First slide

Ellipse

Question

$The locus of the point which divides the double ordinates of the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 in the ratio 1 : 2 internally is$

Moderate

A

$\frac{x^{2}}{a^{2}} + \frac{9 y^{2}}{b^{2}} = 1$
B

$\frac{x^{2}}{a^{2}} + \frac{9 y^{2}}{b^{2}} = \frac{1}{9}$
C

$\frac{9 x^{2}}{a^{2}} + \frac{9 y^{2}}{b^{2}} = 1$
D

$None of these$

Solution

$\begin{array}{l} Let P \equiv (a \cos θ, b \sin θ), Q \equiv (a \cos θ, - b \sin θ) . \\ P R : R Q = 1 : 2 \end{array}$
$\begin{array}{l} ∴ h = a \cos θ \\ or \cos θ = \frac{h}{a} - - - - (1) \\ and k = \frac{b}{3} \sin θ \\ or \sin θ = \frac{3 k}{b} - - - - (2) \end{array}$
On squaring and adding (i) and (ii), we get
$\frac{x^{2}}{a^{2}} + \frac{9 y^{2}}{b^{2}} = 1$

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