Question

What's the equation of tangent to the parabola $y^2=4ax$ having a slope 'm'.

• A. $y=mx+a{m}^2$
• B. $y=mx-a{m}^2$
• C. $y=mx+\frac{a}{m}$
• D. $y=mx-\frac{a}{m}$

Answer 1

The correct option is C

$y=mx+\frac{a}{m}$

Given parabola is $y^2=4ax$ .........(1)

Let the tangent of the parabola be,

y=mx+c ..............(2)

for finding the tangent we solve the equations

of line and parabola and take the case when the number

of solution is equal to 1.

$(mx+c{)}^2=4ax$

$m^2{x}^2+2mcx+c^2=4ax$

$m^2{x}^2+x(2mc-4a)+c^2=0$

for having a single solution

$△=0$

$\Rightarrow (2mc-4a{)}^2-4m^2{c}^2=0$

$\Rightarrow 4m^2{c}^2+16a^2-16mca-4m^2{c}^2=0$

$\Rightarrow 16a^2=16mca$

$\Rightarrow a=mc$

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

Putting the value of c in (c) we get the equation of tangent as,

$y=mx+\frac{a}{m}$

From solving the line with the parabola we also get the point of contacts as $(\frac{a}{m^2},\frac{2a}{m})$