Question

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

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

Answer 1

Answer: (A)

Equation of tangent of ellipse is
$y=mx\pm \sqrt{a^2{m}^2+b^2}$.........(i)
Given equation is $x-2y+4=0$.........(ii)
Since (i) \& (ii) are same, comparing them, we get
$m=\frac{1}{2}\&\sqrt{a^2{m}^2+b^2}=2$
$\Rightarrow 4\cdot \frac{1}{4}+b^2=4$
$\Rightarrow b=\pm \sqrt{3}$
Equation of tangent of parabola
$y=mx+\frac{1}{m}$.........(iii)
By (i) & (iii)
$\frac{1}{m^2}=a^2{m}^2+b^2$
on solving it we get $m=\pm \frac{1}{2}$
with $m=-\frac{1}{2}$ we get $x+2y+4=0$
which is other equation of common tangent.