Teachoo Quiz

Assertion Reasoning Quiz

Chapter 10 Class 11 Conic Sections | 7 questions | about 6 minutes

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Question 1 of 7
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Question 1 of 7
Assertion (A): A man running notes that the sum of his distances from two static flag posts is always \(10\) m, and the distance between the posts is \(8\) m. His path forms an ellipse with an eccentricity of \(4/5\).
Reason (R): The area of the largest rectangle that can be successfully inscribed inside an ellipse defined by \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is geometrically \(2ab\).
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Question 2 of 7
Assertion (A): The hyperbola strictly defined by \(9y^2 - 4x^2 = 36\) has its transverse axis located entirely along the y-axis.
Reason (R): In the standard layout of a hyperbola, the mathematical variable possessing the positive coefficient designates the specific axis upon which the transverse axis and foci reside.
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Question 3 of 7
Assertion (A): The point \((3, 4)\) lies strictly inside the region bounded by the circle \(x^2 + y^2 = 16\).
Reason (R): A point \((x_1, y_1)\) lies outside the circle \(x^2 + y^2 = r^2\) if the condition \(x_1^2 + y_1^2 > r^2\) is satisfied.
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Question 4 of 7
Assertion (A): An equilateral (or rectangular) hyperbola operates with a fixed, invariant eccentricity exactly equal to \(\sqrt{2}\).
Reason (R): The absolute difference of the focal distances measured from any moving point on a hyperbola is consistently equal to the length of its transverse axis (\(2a\)).
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Question 5 of 7
Assertion (A): The eccentricity of any valid hyperbola is unconditionally always strictly greater than \(1\).
Reason (R): For a hyperbola, the fundamental geometric relationship between the semi-axes and focal distance is defined as \(c^2 = a^2 - b^2\).
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Question 6 of 7
Assertion (A): The equation of a circle passing through the origin and making intercepts \(a\) and \(b\) on the positive coordinate axes is \(x^2 + y^2 - ax - by = 0\).
Reason (R): The center of such a circle is fixed at the coordinates \((a, b)\).
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Question 7 of 7
Assertion (A): The circle \(x^2 + y^2 - 4x + 6y + 4 = 0\) has a radius of \(3\).
Reason (R): The point \((2, 0)\) lies exactly on the boundary of this circle.
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