Question

\(y=\frac{x^{2}}{4}-3 x+10\) For \(y=2-\frac{x^{2}}{4}\) eqn of tangent \(y-2=m x+m^{2}\) Solving (II) \& (I) \(m x+2+m^{2}=\frac{x^{2}}{4}-3 x+10\) \(\Rightarrow \frac{x^{2}}{4}-(m+3) x+\left(8-m^{2}\right)=0\) \(\mathrm{D}=0\) \((m+3)^{2}-\left(8-m^{2}\right)=0\) \(2 m^{2}+6 m+1=0\) \(\mathrm{m}_{1}+\mathrm{m}_{2}=-3\)

Short answer

Answer: The equations of the tangent lines to the given curve that also satisfy the given equation are \({y-2=m_1x+m_1^2}\) and \({y-2=m_2x+m_2^2}\), where \(m_1\) and \(m_2\) are the roots of the quadratic equation \(2m^2+6m+1=0\) and their sum is equal to -3.

Step-by-step solution

Find the derivative of the given curve

Consider the given curve \(y=\frac{x^{2}}{4}-3 x+10\). We will find the derivative of this curve using the power rule. \[ \begin{aligned} y'=\frac{d}{dx}\left(\frac{x^2}{4}-3x+10\right) \end{aligned} \] Note that \(\frac{d}{dx}(x^2/4) = \frac{1}{2}x\), \(\frac{d}{dx}(-3x) = -3\), and \(\frac{d}{dx}(10)=0\). Thus, \[ \begin{aligned} y' = \frac{1}{2}x-3 \end{aligned} \]

Substitute the given values and solve for \(m\)

Recall the given equation, \(y-2=mx+m^2\). Replacing y by 2: \[ \begin{aligned} 2-2=m x+m^{2} \end{aligned} \] So, \(mx+m^2=0\). Now, \[ \begin{aligned} \frac{x^{2}}{4}-(m+3) x+\left(8-m^{2}\right)=0 \end{aligned} \] Insert the given condition for the discriminant D = 0: \[ \begin{aligned} (m+3)^{2}-\left(8-m^{2}\right)=0 \end{aligned} \] By solving the equation, we arrive at the quadratic equation: \(2m^2+6m+1=0\).

Finding the roots of the quadratic equation

Next, find the roots of the quadratic equation \(2m^2+6m+1=0\). The sum of roots is given by: \(m_1+m_2=-\frac{B}{A}=-3\).

Finding the tangent equation

The sum of the roots is equal to -3, and the equation for the tangent line is given by \({y-2=mx+m^2}\). We can use the quadratic equation's roots to find the equation of the tangent line. The tangent lines to the curve for each root are given by the equation \({y-2=m_1x+m_1^2}\) and \({y-2=m_2x+m_2^2}\) respectively.