parabola locus-coordinate geometry

Question

If the two tangents drawn from a point P to the parabola $y^{2}=4x$ are at right angles then find the locus of P.

I have no real idea on what to do here...I tried to compare this with the general form of a parabola that is $y=mx+1/m$ and find its intersection with the line $y=-\frac{-1}{m}x-m$ and solve this but this does not give a matching answer.

Answer 1

Take a general point on the parabola $\;\left(\cfrac{t^4}4\,,\,\,t\right)\;$ . The slope of the tangent line to the parabolla at the above point is:

$$m:=\frac2t\implies \text{ the tangent line is}\;\;y-t=\frac2t\left(x-\frac{t^2}4\right)=\frac2tx-\frac t2\implies\color{red}{y=\frac2tx+\frac t2}$$

If we take another point $\;\left(\cfrac{a^2}4,\,a\right)\;$, the tangent line at this point is, of course

$$\color{red}{y=\frac 2ax+\frac a2}\implies\text{ the intersection point of these two lines is given by}$$

$$\frac2ax+\frac a2=\frac 2tx+\frac t2\implies2\left(\frac1a-\frac1t\right)x=\frac{t-a}2\implies$$

$$\begin{cases}x=\cfrac{at}4\\{}\\y=\cfrac2a\cfrac{at}4+\cfrac a2=\cfrac t2+\cfrac a2=\cfrac{a+t}2\end{cases}\;\;\;\;\implies P=\left(\frac{at}4\,,\,\,\frac{a+t}2\right)$$

But we're given both tangent lines above are perpendicular, so:

$$\frac2a\cdot\frac2t=-1\implies at=-4\implies P=\left(-1,\,\,\frac{a-\frac4a}2\right)=\left(-1\,,\,\,\frac{a^2-4}{2a}\right)$$

Answer 2

Let $P=(h,k)$ and $y=mx+c$ be the a tangent to the parabola from $P$. Then

$$k=mh+c$$

Put $y=mx+c$ into the equation of the parabola,

\begin{align*} (mx+c)^2&=4x\\ m^2x^2+(2mc-4)x+c^2&=0 \end{align*}

This equation in $x$ should have a double root. Hence

\begin{align*} (2mc-4)^2-4m^2c^2&=0\\ 4m^2c^2-16mc+16-4m^2c^2&=0\\ mc-1&=0\\ m(k-mh)-1&=0\\ hm^2-km+1&=0 \end{align*}

The roots of the above equation in $m$ are the slopes of tangents to the parabola from $P$. If the two tangents from $P$ are perpendicular to each other, the two roots of this equation should have their product equal to $-1$. That is

$$\frac{1}{h}=-1$$

So the locus of $P$ is $x=-1$.

Answer 3

Using the standard parametrization of the parabola, let $P(p^2,2p)$ and $Q(q^2,2q)$ be points on the parabola. The tangent gradients are $\frac 1p$ and $\frac 1q$ respectively.

Therefore, due to perpendicularity, we have $$pq=-1$$

The equations of the tangents at $P$ and $Q$ are $$py=x+p^2$$ and $$qy=x+q^2$$

Solving simultaneously, we get $$x=pq=-1$$

So the locus is the directrix of the parabola, the line $x=-1$

Answer 4

The two points $P_1(p^2, 2ap)$ and $P_2(\frac 1{p^2}, -\frac 2p)$ lie on the parabola $y^2=4x$.

Differentiating gives the slopes of tangents at $P_1, P_2$ as $\frac 1p, -p$ respectively, i.e. perpendicular.

The equations of the tangents are:

$$\begin{align} \text{At $P_1$}:\qquad\qquad\;\; py&=x+p^2\tag{1}\\ \text{At $P_2$}:\qquad\qquad -\frac yp&=x+\frac 2{p^2}\\ py&=-p^2x-1 \tag{2}\\ (1)=(2):\hspace{4cm}\\ (1+p^2)x&=-(1+p^2)\\ \color{red}{x}&\color{red}{=-1} \end{align}$$ which is the equation of the locus of the point of intersection. This also happens to be the directrix. The fact that the locus of the point of intersection of two mutually perpendicular tangents is the directrix is a well-known property of the parabola.

Answer 5

Let the Parabola be y^2=4ax Thus it's tangent is of form
y=mx+a/m ( c=a/m for line y=mx +c) If x1 and y1 be coordinates of point of intersection of the mutually perpendicular tangents,then y1=mx1 + a/m Or m^2x1+my1+ a=0 M1 and M2 being slopes of the 2 tangents are roots of this equation. Thus M1M2=(-1)=a/x1 Or x1= -a Or X=-a That is the locus of the point of intersection of the two tangents and it represents the directrix of the Parabola. P.S.::Please reply and give a like if you like my solution.!!!