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STD 11 - 10.2 parabola,ellipse,hyperbola — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 10.2 parabola,ellipse,hyperbola4 Marks
MCQ
The equation of the ellipse whose centre is at origin and which passes through the points $(-3, 1)$ and $(2, -2)$ is
  • A
    $5{x^2} + 3{y^2} = 32$
  • ✓
    $3{x^2} + 5{y^2} = 32$
  • C
    $5{x^2} - 3{y^2} = 32$
  • D
    $3{x^2} + 5{y^2} + 32 = 0$
✓

Answer

Correct option: B.
$3{x^2} + 5{y^2} = 32$
b
(b) $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$.

Since it passes through $(-3, 1)$ and $(2, -2)$,

so $\frac{9}{{{a^2}}} + \frac{1}{{{b^2}}} = 1$ and $\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{4}$

==>${a^2} = \frac{{32}}{3}$, ${b^2} = \frac{{32}}{5}$

Hence required equation of ellipse is $3{x^2} + 5{y^2} = 32$.

Trick : Since only equation $3{x^2} + 5{y^2} = 32$ passes through $(-3, 1)$ and $(2, -2)$. Hence the result.

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