Question

For the curve
\(\begin{array}{l}C : \left(x^2 + y^2 – 3\right) + \left(x^2 – y^2 – 1\right)^5 = 0, \end{array}\)

\(\begin{array}{l}\text{the value of}\ 3y’ – y^3y”, \text{at the point}\end{array}\)

\((α, α), α > 0\), on C is equal to ________.

Answer 1

Correct Answer: 16

\(\begin{array}{l}\because C : \left(x^2 + y^2 – 3\right) + \left(x^2 – y^2 – 1\right)^5 = 0\end{array}\)for point \((α, α)\).
\(\begin{array}{l}\alpha ^2 + \alpha ^2 – 3 + (\alpha^ 2 – \alpha ^2 – 1)^5 = 0\end{array}\)
\(\begin{array}{l} \therefore\ \alpha=\sqrt{2}\end{array}\)
\(\begin{array}{l}\text{On differentiating} \left(x^2 + y^2 – 3\right) + \left(x^2 – y^2 – 1\right)^5 = 0 ~\text{we get}\\ x + yy’ + 5 \left(x^2 – y^2 – 1\right)^4 \left(x – yy’\right) = 0 …\left(i\right)\end{array}\)
When\( \begin{array}{l} x=y=\sqrt{2}\end{array}\)then \(\begin{array}{l} y’=\frac{3}{2} \end{array}\)
Again on differentiating eq. (i) we get :
\(\begin{array}{l}1 + \left(y’\right)^2 + yy” + 20 \left(x^2 – y^2 – 1\right) \left(2x – 2 yy’\right) \end{array}\)
\(\begin{array}{l}\left(x – y’y\right) + 5\left(x^2 – y^2 – 1\right)^4 \left(1 – y’^2 – yy”\right) = 0\end{array}\)
For\( \begin{array}{l} x=y=\sqrt{2} \end{array}\)and \(\begin{array}{l} y’=\frac{3}{2}\end{array}\)
we get \(\begin{array}{l} y”=-\frac{23}{4\sqrt{2}}\end{array}\)
\(\begin{array}{l}\therefore 3y’ – y^3y” =3\cdot\frac{3}{2}-\left(\sqrt{2}\right)^3\cdot\left(-\frac{23}{4\sqrt{2}}\right)\\= 16\end{array}\)