Tangents to parabola $y^{2}=4ax$ meet hyperbola $x^2/a^2-y^2/b^2=1$ at $A$ and $B$. Find the locus of intersections of the tangents at $A$ and $B$.

Question

If tangents to the parabola $y^{2} = 4ax$ intersect the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ at $A$ and $B$, then find the locus of point of intersection of tangents at $A$ and $B$.

I know that tangent to parabola is $y = mx + a/m$ ($m$ being the slope), but I am not able to figure out how to take out point of intersections.

Answer 1

Hint: Given $$y^2=4ax$$ then $$2yy'=4a$$

Answer 2

$$y^2=4ax \tag{1}$$

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \tag{2}$$

Let $P(X,Y)$ be the required locus.

• For hyperbola $(2)$, $(X,Y)$ is the pole of the polar $AB$ (i.e. chord $AB$ for the hyperbola).

  Equation of $AB$ is $$\frac{X x}{a^2}-\frac{Y y}{b^2}=1 \tag{3}$$

• Equation of tangent of $(1)$ at $C(x_1,y_1)$ $$y_1 y=2a(x+x_1)$$

  Rearranging, we have $$-\frac{x}{x_1}+\frac{y_1 y}{2a x_1}=1 \tag{4}$$

• Identifying $(3)$ and $(4)$, we get $$(X,Y)=\left( -\frac{a^2}{x_1}, -\frac{b^2 y_1}{2a x_1} \right)$$

  $$(x_1,y_1)=\left( -\frac{a^2}{X}, \frac{2a^3 Y}{b^2 X} \right)$$

  But $$y_1^2=4a x_1$$

  $$\left( \frac{2a^3 Y}{b^2 X} \right)^2=4a\left( -\frac{a^2}{X} \right)$$

The locus of $P$ is

$$\fbox{$a^3 Y^2+b^4 X=0$}$$

Useful fact:

Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x_1,y_1)$ is given by

$$ax_1 x+h(y_1 x+x_1 y)+by_1 y+g(x+x_1)+f(y+y_1)+c=0$$