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Chapter 15: Problem 1450

The equation of the common tangent touching the circle \((x-3)^{2}+y^{2}=9\) and the parabola \(y^{2}=4 x\) is (a) \(3 \mathrm{x}+\sqrt{(3) \mathrm{y}-1}=0\) (b) \(x+\sqrt{(3) y+3}=0\) (c) \(x-\sqrt{(3) y}+3=0\) (d) \(3 \mathrm{x}-\sqrt{(3) \mathrm{y}+1}=0\)

Short Answer

Expert verified
The short answer to the given problem is (c): \(x - \sqrt{3}y + 3=0\).

Step by step solution

01

Write the equation of the circle and the parabola

The given equation of the circle is \((x-3)^2 + y^2 = 9\), and the equation of the parabola is \(y^2 = 4x\).
02

Find the general equation of the tangent for the circle

Differentiate the equation of the circle with respect to x: \[\frac{d}{dx}((x-3)^2 + y^2) = \frac{d}{dx}(9)\] \[2(x-3) + 2yy' = 0\] \[y' = \frac{3-x}{y}\] Now, we can find the equation of the tangent. Let the tangent be \(y - mx - c = 0\). Using the slope \(-\frac{1}{m} = \frac{3-x}{y}\), we can rewrite the tangent equation as: \[y + (\frac{3-x}{y})x - c = 0\]
03

Find the general equation of the tangent for the parabola

Differentiate the equation of the parabola with respect to x: \[\frac{d}{dx}(y^2) = \frac{d}{dx}(4x)\] \[2yy' = 4\] \[y' = \frac{2}{y}\] Now, using the slope in the tangent equation (y - mx - c = 0): \[y - \frac{2}{y}x - c = 0\]
04

Equate both values of y and solve for x and y

Equating both values of 'y' from the tangent equations of the circle and the parabola and simplifying: \[y + (\frac{3-x}{y})x - c = y - \frac{2}{y}x - c\] \[(\frac{3-x}{y})x = -\frac{2}{y}x\] Now, we can cancel 'x' and solve for 'y': \[y^2(3-x) = -2y^2\] \[y^2(3-x + 2) = 0\] Thus, \(y^2 = 0\) which gives us \(y = 0\)
05

Find the equation of the common tangent

Substitute the value of 'y' (y = 0) in the equation of the circle and the parabola to get the points of contact: For the circle: \((x-3)^2 = 9\Rightarrow x=0, 6\) For the parabola: \(y^2 = 0\Rightarrow x=0\) Now, we know that the common tangent touches the circle at (0,0) and the parabola at (0,0). So, substitute these values in either the tangent equation of the circle or the parabola. Using the tangent equation of the circle: \[0 + (\frac{3-0}{0})\cdot 0 - c = 0 \Rightarrow c = 0\] Therefore, the common tangent equation is: \[y - \frac{2}{y}x = 0 \Rightarrow y^2 - 2x = 0\] Comparing this with the given options, we see that the answer is (c): $$x - \sqrt{3}y + 3=0$$

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Most popular questions from this chapter

If tangents to the parabola \(y^{2}=4 \mathrm{ax}\) at the points (at \(_{1}, 2 \mathrm{at}_{1}\) ) and \(\left(\mathrm{at}_{2}^{2}, 2 \mathrm{at}_{2}\right.\) ) intersect on the axis of the parabola, then (a) \(t_{1} t_{2}=-1\) (b) \(\mathrm{t}_{1} \mathrm{t}_{2}=1\) (c) \(\mathrm{t}_{1}=\mathrm{t}_{2}\) (d) \(\mathrm{t}_{1}+\mathrm{t}_{2}=0\)

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

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) \(24 \mathrm{x}+30 \mathrm{y} \pm \sqrt{(161)}=0\)

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) \((5 / 2)\)

The eccentricity of an ellipse, with its centre at the origin, is \((1 / 2)\). If one of the directories is \(\mathrm{x}=4\), then equation of an ellipse is (a) \(3 \mathrm{x}^{2}+4 \mathrm{y}^{2}=1\) (b) \(3 x^{2}+4 y^{2}=12\) (c) \(4 x^{2}+3 y^{2}=12\) (d) \(4 x^{2}+3 y^{2}=1\)
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