Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# The Equations of the Tangents to the Ellipse 9x2 + 16y2 = 144 from the Point (2, 3) Are - Mathematics

MCQ
Sum

The equations of the tangents to the ellipse 9x2 + 16y2 = 144 from the point (2, 3) are

#### Options

•  y = 3, x = 5

•  x = 2, y = 3

• x = 3, y = 2

•  x + y = 5, y = 3

#### Solution

$x + y = 5 , y = 3$
$9 x^2 + 16 y^2 = 144$
$\Rightarrow \frac{x^2}{16} + \frac{y^2}{9} = 1$
Equation of the tangent in case of an ellipse is given by
$y = mx + \sqrt{a^2 m^2 + b^2}$
$\Rightarrow y = mx + \sqrt{16 m^2 + 9} . . . (1)$
$\text{ Substituting }x=2\text{ and }y=3,\text{ we get: }$
$3 = 2m \pm \sqrt{16 m^2 + 9}$
$\Rightarrow 3 - 2m = \sqrt{16 m^2 + 9}$
On squaring both sides, we get:
$\left( 3 - 2m \right)^2 = \left( 16 m^2 + 9 \right)$
$\Rightarrow 9 + 4 m^2 - 12m = \left( 16 m^2 + 9 \right)$
$\Rightarrow 12 m^2 + 12m = 0$
$\Rightarrow 12m\left( m + 1 \right) = 0$
$\Rightarrow m = 0, - 1$
Substituting values ofmin eq. (1), we get:
$\text{ For }m = 0, y = 3$
$\text{ For }m = - 1, y = - x + 5\text{ or }x + y = 5$

Concept: Introduction of Ellipse
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 26 Ellipse
Q 10 | Page 28