Question

The point of intersection of two tangents at the ends of the latus rectum to the parabola $(y+3{)}^2=8(x-2)$ is

• A. $(0,-4)$
• B. $(0,-3)$
• C. $(-1,-3)$
• D. $(-2,-3)$

Answer 1

The correct option is B $(0,-3)$

The tangents at the end points of the latureactum will meet at the point of intersection of axis and directrix
For given parabola $(y+3{)}^2=8(x-2)...(A)$

Let, y+3= Y & x-2=X....(1)

$\Rightarrow Y^2=4\left(2\right)X....\left(2\right)$

$\Rightarrow$Equation of laturectum is $X=2$

So , from (1) we get, $x=4$

And from (A), we get

$(y+3{)}^2=8(4-2)$

$\Rightarrow y+3=\pm 4$

$\Rightarrow y=1ory=-7$

Diff. Equ (A) w.r.t $x$, we get,

$\frac{d}{dx}(y+3{)}^2=\frac{d}{dx}[8(x-2)]$

$\Rightarrow \frac{dy}{dx}=m=\frac{4}{y+3}$

At $y=1,m=1$ & At $y=-7,m=-1$

Equation of tangent $=y-1=1(x-4)$

$\Rightarrow y=x-3$ & $y+7=-1(x-4)$

So, point of intersection

$=x-3=-x-3\Rightarrow x=0$

And $y=-3$

$\therefore$ Tangents meets at $(0,-3)$