Question Statement

Find the equations of the tangent and normal to the hyperbola $\frac{{\left(y-\frac{7}{4}\right)}^2}{\left(\frac{25}{48}\right)}-\frac{{\left(x-\frac{7}{6}\right)}^2}{\left(\frac{25}{72}\right)}=1$ at the point whose ordinate ($y$) is $3$ and the abscissa ($x$) is an integer.

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Background and Explanation

To find the equations of the tangent and normal, we first determine the coordinates of the point on the hyperbola by substituting the given $y$-value and solving for $x$. We then use implicit differentiation to find the slope of the tangent line ($m=\frac{dy}{dx}$) at that point, and use the negative reciprocal ($-\frac{1}{m}$) for the slope of the normal line.

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Solution

Step 1: Find the point on the hyperbola

Given the equation of the hyperbola: $\frac{{\left(y-\frac{7}{4}\right)}^2}{\left(\frac{25}{48}\right)}-\frac{{\left(x-\frac{7}{6}\right)}^2}{\left(\frac{25}{72}\right)}=1$

We can simplify this by multiplying the numerators by the reciprocals of the denominators: $\frac{48{\left(y-\frac{7}{4}\right)}^2}{25}-\frac{72{\left(x-\frac{7}{6}\right)}^2}{25}=1$ Multiplying the entire equation by $25$: $48{\left(y-\frac{7}{4}\right)}^2-72{\left(x-\frac{7}{6}\right)}^2=25$

Given that the ordinate ($y$) of the point is $3$, we substitute $y=3$ into equation (i): $48{\left(3-\frac{7}{4}\right)}^2-72{\left(x-\frac{7}{6}\right)}^2=25$ $48{\left(\frac{5}{4}\right)}^2-72{\left(\frac{6x-7}{6}\right)}^2=25$ $48\left(\frac{25}{16}\right)-72\frac{(6x-7{)}^2}{36}=25$ $3(25)-2(6x-7{)}^2=25$ $75-2(6x-7{)}^2=25$ $2(6x-7{)}^2=50$ $(6x-7{)}^2=25$

Solving for $x$: $6x-7=\pm 5$

• Case 1: $6x-7=-5\Rightarrow 6x=2\Rightarrow x=\frac{1}{3}$
• Case 2: $6x-7=5\Rightarrow 6x=12\Rightarrow x=2$

Since the problem states the abscissa ($x$) must be an integer, we choose $x=2$. Thus, the point of interest is $P(2,3)$.

Step 2: Find the slope of the tangent

Differentiate equation (i) with respect to $x$: $\frac{d}{dx}\left[48{\left(y-\frac{7}{4}\right)}^2-72{\left(x-\frac{7}{6}\right)}^2\right]=\frac{d}{dx}[25]$ $48\left[2\left(y-\frac{7}{4}\right)\right]\frac{dy}{dx}-72\left[2\left(x-\frac{7}{6}\right)\right]=0$ $96\left(y-\frac{7}{4}\right)\frac{dy}{dx}=144\left(x-\frac{7}{6}\right)$ $\frac{dy}{dx}=\frac{144\left(x-\frac{7}{6}\right)}{96\left(y-\frac{7}{4}\right)}=\frac{3\left(x-\frac{7}{6}\right)}{2\left(y-\frac{7}{4}\right)}$

At the point $P(2,3)$: $m={\frac{dy}{dx}|}_{(2,3)}=\frac{3\left(2-\frac{7}{6}\right)}{2\left(3-\frac{7}{4}\right)}$ $m=\frac{3\left(\frac{5}{6}\right)}{2\left(\frac{5}{4}\right)}=\frac{\frac{5}{2}}{\frac{5}{2}}=1$ The slope of the tangent line is $m=1$.

Step 3: Equation of the tangent line

Using the point-slope form $y-y_1=m(x-x_1)$: $y-3=1(x-2)$ $y-3=x-2$ $x-y+1=0$

Step 4: Equation of the normal line

The slope of the normal line is the negative reciprocal of the tangent slope: $m_{\text{normal}}=\frac{-1}{m}=\frac{-1}{1}=-1$

Using the point-slope form $y-y_1=m_{\text{normal}}(x-x_1)$: $y-3=-1(x-2)$ $y-3=-x+2$ $x+y-5=0$

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Key Formulas or Methods Used

• Implicit Differentiation: Used to find the derivative $\frac{dy}{dx}$ of the hyperbola.
• Point-Slope Form: $y-y_1=m(x-x_1)$.
• Slope of Normal: $m_1\cdot m_2=-1$.
• Integer Constraint: Used to select the correct $x$ coordinate from the quadratic solution.

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Summary of Steps

1. Simplify the hyperbola equation and substitute $y=3$ to find the $x$ coordinates.
2. Select the integer value $x=2$ to identify the point $P(2,3)$.
3. Differentiate the hyperbola equation implicitly to find the expression for the slope $\frac{dy}{dx}$.
4. Evaluate the derivative at $P(2,3)$ to find the tangent slope $m=1$.
5. Apply the point-slope formula to find the tangent equation: $x-y+1=0$.
6. Calculate the normal slope $m=-1$ and apply the point-slope formula to find the normal equation: $x+y-5=0$.