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.

  • 1
    $\begingroup$ 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

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

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

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

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

enter image description here

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 $$2yy'=4a$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.