First slide

Ellipse

Question

Equation of a tangent to the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{25} = 1$ passing through the point where a directrix of the ellipse meets the $positive$ x-axis is

Moderate

A

$5 y + \sqrt{11} x = 36$
B

$6 y + \sqrt{11} x = 25$
C

$6 y + \sqrt{11} x = 36$
D

$5 y - \sqrt{11} x = 36$

Solution

Equation of a tangent to the ellipse is
$y = m x + \sqrt{36 m^{2} + 25} (1)$
Equation of the directrix is
$x = \frac{6 \times 6}{\sqrt{36 - 25}} = \frac{36}{\sqrt{11}}$
which meets the $+ve x$ axis at $(\frac{36}{\sqrt{11}}, 0)$
The tangent ( 1) passes through this point if
$\begin{aligned} {[m (\frac{36}{\sqrt{11}})]}^{2} = 36 m^{2} + 25 \\ \Rightarrow m^{2} = \frac{11}{36} \Rightarrow m = \frac{\sqrt{11}}{6} . \end{aligned}$
Since the tangent meets the directrix on positive
a-axis $m < 0$ and the required equation is
$\begin{array}{l} y = - \frac{\sqrt{11}}{6} x + \sqrt{36 \times \frac{11}{36} + 25} \\ 6 y + \sqrt{11} x = 36 \end{array}$

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