I- x 2-16-0II- y 2-9 y -20-0A- xy

The correct option is C x ≤ yEquation I x^2 16=0 ⇒ x^2=16 ⇒ x=√(14) ⇒ x =± 4 Equation II y^2 9y+20=0 ⇒ y^2 4y 5y+20=0 ⇒ y(y 4) 5(y 4)=0 ⇒ y=5,4 ⇒ y ≥ x ...

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Byju's Answer

Standard XII

Mathematics

Double Ordinate of Ellipse

I. x 2-16=0II...

Question

$I . x^{2} - 16 = 0$
$I I . y^{2} - 9 y + 20 = 0$

x > y

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\geq

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x < y

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\leq

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x = y or the relationship cannot be established

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Solution

The correct option is D x $\leq$ y
Equation I
$x^{2} - 16 = 0$
$\Rightarrow x^{2} = 16$
$\Rightarrow x = \sqrt{14}$
$\Rightarrow x = \pm 4$
Equation II
$y^{2} - 9 y + 20 = 0$
$\Rightarrow y^{2} - 4 y - 5 y + 20 = 0$
$\Rightarrow y (y - 4) - 5 (y - 4) = 0$
$\Rightarrow y = 5, 4$
$\Rightarrow y \geq x$

Suggest Corrections

I. x2−16=0 II. y2−9y+20=0

The correct option is D x ≤ yEquation I x2−16=0 ⇒x2=16 ⇒x=√14 ⇒x=±4 Equation II y2−9y+20=0 ⇒y2−4y−5y+20=0 ⇒y(y−4)−5(y−4)=0 ⇒y=5,4 ⇒y≥x

$I . x^{2} - 16 = 0$
$I I . y^{2} - 9 y + 20 = 0$