Point on parabola satisfying a condition

Question

On parabola $p:y^2=4x$ find a point at which tangent and its normal at that point create a triangle of area $20$ with $x$-axis. Find equations of that tangent and its normal.

(Normal is a perpendicular line)

Here is a rough idea drawn in GeoGebra:

We can define tangent and a normal as $t:y=k_1x+n_1$ and $n:y=\frac{-1}{k_1}x+n_2$, respectively. We know that the condition of line being a parabola is $p=2k_1n_1$, where in our case $p=2$, hence $k_1n_1=1$. But after this I'm stuck. How to connect this information with the fact that $A_{\triangle ABC}=20$? Any hints would be appreciated.

Edit: if possible without the use of calculus

Answer 1

As requested by the author, let's solve it WITHOUT using calculus.

Use those labeled points as shown in OP. Let $A(x_0, y_0)$, the equation for tangent line $t$ is $y=kx+b$, combine the tangent line equation with the parabola $y^2=4x$, eliminate $x$, we get

$$y^2=\frac{4}ky-\frac{4b}k\Longleftrightarrow ky^2-4y+4b=0$$

We know $t$ is the tangent line to the parabola, which implies above quadratic equation has two identical roots, namely

$$\Delta=16-16kb=0\Longrightarrow b=\frac1k$$

Hence, tangent line $t$ is: $$ y=kx+\frac1k\tag{1}$$ Since point $A(x_0, y_0)$ is the tangent point, it is on both line $t$ and parabola, we get

$$y_0=kx_0+\frac1k,~~~y_0^2=4x_0$$

Solve them and we get the coordinate of $A$: $$x_0=\frac1{k^2},~~~y_0=\frac2k\tag{2}$$

Let the normal line $n$ be: $\displaystyle y=-\frac1kx+c$. Since the point $A$ is on the normal line $n$, plug in eq.(2) and we get

$$c=\frac2k+\frac1{k^3}\Longrightarrow y=-\frac1kx+\frac2k+\frac1{k^3}\tag{3}$$

From eq.(1), set $y=0$ and we get the coordinate of point $B(-\frac1{k^2},0)$.

From eq.(3), set $y=0$ and we get the coordinate of point $C(2+\frac1{k^2},0)$.

The base of the triangle is $$\text{base}=x_C-x_B=2+\frac2{k^2}$$ The height can be found from eq.(2) $$\text{height}=y_0=\frac2{k}$$

The area is

$$\frac12\cdot \left( 2+\frac2{k^2}\right)\cdot \frac2{k}=20\Longrightarrow k=\frac12$$

From eq.(1) and eq.(3) we get the equations for tangent line and normal line:

$$\boxed{y=\frac12x+2,~~~~~y=-2x+12}$$

Answer 2

HINT.-The derivative is $y'=\dfrac{4}{2\sqrt{4x}}=\dfrac 1x$ so the tangent at point $A=(x_0,y_0)$ is given by $y=\dfrac{x+x_0(\sqrt{x_0}-1)}{\sqrt{x_0}}$ and the normal is $y=\dfrac{-x_0x+x_0(\sqrt{x_0}+1)}{\sqrt{x_0}}$. It follows that $$B=(-x_0(\sqrt{x_0}-1),0);\space C=(\sqrt{x_0}+1,0);\space A=(x_0,2\sqrt{x_0})$$ Finally find the area which is function of $x_0$ and equating it to $20$ you have an equation in $x_0$ so you end the problem.