$9 x^{2} + 2 h x y + 4 y^{2} + 6 x + 2 f y - 3 = 0$ represents two parallel lines if

h = 6, f = 2

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h = - 6, f = - 2

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h = - 6, f = 2

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h = 6, f = - 2

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Solution

The correct options are
A $h = 6, f = 2$
B $h = - 6, f = - 2$
Since the given equation represents a pair of parallel lines, we have
$h^{2} = 9 \times 4 \Rightarrow h = \pm 6$
and $∣ ∣ ∣ ∣ \begin{matrix} 9 & h & 3 h & 4 & f 3 & f & - 3 \end{matrix} ∣ ∣ ∣ ∣ = 0$
$\Rightarrow 9 (- 12 - f^{2}) - h (- 3 h - 3 f) + 3 (h f - 12) = 0$
$\Rightarrow 3 h^{2} + 6 h f - 9 f^{2} - 144 = 0$
$\Rightarrow 108 \pm 36 f - 9 f^{2} - 144 = 0 (∵ h = \pm 6)$
$\Rightarrow 9 f^{2} \mp 36 f + 36 = 0 (i f h = \pm 6)$
$\Rightarrow f = 2 i f (h = 6)$
and $\Rightarrow f = - 2 i f (h = - 6)$

Suggest Corrections

Similar questions

9 x^{2} + 2 h x y + 4 y^{2} + 6 x + 2 f y + c = 0

(h, f > 0)

{lim}_{h \to 0} \frac{f (2 h + 2 + h^{2}) - f (2)}{f (h - h^{2} + 1) - f (1)}

f^{'} (2) = 6

f^{'} (1) = 4

lim h \to 0 \frac{f (2 h + 2 + h^{2}) - f (2)}{f (h - h^{2} + 1) - f (1)}

f^{'} (2) = 6

f^{'} (1) = 4

f (x)

f^{'} (1) = 4, f^{'} (2) = 6, f^{'} (c)

f

c

lim h \to 0 \frac{f (2 + 2 h + h^{2}) - f (2)}{f (1 + h - h^{2}) - f (1)}

lim h \to 0 \frac{f (2 h + 2 + h^{2}) - f (2)}{f (h - h^{2} + 1) - f (1)}

f^{'} (2) = 6

f^{'} (1) = 4

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