A variable parabola is drawn to pass through A & B, the ends of a
diameter of a given circle with centre at the origin and radius c & to
have as directrix a tangent to a concentric circle of radius 'a' (a>c) ; the axes being AB & a perpendicular diameter, prove that the locus of the focus of the parabola is the standard ellipse
$$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$ where $b^2= a^2– c^2$.
Doubt:
I am not being able to frame to equation of the variable parabola. I feel that any parabola can have a double ordinate whose length is equal to $2c$, which can pass through the ends of diameter of the smaller circle. However, one catch is that the directrix must be a tangent to the bigger circle. But how to write the equation of such a variable parabola, whose locus of focus I need to find ?
Best Answer
We don't need to know the equation of the parabola.
We may assume that $A(c,0),B(-c,0)$.
Let $F(X,Y)$ be the coordinates of the focus of the parabola.
The equation of the directrix is given by $mx+ny=a^2$ where $P(m,n)$ is the tangent point such that $m^2+n^2=a^2$.
Now, by the definition of the parabola, the distance from the focus to $(\pm c,0)$ is equal to the distance from $(\pm c,0)$ to the directrix.
So, we get $$\sqrt{(X-c)^2+(Y-0)^2}=\frac{|mc+n\cdot 0-a^2|}{\sqrt{m^2+n^2}},$$ i.e. $$a^2(X-c)^2+a^2Y^2=m^2c^2-2mca^2+a^4\tag1$$ Also, $$\sqrt{(X+c)^2+(Y-0)^2}=\frac{|m(-c)+n\cdot 0-a^2|}{\sqrt{m^2+n^2}},$$ i.e. $$a^2(X+c)^2+a^2Y^2=m^2c^2+2mca^2+a^4\tag2$$
From $(1)-(2)$, $$a^2(-4cX)=-4mca^2\implies X=m$$ From $(1)+(2)$, $$a^2X^2+a^2c^2+a^2Y^2=m^2c^2+a^4\implies a^2X^2+a^2c^2+a^2Y^2=X^2c^2+a^4$$ which can be written as $$\frac{X^2}{a^2}+\frac{Y^2}{a^2-c^2}=1$$