Find the sum of minimum and maximum values of $y$ in $4 x^{2} + 12 x y + 10 y^{2} - 4 y + 3 = 0$ .

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Solution

$4 x^{2} + 12 x y + 10 y^{2} - 4 y + 3 = 0$
Let us differentiate w.r.t $^{'} x^{'}$
$\Rightarrow 8 x + 12 y + 20 y \frac{d y}{d x} + 12 x \frac{d y}{d x} - 4 \frac{d y}{d x} = 0$
$\Rightarrow 2 x + 3 y + 5 y \frac{d y}{d x} + 3 x \frac{d y}{d x} . \frac{d y}{d x} = 0$
$\Rightarrow (3 x + 5 y - 1) \frac{d y}{d x} = - (2 x + 3 y) \Rightarrow \frac{d y}{d x} = \frac{- (2 x - 3 y)}{(3 x + 5 y - 1)}$
To find critical points, $\frac{d y}{d x} = 0$
$\Rightarrow \frac{- (2 x - 3 y)}{(3 x + 5 y - 1)} = 02 x + 3 y = 0$
The line $2 x + 3 y = 0$ will pass through critical point.
$x = - \frac{3}{2} y$ . We will not substitute this value of $x$ in the curve & solve the equation in terms of $y$ .
$\Rightarrow 4 {(\frac{- 3 y}{2})}^{2} + 12 (\frac{- 3 y}{2}) y + 10 y^{2} - 4 y + 3 = 0 \Rightarrow 9 y^{2} - 18 y^{2} + 10 y^{2} - 4 y + 3 = 0 \Rightarrow y^{2} - 4 y + 3 = 0 \Rightarrow (y - 1) (y - 3) = 0$
Clearly, min value of $y = 1$
max value of $y = 3$
$∴$ Sum of min & max values $= 1 + 3 = 4$

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