Question

Let the normals at all the points on a given curve pass through a fixed point $\left(a,b\right)$. If the curve passes through $\left(3,-3\right)$ and $\left(4,-2\sqrt{2}\right)$, and given that $a-2\sqrt{2}b=3$, then $\left(a^2+b^2+ab\right)$ is equal to?

Answer 1

Determine the value of $\left(a^2+b^2+ab\right)$

Given that all normals passes through a fixed point $\left(a,b\right)$ on the given curve.

So the curve will be a circle with the fixed point $\left(a,b\right)$ as center.

The given points are $\left(3,-3\right)$ and $\left(4,-2\sqrt{2}\right)$ on the curve.

Let us use distance formula to find $\left(a,b\right)$.

${\left(a-3\right)}^2+{\left(b+3\right)}^2={\left(a-4\right)}^2+{\left(b+2\sqrt{2}\right)}^2$ (Radius )

Use the algebraic identity and expand the above equation.

$$\begin{gathered} \Rightarrow a^2-6a+9+b^2+6b+9=a^2-8a+16+b^2+4\sqrt{2}b+8 \\ \Rightarrow 8a-6a-4\sqrt{2}b+6b=16+8-18 \\ \Rightarrow 2a-4\sqrt{2}b+6b=6 \\ \Rightarrow a-2\sqrt{2}b+3b=3 \end{gathered}$$

We are given $a-2\sqrt{2}b=3$,

Substitute $a-2\sqrt{2}$ as $3$ in the obtained equation.

$$\begin{gathered} \Rightarrow 3+3b=3 \\ \Rightarrow 3b=0 \\ \Rightarrow b=0 \end{gathered}$$

Now substitute $b$ as $0$ in the equation of $a-2\sqrt{2}b=3$.

$$\begin{gathered} \Rightarrow a-2\sqrt{2}\left(0\right)=3 \\ \Rightarrow a-0=3 \\ \Rightarrow a=3 \end{gathered}$$

substitute $b$ as $0$ and $a$ as $3$ in $\left(a^2+b^2+ab\right)$.

$\begin{array}{rcl}\left(a^2+b^2+ab\right)& =& 3^2+0^2+3\times 0\\ & =& 9+0+0\\ & =& 9\end{array}$

Therefore, the value of $\left(a^2+b^2+ab\right)$ is equal to $9$.