Question

A common tangent T to the curves
$C_1:\frac{x^2}{4}+\frac{y^2}{9} = 1$
and
$C_2:\frac{x^2}{4^2}\frac{-y^2}{143} = 1$
does not pass through the fourth quadrant. If T touches C 1 at (x 1, y 1) and C 2 at (x 2, y 2), then |2 x 1 + x 2| is equal to ______.

Answer 1

Equation of tangent to ellipse
$\frac{x^2}{4}+\frac{y^2}{9} = 1$
and given slope _m _is :
$y = mx + \sqrt{4m^2+9}...(i)$
For slope m equation of tangent to hyperbola is :
$y = mx+\sqrt{42m^2-143}...(ii)$
Tangents from (i) and (ii) are identical then
4 m 2 + 9 = 42 m 2 – 143
∴ m = ±2 (+2 is not applicable)
∴ m = -2
Hence
x1 = $\frac{8}{5}$
and
x2 = $\frac{84}{5}$
$∴ |2x_1+x_2| = |\frac{16}{5}+\frac{84}{5}|$
= 20