The point $P \left( 2, \frac{1}{p^{2}} \right)$, where $p \neq 0$, lies on the parabola $x^{2} = 4y$ - HSC - SSCE Mathematics Extension 1 - Question 6 - 2017 - Paper 1
Question 6
The point $P \left( 2, \frac{1}{p^{2}} \right)$, where $p \neq 0$, lies on the parabola $x^{2} = 4y$.
What is the equation of the normal at $P$?
Worked Solution & Example Answer:The point $P \left( 2, \frac{1}{p^{2}} \right)$, where $p \neq 0$, lies on the parabola $x^{2} = 4y$ - HSC - SSCE Mathematics Extension 1 - Question 6 - 2017 - Paper 1
Step 1
Determine the coordinates of point P
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Answer
The point P(2,p21) lies on the parabola, so we can substitute x=2 into the equation of the parabola to find y:
x^{2} &= 4y \
2^{2} &= 4y \
4 &= 4y \
y &= 1.
\end{align*}$$
Thus, the coordinates of point $P$ are $\left( 2, 1 \right)$.
Step 2
Find the slope of the tangent at P
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Answer
The equation of the parabola is x2=4y. The derivative with respect to x gives:
dxdy=2y1.
Substituting y=1:
dxdy=211=21.
So the slope of the tangent line at point P is 21. The slope of the normal line is then the negative reciprocal: −2.
Step 3
Write the equation of the normal line at P
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Answer
Using the point-slope form of the line equation, we have:
y−y1=m(x−x1)
where (x1,y1)=(2,1) and m=−2:
y−1=−2(x−2).
Distributing,
y = -2x + 5.$$
Rearranging this gives:
$$2x + y - 5 = 0.$$
To express it in the required form, we can recognize that it is equivalent to:
$$p^{2}y + p^{3}x = 1 + 2p^{2}$$
where $p$ correlates with the coefficients in the standard form.
