Question

What is the condition for a line $y=mx+c$ to be a tangent to the parabola$(y-k{)}^2=4a(x-h)$.

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

In all such questions related to a line and a parabola having certain number of solutions, we first solve their equations. First on solving we get a quadratic equation and we can put the required condition for determinant.

Here number of solution = 1

$\therefore △=0$

On solving the equations

$(y-k{)}^2=4a(x-h)andy=mx+c$

$y^2+k^2-2yk=4a[\frac{y-c}{m}-h]$

$m{y}^2+m{k}^2-2myk=4a[y-c-mh]$

$y^2\left[m\right]+y[-2mk-4a]+m{k}^2+4ac+4mah=0$

$△=0$

$\Rightarrow (2mk+4a{)}^2-4m(m{k}^2+4ac+4mah)=0$

$4m^2{k}^2+16a^2+16mak-4m^2{k}^2-16mac-16m^{2}ah=0$

$a+mk-mc-m^{2}h=0$

$mc+m^{2}h=a+mk$

i.e., $mh+c=\frac{a}{m}+k$ is the required condition