Question

The point intersection of the two tangents at the ends of the latus rectum of 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 $(0,-3)$

Now let us find the point of intersection of two tangent at the ends of latureactum of

$y^2=8x$ then after that we will shift the point by $(2,3)$ to obtain result for $(y+3{)}^2=8(x-2)$

$L.R.$ end point $\to (2,4)$ $(2,-4)$

$T:4y=4(x+2)T:-4y=4(x+2)$

$y=x+2-y=x+2$

Intersection point is $(-2,0)$

Now shift by $(2,-3)\to (0,-3)$

$B$ is correct.