First slide

Ellipse

Question

$x - 2 y + 4 = 0$ is a common tangent to $y^{2} = 4 x & \frac{x^{2}}{4} + \frac{y^{2}}{b^{2}} = 1$ Then the value of b and the other common tangent are given by:

Easy

A

$b = \sqrt{3}$
B

x+2y + 4 = 0
C

b=3
D

x- 2y -4 =0

Solution

$x - 2 y + 4 = 0$ is a tangent to $\frac{x^{2}}{4} + \frac{y^{2}}{b^{2}} = 1$
$\begin{array}{l} \Rightarrow a^{2} l^{2} + b^{2} m^{2} = n^{2} \\ \Rightarrow 4 + b^{2} (4) = 16 \Rightarrow b^{2} = 3 \end{array}$
The other tangent must be reflection of x-2y+4 = 0 in the x-axis since both curves are symmetrical about x-axis.
$x + 2 y + 4 = 0$ is the other common tangent.

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