If the points of intersection of the ellipse x 2 16 + y 2 b 2 = 1 and the circle x 2 + y 2 = 4 b , b > 4 lie on the curve y 2 = 3 x 2 , then b is equal to :

y 2 = 3 x 2 and x 2 + y 2 = 4 b Solving both the equations we get, x 2 = b As x 2 16 + y 2 b 2 = 1 ⇒ x 2 16 + 3 x 2 b 2 = 1 ⇒ b 16 + 3 b = 1 ⇒ b 2 - 16 b + 48 = 0 ⇒ b - 12 b - 4 = 0 ⇒ b = 12 ∵ b > 4

Mathematics - Ellipse Question with Solution | TestHub

MathematicsEllipseGeneralEasy2 minPYQ_2021

MathematicsEasystatement

If the points of intersection of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$ and the circle $x^{2} + y^{2} = 4 b, b > 4$ lie on the curve $y^{2} = 3 x^{2}$ , then $b$ is equal to :

Options:

Answer:

Solution:

$y^{2} = 3 x^{2}$ and $x^{2} + y^{2} = 4 b$

Solving both the equations we get,

$x^{2} = b$

As $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$

$\Rightarrow \frac{x^{2}}{16} + \frac{3 x^{2}}{b^{2}} = 1$

$\Rightarrow \frac{b}{16} + \frac{3}{b} = 1$

$\Rightarrow b^{2} - 16 b + 48 = 0$

$\Rightarrow (b - 12) (b - 4) = 0$

$\Rightarrow b = 12 \{∵ b > 4\}$

Stream:JEESubject:MathematicsTopic:EllipseSubtopic:General

⏱ 2mℹ️ Source: PYQ_2021

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