Question

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

• A. $b=\sqrt{3}$
• B. $x+2y+4=0$
• C. $b=3$
• D. $x-2y-4=0$.

Answer 1

The correct option is A $b=\sqrt{3}$

$y^2=4x............\left(1\right)(a=1)$

$Tangent$ of equation $\left(1\right)$ be $y=mx+\frac{1}{m}..................\left(2\right)$

Now, $\frac{x^2}{4}+\frac{y^2}{b^2}=1...............\left(3\right)$

$Tangent$ of equation $\left(3\right)$ be $y=mx+\sqrt{4m^2+b^2}.......................\left(4\right)$

As given, $x-2y+4=0$ is tangent to both the curves, $m=\frac{1}{2}$ and

$c=2$

$=>\sqrt{4(\frac{1}{2}{)}^2+b^2}=2$

$=>b^2+1=4$

$=>b^2=3$

$=>b=\pm \sqrt{3}.$