Question Statement

Find the equation of the tangent and the normal to the parabola $y^2=18x$ at the end points of its latus rectum.

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Background and Explanation

To solve this problem, we need to understand the properties of a parabola in the form $y^2=4ax$. The latus rectum is a chord passing through the focus perpendicular to the axis of symmetry, with endpoints at $(a,2a)$ and $(a,-2a)$. The slope of the tangent at any point is found using the derivative $\frac{dy}{dx}$, and the normal is the line perpendicular to the tangent at that same point.

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Solution

1. Finding the Endpoints of the Latus Rectum

The given equation of the parabola is: $y^2=18x$

By comparing this with the standard form $y^2=4ax$, we can find the value of $a$: $\begin{array}{rl}4a& =18\\ a& =\frac{18}{4}=\frac{9}{2}\end{array}$

The endpoints of the latus rectum for the parabola $y^2=4ax$ are given by the coordinates $A(a,2a)$ and $B(a,-2a)$. Substituting $a=\frac{9}{2}$:

• For point $A$: $A\left(\frac{9}{2},2\left(\frac{9}{2}\right)\right)=A\left(\frac{9}{2},9\right)$
• For point $B$: $B\left(\frac{9}{2},-2\left(\frac{9}{2}\right)\right)=B\left(\frac{9}{2},-9\right)$

2. Finding the Slope of the Tangent

To find the slope of the tangent at any point $(x,y)$, we differentiate the equation $y^2=18x$ with respect to $x$: $\begin{array}{rl}2y\frac{dy}{dx}& =18\\ \frac{dy}{dx}& =\frac{18}{2y}\\ \frac{dy}{dx}& =\frac{9}{y}\end{array}$ The slope of the tangent line at any point on the parabola is $m=\frac{9}{y}$.

3. Equations at Point $A\left(\frac{9}{2},9\right)$

Slope of Tangent at $A$: $m_1=\frac{9}{9}=1$

Equation of Tangent at $A$: Using the point-slope form $y-y_1=m(x-x_1)$: $\begin{array}{rl}y-9& =1\left(x-\frac{9}{2}\right)\\ 2(y-9)& =2x-9\\ 2y-18& =2x-9\\ 2x-2y+9& =0\end{array}$

Slope of Normal at $A$: Since the normal is perpendicular to the tangent: $m_{normal}=\frac{-1}{m_{tangent}}=\frac{-1}{1}=-1$

Equation of Normal at $A$: $\begin{array}{rl}y-9& =-1\left(x-\frac{9}{2}\right)\\ 2(y-9)& =-2x+9\\ 2y-18& =-2x+9\\ 2x+2y-27& =0\end{array}$

4. Equations at Point $B\left(\frac{9}{2},-9\right)$

Slope of Tangent at $B$: $m_2=\frac{9}{-9}=-1$

Equation of Tangent at $B$: $\begin{array}{rl}y-(-9)& =-1\left(x-\frac{9}{2}\right)\\ y+9& =-x+\frac{9}{2}\\ 2(y+9)& =-2x+9\\ 2y+18& =-2x+9\\ 2x+2y+9& =0\end{array}$

Slope of Normal at $B$: $m_{normal}=\frac{-1}{-1}=1$

Equation of Normal at $B$: $\begin{array}{rl}y-(-9)& =1\left(x-\frac{9}{2}\right)\\ y+9& =x-\frac{9}{2}\\ 2(y+9)& =2x-9\\ 2y+18& =2x-9\\ 2x-2y-27& =0\end{array}$

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Key Formulas or Methods Used

• Standard Parabola: $y^2=4ax$
• Latus Rectum Endpoints: $(a,2a)$ and $(a,-2a)$
• Slope of Tangent (Calculus): $\frac{dy}{dx}$
• Point-Slope Form: $y-y_1=m(x-x_1)$
• Perpendicular Slopes: $m_1\cdot m_2=-1$

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Summary of Steps

1. Identify $a$ by comparing $y^2=18x$ to $y^2=4ax$.
2. Determine the coordinates of the endpoints of the latus rectum, $A(a,2a)$ and $B(a,-2a)$.
3. Differentiate the parabola equation to find the general expression for the slope of the tangent ($\frac{dy}{dx}=\frac{9}{y}$).
4. Calculate the specific tangent slope for each point and use the point-slope formula to find the tangent equations.
5. Calculate the negative reciprocal of the tangent slopes to find the normal slopes.
6. Use the point-slope formula to find the equations of the normals at both points.