Question

Find the equation of the tangent to the parabola $y^2=12x$, which passes through the point $(2,5).$ Find also the co-ordinates of their points of contact.

• A. $x-y-3=0,(3,6);3x-2y-4=0$, $\left(\frac{4}{3},4\right)$
• B. $x-y-3=0,(3,6);3x-2y-4=0$, $\left(\frac{2}{3},8\right)$
• C. $x-y+3=0,(3,6);3x-2y+4=0$, $\left(\frac{2}{3},8\right)$
• D. $x-y+3=0,(3,6);3x-2y+4=0$, $\left(\frac{4}{3},4\right)$

Answer 1

The correct option is D $x-y+3=0,(3,6);3x-2y+4=0$, $\left(\frac{4}{3},4\right)$

Given parabola is $y^2=12x$ ...... $\left(i\right)$

Equation of tangent is $y=mx+\frac{a}{m}$ where $4a=12$ i.e. $a=3$
Then, the equation of tangent becomes $y=mx+\frac{3}{m}$ ..... $\left(ii\right)$

It passes through the point $(2,5)$
$\Rightarrow 2m^2-5m+3=0$
$\Rightarrow (2m-3)(m-1)=0$
$\Rightarrow m=1,\frac{3}{2}$
On putting the value of $m$ in the equation $\left(i\right)$, we get

$y=x+3$ and $y=\frac{3}{2}x+\frac{3\times 2}{3}$

$y=x+3$ and $y=\frac{3}{2}x+2$

$x-y+3=0$ ..... $\left(iii\right)$ and $3x-2y+4=0$ ........ $\left(iv\right)$

Point of contact of $\left(i\right)$ and $\left(iii\right)$ is $(3,6)$

Point of contact of $\left(i\right)$ and $\left(iv\right)$ is $(\frac{4}{3},4)$

Hence, option D is correct.