The curve for which the normal at any point $(x, y)$ and the line joining the origin to that point form an isosceles triangle with the $x$ -axis as the base is

Answer (Detailed Solution Below)View answer ▾

Answer: Option B — A rectangular hyperbola

Detailed Solution

Let $P = (x, y)$ and let the normal at $P$ meet the $x$ -axis at $N$ . The normal has slope $- \frac{1}{y '}$ , so $N = (x + y y', 0)$ . The triangle is $O P N$ with base $O N$ on the $x$ -axis, so the equal sides are $O P$ and $P N$ .

$O P^{2} = x^{2} + y^{2}$ and $P N^{2} = (y y')^{2} + y^{2}$ . Setting $O P = P N$ gives $(y y')^{2} = x^{2}$ , so $y y' = x$ or $y y' = - x$ .

$y y' = x$ integrates to $y^{2} = x^{2} + c$ , i.e. $x^{2} - y^{2} = c$ , a rectangular hyperbola. The case $y y' = - x$ gives $x^{2} + y^{2} = c$ , but then $N = (x + y y', 0) = (0, 0)$ is the origin itself, so no triangle is formed and the circle is rejected.

✓ Check

x^{2} - y^{2} = 1

take

P = (2, 1)

. Then

y' = \frac{x}{y} = 2

, so

N = (2 + 2, 0) = (22, 0)

, and

O P^{2} = 3

and

P N^{2} = (2)^{2} + 1^{2} = 3

, which are equal.

⚠ Common trap

Option (C), the circle, also satisfies

(y y')^{2} = x^{2}

through the case

y y' = - x

, but the normal then passes through the origin, so

O

and

N

coincide and there is no triangle.

Quick tip

$O P = P N \Rightarrow (y y^{'})^{2} = x^{2}$ ; $y y^{'} = x$ gives $x^{2} - y^{2} = c$

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Hyperbola — Preparation

This question tests your understanding of Hyperbola, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

A hyperbola is the set of points where the difference of the distances from two fixed points (the foci) is constant. Its standard form is x^2/a^2 - y^2/b^2 = 1, with eccentricity given by b^2 = a^2(e^2 - 1), where e > 1. The foci are at (ae, 0) and (-ae, 0), and the asymptotes are the lines y = (b/a)x and y = -(b/a)x. Exam questions test finding the foci, eccentricity, transverse and conjugate axes, latus rectum, asymptotes, and equations of tangents. A special case is the rectangular hyperbola, where a = b and e = sqrt 2. Many students confuse the hyperbola with the ellipse, so compare the signs in the equation and remember that the difference of focal distances equals 2a. Practise the hyperbola questions below, each with a step-by-step solution.

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The curve for which the normal at any point $(x, y)$ and the line joining the origin to that point form an isosceles triangle with the $x$ -axis as the base is

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The curve for which the normal at any point (x,y) and the line joining the origin to that point form an isosceles triangle with the x -axis as the base is

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The curve for which the normal at any point $(x, y)$ and the line joining the origin to that point form an isosceles triangle with the $x$ -axis as the base is