# 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 $B$.

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.

• Since you have a tangent line equation, you can find $A$ and $B$ by the standard substitution method: replace $y$ with $mx+a/m$ for $y$ in the hyperbola equation, and solve for $x$; then put the resulting $x$-values into the line equation to get the corresponding $y$-values. For the tangents at $A$ and $B$ ... There are many ways to find tangent line equations. Your comment to Dr. Graubner's answer indicates that you don't know calculus-based approaches. What approaches do you know? How did you get the formula for the tangent to the parabola? Help us help you.
– Blue
Dec 29 '18 at 23:30
• I dont know the method of calculus but the basic method that tangent at any point on a curve is T=0. Dec 30 '18 at 8:47
• What is "$T$" ?
– Blue
Dec 30 '18 at 8:50

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

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

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

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

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

• Not understood. Dec 29 '18 at 16:22
• I have neither studied nor seen what youve written here Dec 29 '18 at 16:23
• @Badguy It's notation from (differential) calculus. $y'$ here is known as the derivative of $y(x)$ with respect to $x$, sometimes it's written as $\dfrac{dy}{dx}$. $y'$ is equivalent to the slope at $(x,y)$ of the curve. Jan 2 '19 at 8:46