Question

The line $y=m(x+a)+\frac{a}{m}$ touches the parabola $y^2=4a(x+a)$ when $m$

• A. is equal to zero
• B. is any positive real number
• C. is any negative number
• D. is any real number, $m\ne 0$

Answer 1

Answer: (D)

is any real number, $m\ne 0$

Solving line with the parabola
${\left(m\right(x+a)+\frac{a}{m})}^2=4a(x+a)$
$m^2(x^2+a^2+2ax)+\frac{a^2}{m^2}+2\times a(x+a)=4ax+4a^2$
$m^2{x}^2+x(2a{m}^2-2a)+m^2{a}^2+\frac{a^2}{m^2}-2a^2=0$
$D=0$ (Discriminant ) ........... [Since the line and parabola intersects]
$4(a{m}^2-a{)}^2-4(m^2\left)\right(m^2{a}^2+\frac{a^2}{m^2}-2)=0$
$(m^2-1{)}^2-(m\left(m-\frac{1}{m}\right){)}^2=0$
$(m^2-1{)}^2-(m^2-1{)}^2=0$
$\therefore$ For all value of $m$ accept $m\ne 0$.