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]