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]