Standard +0.3 This is a straightforward implicit differentiation question requiring students to differentiate implicitly, substitute a point to find the gradient, then find the normal equation. While it involves multiple steps, each is routine and the question type is standard C4 fare with no conceptual challenges beyond applying the product rule and implicit differentiation formula.
6 The equation of a curve is \(x ^ { 2 } + 3 x y + 4 y ^ { 2 } = 58\). Find the equation of the normal at the point \(( 2,3 )\) on the curve, giving your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.
Solve for \(\frac{dy}{dx}\) & subst \((x, y) = (2, 3)\)
M1
or v.v. Subst now or at normal eqn stage; (\(M1\) dep on either/both B1 M1 earned)
\(\frac{dy}{dx} = -\frac{13}{30}\)
A1
Implied if grad normal \(= \frac{30}{13}\)
Grad normal \(= \frac{30}{13}\) follow-through
√B1
This f.t. mark awarded only if numerical
Find equ any line thro \((2,3)\) with any num grad
M1
\(30x - 13y - 21 = 0\) AEF
A1
No fractions in final answer
$\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ | B1 |
Using $d(uv) = u\,dv + v\,du$ for the $(3)xy$ term | M1 |
$\frac{d}{dx}(x^2 + 3xy + 4y^2) = 2x + 3x\frac{dy}{dx} + 3y + 8y\frac{dy}{dx}$ | A1 |
Solve for $\frac{dy}{dx}$ & subst $(x, y) = (2, 3)$ | M1 | or v.v. Subst now or at normal eqn stage; ($M1$ dep on either/both B1 M1 earned)
$\frac{dy}{dx} = -\frac{13}{30}$ | A1 | Implied if grad normal $= \frac{30}{13}$
Grad normal $= \frac{30}{13}$ follow-through | √B1 | This f.t. mark awarded only if numerical
Find equ any line thro $(2,3)$ with any num grad | M1 | | 8
$30x - 13y - 21 = 0$ AEF | A1 | No fractions in final answer
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6 The equation of a curve is $x ^ { 2 } + 3 x y + 4 y ^ { 2 } = 58$. Find the equation of the normal at the point $( 2,3 )$ on the curve, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\hfill \mbox{\textit{OCR C4 2007 Q6 [8]}}