Question

The equation to the parabola whose focus is (1, –1) and vertex is (2, 1).


Answer:
4x² – 4xy + y² + 8x + 46y – 71 =0 ​

Answer 1

We know the parabola is the locus of a point which is equidistant from a fixed point called focus and a straight line called directrix.

Let (h, k) be a point on the parabola. The focus and vertex of this parabola are (1, -1) and (2, 1) respectively. Let the equation of the directrix be $Ax+By+C =0.$

The line joining focus and vertex will be perpendicular to the directrix.

The slope of this line is,

$\longrightarrow m=\dfrac{1+1}{2-1}$

$\longrightarrow m=2$

Then the slope of directrix will be,

$\longrightarrow -\dfrac{A}{B}= -\dfrac{1}{2}$

$\longrightarrow \dfrac{A}{B}=\dfrac{1}{2}$

The distance between (h, k) and focus (1, -1) is,

$\longrightarrow p=\sqrt{(h-1)^2+(k+1)^2}\quad\quad\dots (1)$

The distance between (h, k) and directrix is,

$\longrightarrow p=\dfrac{|Ah+Bk+C|}{\sqrt{A^2+B^2} }$

$\longrightarrow p=\dfrac{|B|\left|\dfrac{A}{B}h+k+\dfrac{C}{B}\right|}{|B|\sqrt{\left(\dfrac{A}{B}\right)^2+1}}$

$\longrightarrow p=\dfrac{\left|\dfrac{h}{2}+k+\dfrac{C}{B}\right|}{\sqrt{\left(\dfrac{1}{2}\right)^2+1}}$

$\longrightarrow p=\dfrac{\left|h+2k+\dfrac{2C}{B}\right|}{\sqrt5 }\quad\quad\dots(2)$

The distance between focus (1, -1) and vertex (2, 1),

$\longrightarrow a=\sqrt{(2-1)^2+(1+1)^2}$

$\longrightarrow a=\sqrt{5}$

This is the same as the distance between vertex and directrix, because vertex is a point on the parabola which is equidistant from focus and directrix.

$\longrightarrow\dfrac{|2A+B+C|}{\sqrt{A^2+B^2}} =\sqrt 5$

$\longrightarrow\dfrac{\left|\dfrac{2A}{B}+1+\dfrac{C}{B}\right|}{\sqrt{\left(\dfrac{A}{B}\right)^2+1}} =\sqrt 5$

$\longrightarrow\dfrac{\left|1+1+\dfrac{C}{B}\right|}{\sqrt{\left(\dfrac{1}{2}\right)^2+1}} =\sqrt 5$

$\longrightarrow\dfrac{2\left|\dfrac{C}{B}+2\right|}{\sqrt{5}} =\sqrt 5$

$\longrightarrow\left|\dfrac{C}{B}+2\right|=\dfrac{5}{2}$

Let,

$\longrightarrow\dfrac{C}{B}+2=\dfrac{5}{2}$

$\longrightarrow\dfrac{C}{B}=\dfrac{1}{2}$

Then (2) becomes,

$\longrightarrow p=\dfrac{\left|h+2k+1\right|}{\sqrt5}\quad\quad\dots(3.1 )$

From (1) and (3.1),

$\longrightarrow\sqrt{(h-1)^2+(k+1)^2}=\dfrac{|h+2k+1|}{\sqrt5 }$

$\longrightarrow(h-1)^2+(k+1)^2=\dfrac{(h+2k+1)^2}{5}$

$\longrightarrow 5h^2+5k^2-10h+10k+10=h^2+4k^2+1+4hk+2h+4k$

$\longrightarrow 4h^2+k^2-12h+6k-4hk+9=0$

Replacing $(h,\ k)$ by $(x,\ y)$ we get the equation.

$\longrightarrow 4x^2+y^2+9+6y-12x-4xy=0$

But, before concluding, take,

• $a=4$

• $b=1$

• $c=9$

• $f=\dfrac{6}{2}=3$

• $g=-\dfrac{12}{2}=-6$

• $h=-\dfrac{4}{2}=-2$

We see,

$\longrightarrow\left|\begin{array}{ccc}a&h&g\\h&b&f\\g&f&c\end{array}\right|=\left|\begin{array}{ccc}4&-2&-6\\-2&1&3\\-6&3&9\end{array}\right|$

$\longrightarrow\left|\begin{array}{ccc}a&h&g\\h&b&f\\g&f&c\end{array}\right|=4(9-9)-2(-18+18)-6(-6+6 )$

$\longrightarrow\left|\begin{array}{ccc}a&h&g\\h&b&f\\g&f&c\end{array}\right|=0$

It means the equation represents a pair of straight lines, not parabola.

Let,

$\longrightarrow\dfrac{C}{B}+2=-\dfrac{5}{2 }$

$\longrightarrow\dfrac{C}{B}=-\dfrac{9}{2 }$

Then (2) becomes,

$\longrightarrow p=\dfrac{\left|h+2k-9\right|}{\sqrt5 }\quad\quad\dots(3.2)$

From (1) and (3.1),

$\longrightarrow\sqrt{(h-1)^2+(k+1)^2}=\dfrac{|h+2k-9|}{\sqrt5 }$

$\longrightarrow(h-1)^2+(k+1)^2=\dfrac{(h+2k-9)^2}{5}$

$\longrightarrow 5h^2+5k^2-10h+10k+10=h^2+4k^2+81+4hk-18h-36k$

$\longrightarrow 4h^2+k^2+8h+46k-4hk-71=0$

Replacing $(h,\ k)$ by $(x,\ y)$ we get the equation.

$\longrightarrow 4x^2+y^2-71+46y+8x-4xy= 0$

Before concluding, we also take here,

• $a=4$

• $b=1$

• $c=-71$

• $f=\dfrac{46}{2}=23$

• $g=\dfrac{8}{2}=4$

• $h=-\dfrac{4}{2}=-2$

We see,

$\longrightarrow\left|\begin{array}{ccc}a&h&g\\h&b&f\\g&f&c\end{array}\right|=\left|\begin{array}{ccc}4&-2&4\\-2&1&23\\4&23&-71\end{array}\right|$

$\longrightarrow\left|\begin{array}{ccc}a&h&g\\h&b&f\\g&f&c\end{array}\right|=4(-71-529)-2(92-142)+4(-46-4)$

$\longrightarrow\left|\begin{array}{ccc}a&h&g\\h&b&f\\g&f&c\end{array}\right|=4(-71-529)-2(92-142)+4(-46-4)$

$\longrightarrow\left|\begin{array}{ccc}a&h&g\\h&b&f\\g&f&c\end{array}\right|=-2500\neq 0$

This means the equation does not represent a pair of straight lines.

But it does not mean if the equation represents a parabola!

For this we need to check if $h^2=ab.$

$\longrightarrow h^2=(-2)^2=4$

$\longrightarrow ab=4\times1=4$

So we see,

$\longrightarrow h^2= ab$

This means the equation represents a parabola.

Hence the answer is,

$\longrightarrow\underline{\underline{4x^2-4xy+y^2+8x+46y-71 =0}}$