The point of intersection of the tangents to the parabola $y^2=4x$ at the points where the circle $(x-3)^2+y^2=9$

Question

Problem :

The point of intersection of the tangents to the parabola $y^2=4x$ at the points where the circle $(x-3)^2+y^2=9$ meets the parabola, other than the origin, is ..

Solution :

Point of the intersection of the parabola and circle is given by $(x-3)^2+(4x)^2=9$

$\Rightarrow x^2-6x +9 +16x^2=9$ $\Rightarrow 17x^2-6x=0 $ $\Rightarrow x(17x-6)=0$ $\Rightarrow x =0 ; x = 6/17$

$y =0, $ and $y =\sqrt{\frac{24}{17}}$

How to proceed to find the point of intersection of tangents please suggest. Thanks.

Answer 1

Going from where you ended:

$x=0$ gives $y=0$ and $x=2$ gives $y^2=4\cdot2=8 \Leftrightarrow y=\pm2\sqrt2$.

Let's begin with the tangent line to the parabola at $(2,+2\sqrt2)$, and then exploit the symmetry:

$y=\sqrt{4x}$ gives $y'=\frac{2}{\sqrt{4x}}$. $y'(2)=\frac{2}{\sqrt{4\cdot2}}=\frac{1}{\sqrt{2}}=\frac12\sqrt2$.

The equation of the tangent line at $y=+2\sqrt2$ becomes $y=\dfrac{x}{\sqrt2}+\sqrt2$. By symmetry, the equation of the tangent line at $y=-2\sqrt2$ is $y=-\dfrac{x}{\sqrt2}-\sqrt2$.

$\dfrac{x}{\sqrt2}+\sqrt2=-\dfrac{x}{\sqrt2}-\sqrt2$ gives $x=-2$. The point of intersection is $(-2,0)$.

Answer 2

You can proceed via pole-polar relationships: the intersection of the tangents is the pole of their chord of contact. Once you correct your calculation of the intersection points so that you have two points besides the origin, you’ll find that the equation of the chord of contact will be of the form $x=a$. Now, the polar of the point $(x_0,y_0)$ with respect to this parabola is the line with equation $yy_0=2(x+x_0)$. Comparing this with $x=a$, we can see that $y_0=0$ and $x_0=-a$.