Show that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{5y - 2x}}{{2y - 5x}}\).

Find the equation of the normal to the curve at the point \((6,{\text{ }}1)\).

Find the distance between the two points on the curve where each tangent is parallel to the line \(y = x\).

attempt at implicit differentiation     M1

\(2x - 5x\frac{{{\text{d}}y}}{{{\text{d}}x}} - 5y + 2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)     A1A1

Note:     A1 for differentiation of \({x^2} - 5xy\), A1 for differentiation of \({y^2}\) and \(7\).

\(2x - 5y + \frac{{{\text{d}}y}}{{{\text{d}}x}}(2y - 5x) = 0\)

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{5y - 2x}}{{2y - 5x}}\)     AG

[3 marks]

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{5 \times 1 - 2 \times 6}}{{2 \times 1 - 5 \times 6}} = \frac{1}{4}\)     A1

gradient of normal \( =  - 4\)     A1

equation of normal \(y =  - 4x + c\)     M1

substitution of \((6,{\text{ }}1)\)

\(y =  - 4x + 25\)     A1

Note:     Accept \(y - 1 =  - 4(x - 6)\)

[4 marks]

setting \(\frac{{5y - 2x}}{{2y - 5x}} = 1\)     M1

\(y =  - x\)     A1

substituting into original equation     M1

\({x^2} + 5{x^2} + {x^2} = 7\)     (A1)

\(7{x^2} = 7\)

\(x =  \pm 1\)     A1

points \((1,{\text{ }} - 1)\) and \(( - 1,{\text{ }}1)\)     (A1)

distance \( = \sqrt 8 \;\;\;\left( { = 2\sqrt 2 } \right)\)     (M1)A1

[8 marks]

Total [15 marks]