Question

If $y=mx+c$ touches the parabola $y^2=4a(x+a)$, then $am+\frac{a}{m}$ is equal to

• A. $y$
• B. $x$
• C. $c$
• D. None of these.

Answer 1

The correct option is C $c$

Given: Prabola:$y^2=4a(x+a)$ $⟶1$

Tangent: $y=mx+c$ $⟶2$

Condition of tangency $c+m\times h=k+\frac{a}{m},$ where$h=-a,k=0,$

$\Rightarrow c+m(-a)=0+\frac{a}{m}$

$\Rightarrow c=m\times a+\frac{a}{m}$

So,$a\times m+\frac{a}{m}=c$.