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Chapter 15: Problem 1438

The locus of a point \(\mathrm{P}(\alpha, \beta)\) moving under the condition that the line \(\mathrm{y}=\alpha \mathrm{x}+\beta\) is a tangent to the hyperbola \(\left(x^{2} / a^{2}\right)-\left(y^{2} / b^{2}\right)=1\) is (a) a circle (b) a parabola (c) an ellipse (d) a hyperbola

Short Answer

Expert verified
The locus of the point P(α, β) moving under the given conditions is a parabola.

Step by step solution

01

Identify the equation of the given line and hyperbola

The given line is y = αx + β, and the given hyperbola is \(\left(\frac{x^2}{a^2}\right) - \left(\frac{y^2}{b^2}\right) = 1\).
02

Find the point of intersection between the line and hyperbola

To find the point of intersection, we need to simultaneously solve these two equations. Let the point of intersection be Q(x, y). We can write y = αx + β as \(y = αx + \beta\) and substitute this into the hyperbola equation: \[\left(\frac{x^2}{a^2}\right) - \left(\frac{(αx + β)^2}{b^2}\right) = 1\]
03

Differentiate the hyperbola equation and find the condition of tangency

To find the slope of the tangent line at the point Q(x, y), we will differentiate the hyperbola equation with respect to x: \[\frac{d}{dx} \left(\frac{x^2}{a^2} - \frac{y^2}{b^2}\right) = \frac{d}{ dx}(1)\] \[2 \frac{x}{a^2} - 2 \frac{yy'}{b^2} = 0\] Now, we can solve for the slope y': \[y' = \frac{b^2 x}{a^2 y}\] This is the slope of the hyperbola at the point of intersection Q(x, y).
04

Equate the slopes for the condition of tangency

As the tangent line and the hyperbola have the same slope at the point of intersection, we can equate their slopes: \[\frac{b^2 x}{a^2 y} = α\]
05

Eliminate x to find the equation of the locus

From the condition of tangency, we can find x in terms of α and y: \[x = \frac{a^2 yα}{b^2}\] Substitute this value of x into the equation of the tangent line: \[y = α\left(\frac{a^2 yα}{b^2}\right) + β\] Simplifying and rearranging, we get: \[\frac{a^2 α^2 y^2}{b^2} - y + β = 0\] This represents the equation of the locus of the point P(α, β) under the given conditions.
06

Compare the locus equation with the standard equation of conic sections

The locus equation that we derived in the previous step can be written as: \[\frac{a^2 α^2}{b^2} y^2 - y + β = 0\] This equation is similar to the standard equation of a parabola, which is of the form Ay^2 + By + C = 0 if A > 0 and B ≠ 0. In our case, A = \(\frac{a^2 α^2}{b^2}\) > 0, B = -1 ≠ 0, and C = β. So, the locus of point P(α, β) moving under the given conditions is a parabola. Thus, the correct answer is (b) a parabola.

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Most popular questions from this chapter

If the line \(\mathrm{lx}+\mathrm{my}+\mathrm{n}=0\) cuts an ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\) in points whose eccentric angles differ by \((\pi / 2)\), then \(\left[\left(a^{2} \ell^{2}+b^{2} \mathrm{~m}^{2}\right) / \mathrm{n}^{2}\right]=\ldots \ldots .\) (a) 1 (b) \((3 / 2)\) (c) 2 (d) \((5 / 2)\)

If angle between asymptote of hyperbola \(\left(\mathrm{x}^{2} / \mathrm{a}^{2}\right)-\left(\mathrm{y}^{2} / \mathrm{b}^{2}\right)=1\) is \(45^{\circ} .\) The value of eccentricity is \(\ldots \ldots\) (a) \(\sqrt{(4 \pm 2 \sqrt{2})}\) (b) \(\sqrt{(4+2 \sqrt{2})}\) (c) \(\sqrt{(4-2 \sqrt{2})}\) (d) none of these

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