + 10 = 0
- 5x^2 - y^2 - 6xy - 10x + 14y + 10 = 0
- x^2 - y^2 - 4xy - 10x + 14y + 10 = 0
- x^2 - 5y^2 - 2xy - 10x + 14y + 10 = 0
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Answer: A. x^2 - 5y^2 - 6xy - 10x + 14y + 10 = 0
Let the moving point be (h, k). The distance from (2, 3) is sqrt((h-2)^2 + (k-3)^2). The distance from the line x + y - 1 = 0 is |h + k - 1|/sqrt(1^2 + 1^2). According to the condition, sqrt((h-2)^2 + (k-3)^2) = 2 * |h + k - 1|/sqrt(2). Squaring both sides: (h-2)^2 + (k-3)^2 = 4 * (h + k - 1)^2 / 2 = 2(h + k - 1)^2. Expanding and simplifying, we get the locus equation.
Find the equation of the normal to the parabola y^2 = 4x at the point (1, 2).
- x + y - 3 = 0
- x - y + 1 = 0
- x + y - 1 = 0
- x - y - 1 = 0
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Answer: A. x + y - 3 = 0
The equation of the parabola is y^2 = 4x. This is in the form y^2 = 4ax, where 4a = 4, so a = 1. The point is (1, 2). The slope of the tangent at a point (x, y) on the parabola y^2 = 4ax is given by dy/dx = 2a/y. For this parabola, dy/dx = 2/y. At the point (1, 2), the slope of the tangent (m_t) is 2/2 = 1. The normal is perpendicular to the tangent, so the slope of the normal (m_n) is -1/m_t = -1/1 = -1. The equation of the normal to the parabola at the point (x1, y1) is y - y1 = m_n(x - x1). Substituting the point (1, 2) and m_n = -1, we get y - 2 = -1(x - 1). This simplifies to y - 2 = -x + 1. Rearranging the terms, we get x + y - 3 = 0. This matches option A.
Determine the angle between the asymptotes of the hyperbola x^2/a^2 - y^2/b^2 = 1.
- 2 tan^{-1}(b/a)
- 2 tan^{-1}(a/b)
- 2 sin^{-1}(b/a)
- 2 sin^{-1}(a/b)
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Answer: A. 2 tan^{-1}(b/a)
The asymptotes of the hyperbola x^2/a^2 - y^2/b^2 = 1 are given by the equations y = +/- (b/a)x. Let theta be the angle that the asymptote y = (b/a)x makes with the positive x-axis. Then tan(theta) = b/a. The angle of the other asymptote, y = -(b/a)x, with the positive x-axis is pi - theta. The angle between the two asymptotes is (pi - theta) - theta = pi - 2*theta. However, we are usually interested in the acute angle between them. The angle between two lines y = m1x + c1 and y = m2x + c2 is given by tan(phi) = |(m1 - m2) / (1 + m1*m2)|. Here, m1 = b/a and m2 = -b/a. So, tan(phi) = |(b/a - (-b/a)) / (1 + (b/a)(-b/a))| = |(2b/a) / (1 - b^2/a^2)| = |(2b/a) / ((a^2 - b^2)/a^2)| = |(2ab) / (a^2 - b^2)|. This is the angle between the asymptotes. However, let's consider the angle made by each asymptote with the transverse axis. Let alpha be the angle y = (b/a)x makes with the x-axis. Then tan(alpha) = b/a. The angle of the other asymptote is pi - alpha. The angle between them is (pi - alpha) - alpha = pi - 2*alpha. The question asks for 'the angle', usually implying the acute angle. If tan(alpha) = b/a, then alpha = tan^{-1}(b/a). The angle between the asymptotes is 2*alpha if the angle is interpreted as the vertex angle of the region bounded by them. The angle between the asymptotes is 2 * alpha, where tan(alpha) = b/a. So, the angle is 2 * tan^{-1}(b/a).
An ellipse has its center at the origin, passes through (4, 0) and (0, 3). Find its equation.
- x^2/16 + y^2/9 = 1
- x^2/9 + y^2/16 = 1
- x^2/25 + y^2/25 = 1
- x^2/16 + y^2/25 = 1
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Answer: A. x^2/16 + y^2/9 = 1
The standard equation of an ellipse centered at the origin is x^2/a^2 + y^2/b^2 = 1. Since the ellipse passes through (4, 0), substituting these coordinates gives 4^2/a^2 + 0^2/b^2 = 1, which means 16/a^2 = 1, so a^2 = 16. Since the ellipse passes through (0, 3), substituting these coordinates gives 0^2/a^2 + 3^2/b^2 = 1, which means 9/b^2 = 1, so b^2 = 9. Substituting the values of a^2 and b^2 into the standard equation, we get x^2/16 + y^2/9 = 1.
Find the equation of the tangent to the parabola y^2 = 4x at the point where the line y = x makes an angle of 45 degrees with the axis of the parabola.
- y = x + 1
- y = x - 1
- y = -x + 1
- y = -x - 1
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Answer: A. y = x + 1
The parabola is y^2 = 4x, so 4a = 4, which means a = 1. The line y = x makes an angle of 45 degrees with the x-axis (which is the axis of the parabola). The point of intersection of the line y = x and the parabola y^2 = 4x is found by substituting y = x into the parabola equation: x^2 = 4x. This gives x^2 - 4x = 0, so x(x - 4) = 0. The solutions are x = 0 or x = 4. If x = 0, y = 0. If x = 4, y = 4. The point (0,0) is the vertex, and a chord through the vertex with slope 1 would be y=x. The question asks for the tangent at a point where y=x makes an angle of 45 degrees. This means the line y=x itself is not the tangent. The wording implies we need to find a point on the parabola such that the tangent at that point has a slope related to the line y=x. The line y=x has a slope of 1. The tangent to y^2 = 4ax has slope m = 2a/y. If the tangent has slope m=1, then 1 = 2(1)/y, so y = 2. When y = 2, from y^2 = 4x, 4 = 4x, so x = 1. The point of tangency is (1, 2). The equation of the tangent at (x1, y1) is yy1 = 2a(x + x1). Substituting (1, 2) and a=1: y(2) = 2(1)(x + 1) => 2y = 2(x + 1) => y = x + 1. This matches option A.