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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.

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    $\begingroup$ 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. $\endgroup$
    – Blue
    Dec 29 '18 at 23:30
  • $\begingroup$ I dont know the method of calculus but the basic method that tangent at any point on a curve is T=0. $\endgroup$
    – Badguy
    Dec 30 '18 at 8:47
  • $\begingroup$ What is "$T$" ? $\endgroup$
    – Blue
    Dec 30 '18 at 8:50
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$$y^2=4ax \tag{1}$$

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

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

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

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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

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

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

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  • $\begingroup$ Not understood. $\endgroup$
    – Badguy
    Dec 29 '18 at 16:22
  • $\begingroup$ I have neither studied nor seen what youve written here $\endgroup$
    – Badguy
    Dec 29 '18 at 16:23
  • $\begingroup$ @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. $\endgroup$ Jan 2 '19 at 8:46

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