1

I'm sitting an exam soon and have been revising for a while but I sometimes get stuck with questions that I know are simple but still have no idea about how to do them at all! Please explain to me how this is done. (Note: this is all the info I have about the question):

Let $L$ be the line with equation $x + y = 0$, $F = (1,1)$ and $P = (a,b)$.

(a) Find the equation of the line perpendicular to $L$ which passes through $P$, the intersection of the two lines and hence the least distance between $P$ and $L$.

(b) Determine the Cartesian equation of the parabola with focus $F$ and directrix $L$.

(c) Show that the tangent to the parabola at the point $(x_1,y_1)$ has equation $(x_1 - y_1 + 2)y + (y_1 - x_1 + 2)x - 2(x_1 + x_1y_1 + y_1) + x_1^2 + y_1^2 = 0$.

Librecoin
  • 2,740
F Addy
  • 11
  • Your question is phrased as an isolated problem, without any further information or context. This does not match MSE quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and *include your work and thoughts on the problem*. Making these improvements will attract more appropriate answers and make the question more valuable for future MSE visitors. – apnorton Apr 28 '13 at 17:09
  • For (a), recall that the product of the slopes of perpendicular lines is $-1$; you will also need the point-slope form for a line. For (b), a parabola is the locus of points such that the distance from the focus to a point on the curve is equal to the perpendicular distance of that point to the directrix. Once you have the parabola's equation, (c) will involve finding the slope of a tangent line to any point on the curve. – colormegone Apr 28 '13 at 17:11
  • @anorton noted. Please excuse me as I am new to MSE, I shall do as you ask in future questions, and will try to improve this one. Thank you for your advice. – F Addy Apr 28 '13 at 17:17
  • @RecklessReckoner Hmm I see. It's starting to make sense now, thank you for your help! – F Addy Apr 28 '13 at 17:18

1 Answers1

0

a) The line $y_L=-x$ has a slope of $-1$, so a perpendicular line must have a slope of $-\frac{-1}{1}=1$; hence $y_P=x+C$, where $b=y_P(a)=a+C$ implies $C=b-a$. The lines intersect at $y_L(x_I)=-x_I=x_I+b-a=y_P(x_I)$; that is, at $x_I=\frac{a-b}{2}$. Thus the distance between the point $P=(a,b)$ and the line $L$ is given by $\sqrt{\left(a-\frac{a-b}{2}\right)^2+\left(b-\frac{b-a}{2}\right)^2}=\sqrt{2}\left(\frac{a+b}{2}\right)$.

b) Let $(x,y)$ be any point on the parabola. The distance between the point and the focus $F=(1,1)$ is $\sqrt{(x-1)^2+(y-1)^2}$ and the distance between the point and the directrix $L$ is $\sqrt{2}\left(\frac{x+y}{2}\right)$, which we calculated above. Now, we equate and square the two expressions:

$$\begin{align*} &\quad(x-1)^2+(y-1)^2=2\left(\frac{x+y}{2}\right)^2=\frac{(x+y)^2}{2}\\ &\implies x^2-2x+1+y^2-2y+1=\frac{x^2+2xy+y^2}{2}\\ &\implies \frac{x^2}{2}-2x+\frac{y^2}{2}-2y-xy+2=0 \end{align*}$$

This is the equation of the parabola with focus $F$ and directrix $L$.

c) We first find the slope of the tangent line. To do this, we consider $y$ as a function of $x$ and differentiate with respect to $x$:

$$\begin{align*} &\quad\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x^2}{2}-2x+\frac{y(x)^2}{2}-2y(x)-xy(x)+2\right)=\\ &=x-2+y(x)y'(x)-2y'(x)-y(x)-xy'(x)=0\\ &\implies \frac{\mathrm{d}y}{\mathrm{d}x}=y'(x)=\frac{2-x+y}{y-x-2} \end{align*}$$

At the point $(x_1,y_1)$, we obtain the slope $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2-x_1+y_1}{y_1-x_1-2}$. Hence the equation of the tangent line is given by

$$y-y_1=\frac{2-x_1+y_1}{y_1-x_1-2}(x-x_1)\\ \implies yy_1-yx_1-2y-y_1^2+x_1y_1+2y_1=2x-xx_1+xy_1-2x_1+x_1^2-x_1y_1\\ \implies x(y_1-x_1+2)+y(x_1-y_1+2)-2(x_1+x_1y_1+y_1)+x_1^2+y_1^2=0$$

This is what we wanted to show.

Librecoin
  • 2,740