Find the equation of the normal to the curve at the point $\left( {1,{\text{ }}\sqrt 3 } \right)$.

Find the volume of the solid formed when the region bounded by the curve, the $x$-axis for $x \geqslant 0$ and the $y$-axis for $y \geqslant 0$ is rotated through $2\pi$ about the $x$-axis.

METHOD 1

$4{x^2} + {y^2} = 7$

$8x + 2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$     (M1)(A1)

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{{4x}}{y}$

Note:     Award M1A1 for finding $\frac{{{\text{d}}y}}{{{\text{d}}x}} = - 2.309 \ldots$ using any alternative method.

hence gradient of normal $= \frac{y}{{4x}}$     (M1)

hence gradient of normal at $\left( {1,{\text{ }}\sqrt 3 } \right)$ is $\frac{{\sqrt 3 }}{4}\,\,\,( = 0.433)$     (A1)

hence equation of normal is $y - \sqrt 3 = \frac{{\sqrt 3 }}{4}(x - 1)$     (M1)A1

$\left( {y = \frac{{\sqrt 3 }}{4}x + \frac{{3\sqrt 3 }}{4}} \right)\,\,\,(y = 0.433x + 1.30)$

METHOD 2

$4{x^2} + {y^2} = 7$

$y = \sqrt {7 - 4{x^2}}$     (M1)

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{{4x}}{{\sqrt {7 - 4{x^2}} }}$     (A1)

Note:     Award M1A1 for finding $\frac{{{\text{d}}y}}{{{\text{d}}x}} = - 2.309 \ldots$ using any alternative method.

hence gradient of normal $= \frac{{\sqrt {7 - 4{x^2}} }}{{4x}}$     (M1)

hence gradient of normal at $\left( {1,{\text{ }}\sqrt 3 } \right)$ is $\frac{{\sqrt 3 }}{4}\,\,\,( = 0.433)$     (A1)

hence equation of normal is $y - \sqrt 3 = \frac{{\sqrt 3 }}{4}(x - 1)$     (M1)A1

$\left( {y = \frac{{\sqrt 3 }}{4}x + \frac{{3\sqrt 3 }}{4}} \right)\,\,\,(y = 0.433x + 1.30)$

[6 marks]

Use of $V = \pi \int\limits_0^{\frac{{\sqrt 7 }}{2}} {{y^2}{\text{d}}x}$

$V = \pi \int\limits_0^{\frac{{\sqrt 7 }}{2}} {\left( {7 - 4{x^2}} \right){\text{d}}x}$    (M1)(A1)

Note:     Condone absence of limits or incorrect limits for M mark.

Do not condone absence of or multiples of $\pi$.

$= 19.4\,\,\,\left( { = \frac{{7\sqrt 7 \pi }}{3}} \right)$     A1

[3 marks]