The equation of the common tangent with positive slope to the circle $x^{2} + y^{2} - 8 x = 0$ and the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ is

Answer (Detailed Solution Below)View answer ▾

Answer: Option B — $2 x - 5 y + 4 = 0$

Detailed Solution

A tangent $y = m x + c$ to the hyperbola needs $c^{2} = 9 m^{2} - 4$ .

The circle has centre $(4, 0)$ and radius 4, so $\frac{∣ 4 m + c ∣}{1 + m ^{2}} = 4$ , i.e. $8 m c + c^{2} = 16$ .

Eliminating c gives $495 m^{4} + 104 m^{2} - 400 = 0$ , so $m^{2} = \frac{4}{5}$ . With $m = \frac{2}{5} > 0$ , $c = \frac{4}{5}$ , and the tangent is $5 y = 2 x + 4$ , i.e. $2 x - 5 y + 4 = 0$ .

✓ Check

Distance from

(4, 0)

2 x - 5 y + 4 = 0

\frac{∣ 8 + 4 ∣}{4 + 5} = 4

, and

c^{2} = \frac{16}{5} = 9 \cdot \frac{4}{5} - 4

⚠ Common trap

(A) is tangent to the circle (distance

\frac{∣ 8 - 20 ∣}{3} = 4

) but not to the hyperbola, since its

c^{2} = 80 \neq = \frac{16}{5}

Quick tip

Hyperbola tangent $c^{2} = a^{2} m^{2} - b^{2}$ ; circle: distance from centre $= 4 \Rightarrow m = \frac{2}{5}, c = \frac{4}{5}$

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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 equation of the common tangent with positive slope to the circle $x^{2} + y^{2} - 8 x = 0$ and the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ is

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The equation of the common tangent with positive slope to the circle $x^{2} + y^{2} - 8 x = 0$ and the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ is