The equation of the normal to the curve $2x^{2} + 3y^{2} - 5 = 0$ at the point $P(1, 1)$ is:

The area of the triangle formed by the coordinate axes and a tangent to the curve $xy = a^2$ at the point $(x_1, y_1)$ on it is

The sum of intercepts on coordinate axes made by the tangent to the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}$ is

If the normal to the curve $y = f(x)$ at the point $(4, 6)$ makes an angle $\frac{2\pi}{3}$ with the positive $x$-axis in the anticlockwise direction,then $f'(4)$ is:

The normal to the curve $y=f(x)$ at the point $(3,4)$ makes an angle $\frac{3 \pi}{4}$ with the positive $X$-axis. Then $f^{\prime}(3)$ is equal to:

If $\theta$ denotes the acute angle between the curves $y=10-x^2$ and $y=2+x^2$,at a point of the intersection,then $|\tan \theta|$ is equal to