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Chapter 15: Problem 1448

A common tangent to \(9 \mathrm{x}^{2}-16 \mathrm{y}^{2}=144\) and \(\mathrm{x}^{2}+\mathrm{y}^{2}=9\) is (a) \(\mathrm{y}=3 \sqrt{(2 / 7) \mathrm{x}+(15 / \sqrt{7})}\) (b) \(\mathrm{y}=2 \sqrt{(3 / 7) \mathrm{x}-15 \sqrt{7}}\) (c) \(\mathrm{y}=(3 / \sqrt{7}) \mathrm{x}+(15 / \sqrt{7})\) (d) \(\mathrm{y}=2 \sqrt{(3 / 7) \mathrm{x}+15 \sqrt{7}}\)

Short Answer

Expert verified
The common tangent to the given curves is: \(y = 2\sqrt{\frac{3}{7}x - 15\sqrt{7}}\)

Step by step solution

01

(Step 1: Rewrite the equations in standard form)

Divide the equation of the hyperbola by 144 to get the standard form: 1) \(\frac{x^2}{4^2} - \frac{y^2}{3^2} = 1\) and the equation for circle is already in standard form: 2) \(x^2 + y^2 = 9\)
02

(Step 2: Find the tangent of the hyperbola using parameter t)

The equation of the tangent of the hyperbola in the parametric form is : \(y = tx + \frac{3}{4t}\)
03

(Step 3: Check the same tangent on the circle, by substituting the point and slope from the tangent equation of a hyperbola)

Substitute the point on the hyperbola tangent: x = 4\(\sqrt{\frac{1 + t^2}{t^2}}\); y = 3\(\sqrt{\frac{1 + t^2}{1 - t^2}}\) Now rewrite the tangent equation as : \(y = tx - 3t^3 + 3t\)
04

(Step 4: Compare the resulting equation with the given options to get the correct answer)

Comparing the resulting equation with the given options, we find that the correct answer is: (b) \(y = 2\sqrt{\frac{3}{7}x - 15\sqrt{7}}\) Hence, the common tangent to the given curves is: \(y = 2\sqrt{\frac{3}{7}x - 15\sqrt{7}}\)

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Most popular questions from this chapter

The length of the common chord of the ellipse \(\left[(x-1)^{2} / 9\right]+\left[(y-2)^{2} / 4\right]=1\) and the circle \((x-1)^{2}+(y-2)^{2}=1\) (a) \(\sqrt{2}\) (b) \(\sqrt{3}\) (c) 4 (d) None of these

The curve represented by \(\mathrm{x}=3(\cos \mathrm{t}+\sin \mathrm{t})\); \(\mathrm{y}=4(\cos \mathrm{t}-\sin \mathrm{t})\) is (a) circle (b) parabola (c) ellipse (d) hyperbola

If \(\mathrm{AB}\) is a double ordinates of the hyperbola \(\left(\mathrm{x}^{2} / \mathrm{a}^{2}\right)-\left(\mathrm{y}^{2} / \mathrm{b}^{2}\right)=1\) such that \(\mathrm{OAB}\) is an equilateral triangle, \(O\) being the centre of the hyperbola, then the eccentricity e of the hyperbola satisfies. (a) \(1<\mathrm{e}<(2 / \sqrt{3})\) (b) \(\mathrm{e}<(1 / \sqrt{3})\) (c) \(\mathrm{e}>(\sqrt{3} / 2)\) (d) \(e>(2 / \sqrt{3})\)

One of the diameters of the circle circumscribing the rectangle \(\mathrm{ABCD}\) is \(\mathrm{x}-4 \mathrm{y}+7=0 .\) If \(\mathrm{A}\) and \(\mathrm{B}\) are points \((-3,4)\) and \((5,4)\) respectively, then the area of the rectangle is ... (a) 32 sq. units (b) 16 sq. units (c) 64 sq. units (d) 8 sq. units

The equation of the tangent to the curve \(4 \mathrm{x}^{2}-9 \mathrm{y}^{2}=1\). Which is parallel to \(5 \mathrm{x}-4 \mathrm{y}+7=0\) is (a) \(30 \mathrm{x}-24 \mathrm{y}+17=0\) (b) \(24 \mathrm{x}-30 \mathrm{y} \pm \sqrt{(161)}=0\) (c) \(30 \mathrm{x}-24 \mathrm{y} \pm \sqrt{(161)}=0\) (d) \(24 \mathrm{x}+30 \mathrm{y} \pm \sqrt{(161)}=0\)
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