Question

Find the equation of the tangents to the parabola $y^2=16x$, which are parallel and perpendicular respectively to the line $2x-y+5=0$. Find also the co-ordinates of their points of contact.

• A. $2x-y+2=0;x+2y+16=0,(16,-16)$
• B. $2x+y-2=0,;x=2y+16=0,(-16,16)$
• C. $2x+y+2=0,;x=2y+16=0,(-16,16)$
• D. $2x+y-2=0;x=2y+16=0,(16,-16)$

Answer 1

The correct option is A $2x-y+2=0;x+2y+16=0,(16,-16)$

Equation of the parabola is $y^2=16x$
The tangent is parallel to $2x-y+5=0$
Equation of the tangent can be taken as $y=2x+c$
Condition for tangency is $c$ $=\frac{a}{m}=\frac{4}{2}=2$
Equation of the tangent can be taken as $y=2x+2$
$\Rightarrow 2x-y+2=0$
Point of contact is $(\frac{a}{m^2},\frac{2a}{m})=(\frac{4}{4},\frac{8}{2})=(1,4)$
Slope of the perpendicular tangent is $m^{\prime }$$=-\frac{1}{m}=\frac{-1}{2}$
Equation of the perpendicular tangent is $y=m^{\prime }x+c^{\prime }$$=(-\frac{1}{2})x+c^{\prime }$
where $c^{\prime }$$=\frac{a}{m^{\prime }}=\left(\frac{4}{(-\frac{1}{2})}\right)=-8$
Equation of the perpendicular tangent is $y$$=-\frac{1}{2}x-8\Rightarrow 2y=-x+16\Rightarrow x+2y+16=0$
Point of contact is $(\frac{4}{\left(\frac{1}{4}\right)},\frac{8}{(-\frac{1}{2})})=(16,-16)$