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Chapter 8: Problem 3

Find the equations of the tangents drawn to the curve \(y^{2}-2 x^{3}-4 y+8=0\) from the point \((1,2)\).

Short Answer

Expert verified
The equation of the tangent line drawn to the given curve from the point \((1,2)\) is \(x - 2y + 3 = 0\).

Step by step solution

01

Differentiate the curve equation

Differentiation of the curve equation in order to find the slope of the tangential line yields: \[\frac{dy}{dx} = \frac{6x^{2} - 4y}{2y}\] and simplify to \[\frac{dy}{dx} = \frac{3x^{2}}{y} - 2\].
02

Find the slope of the tangential line

Substitute the given point \((1,2)\) in the differentiated equation to find \(\frac{dy}{dx}\). This results in: \[\frac{dy}{dx} = \frac{3(1)^{2}}{2} - 2 = \frac{1}{2}\]. This is the slope of the tangent line at the given point.
03

Identify the equation of the tangent line

We can find the equation of the tangent line using the point-slope form of a linear equation, \[y - y_{1} = m(x - x_{1})\], where \(m\) is the slope and \((x_{1}, y_{1})\) is the given point. Substituting for \(m\), \(x_{1}\), and \(y_{1}\) in terms of the given point and slope, we get \[y - 2 = \frac{1}{2}(x - 1)\] or \[2y - 4 = x - 1\] which can be rearranged to be \[x - 2y + 3 = 0\nonumber\]. This is the equation of one tangent line.
04

Verify the point is on the tangent line

Check that the given point \((1,2)\) satisfies the equation of the tangent line, \[x - 2y + 3 = 0\], by substituting \(x\) and \(y\). This gives: \[1 - 2(2) + 3 = 0\nonumber\] which is true, thus satisfying the equation of the tangent line.

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Most popular questions from this chapter

The traingle formed by the tangent to the curve \(f(x)=x^{2}+b x-b\) at the point \((1,1)\) and the co-ordinate axes, lies in the first quadrant. If the area is 2 , then find the value of \(b\).

Find the equation of the common tangent to the curves \(y=3 x^{2}\) and \(y=2 x^{3}+1\).

Find the shortest distance between the curves \(y^{2}=x^{3}\) and \(9 x^{2}+9 y^{2}-30 y+16=0 .\)

The area of the triangle formed by the tangent to the curve \(y=\frac{8}{4+x^{2}}\) at \(x=2\) and the co-ordinate axes is (a) 2 sq. units (b) 4 sq. units (c) 8 sq. units (d) \(\frac{7}{2}\) sq. units

Find all the lines that pass through the point \((1,1)\) and are tangent to the curve represented parametrically as \(x=2 t-t^{2}, y=t+t^{2}\) provided \(t \neq 1\).
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