The straight lines x/a+y/b=k and x/a-y/b=1/k (with k≠ 0) meet on:

C. a hyperbola. Step 1: Multiply the two equations: (x/a+y/b)(x/a-y/b)=k⋅1/k=1. Step 2: So every meeting point satisfies x^2a^2-y^2b^2=1, which is a hyperbola. Check: a=2,b=3,k=2: solving x/2+y/3=2 and x/2-y/3=1/2 gives x=5/2,y=9/4, and 25/16-9/16=1. Trap: (B) an ellipse needs a sum of squares; (u+v)(u-v) gives a difference of squares. (The source printed both lines with +, which makes them parallel; the second sign has been restored to −.)

Question 4 of 8 · Hyperbola

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The straight lines $\frac{x}{a} + \frac{y}{b} = k$ and $\frac{x}{a} - \frac{y}{b} = \frac{1}{k}$ (with $k \neq = 0$ ) meet on:

Answer (Detailed Solution Below)View answer ▾

Answer: Option C — a hyperbola

Detailed Solution

Multiply the two equations: $(\frac{x}{a} + \frac{y}{b}) (\frac{x}{a} - \frac{y}{b}) = k \cdot \frac{1}{k} = 1$ .

So every meeting point satisfies $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ , which is a hyperbola.

✓ Check

a = 2, b = 3, k = 2

: solving

\frac{x}{2} + \frac{y}{3} = 2

and

\frac{x}{2} - \frac{y}{3} = \frac{1}{2}

gives

x = \frac{5}{2}, y = \frac{9}{4}

, and

\frac{25}{16} - \frac{9}{16} = 1

⚠ Common trap

(B) an ellipse needs a sum of squares;

(u + v) (u - v)

gives a difference of squares. (The source printed both lines with +, which makes them parallel; the second sign has been restored to −.)

Quick tip

$(\frac{x}{a} + \frac{y}{b}) (\frac{x}{a} - \frac{y}{b}) = k \cdot \frac{1}{k} = 1 \Rightarrow \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$

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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 straight lines $\frac{x}{a} + \frac{y}{b} = k$ and $\frac{x}{a} - \frac{y}{b} = \frac{1}{k}$ (with $k \neq = 0$ ) meet on: