Question

Consider the curve$y^2=2x$ and point$A(2,2)$. If the normal at $A$intersects the curve again at point $B$ and the tangent at $A$ intersects the $x-$axis at $C$, then the area of $∆ABC$is

Answer 1

Step 1. Draw the diagram :

Step 2. equation of tangent and normal at $A$:

Equation of tangent at $A$

$y{y}_1=x+x_1$ where $(x_1,y_1)=(2,2)$

$\Rightarrow 2y=x+2$

Tangent at $A$ intersects the $x-$axis at $C(-2,0)$.

Equation of normal at $A$

$y+2x=6$

Solving normal equation with $y^2=2x$,

$$\begin{gathered} \Rightarrow y+y^2=6 \\ \Rightarrow y^2+y-6=0 \\ \Rightarrow (y+3)(y-2)=0 \\ \Rightarrow y=-3,2 \end{gathered}$$

We get $B=\left(\frac{9}{2},-3\right)$.

Step 3. Find the area of $∆ABC$ :

Area $=\frac{1}{2}\times AC\times AB$

$$\begin{gathered} =\frac{1}{2}\times 2\sqrt{5}\times \sqrt{\left(\frac{25}{4}+25\right)} \\ =\frac{1}{2}\times 2\sqrt{5}\times \frac{5\sqrt{5}}{2} \\ =\frac{25}{2} \end{gathered}$$

Hence the area of $∆ABC$is $\frac{25}{2}$ .