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10-2. Parabola, Ellipse, Hyperbola
medium
दीर्घवृत्त $3{x^2} + 4{y^2} - 12x - 8y + 4 = 0$ की नाभियों के निर्देशांक हैं
A
$(1, 2), (3, 4)$
B
$(1, 4), (3, 1)$
C
$(1, 1), (3, 1)$
D
$(2, 3), (5, 4)$
Solution
(c) $3{x^2} – 12x + 4{y^2} – 8y = – 4$Þ$3{(x – 2)^2} + 4{(y – 1)^2} = 12$
$\frac{{{{(x – 2)}^2}}}{4} + \frac{{{{(y – 1)}^2}}}{3} = 1$
$\frac{{{X^2}}}{4} + \frac{{{Y^2}}}{3} = 1$
$e = \sqrt {1 – \frac{3}{4}} = \frac{1}{2}$.
नाभियाँ $\left( {X = \pm 2 \times \frac{1}{2},\,Y = 0} \right)$ हैं
अर्थात् $(x – 2 = \pm 1,\,\,y – 1 = 0)$$ = (3,\,1)$ व $(1,\,1)$.
Standard 11
Mathematics
