A tangent and a normal are drawn at the point P(2,-4) on the parabola y^2=8x, which meet t…

Question

A tangent and a normal are drawn at the point $P(2,-4)$ on the parabola $y^2=8x$, which meet the directrix of the parabola at the points $A$ and $B$ respectively. If $Q(a,b)$ is a point such that $AQBP$ is a square, then $2a+b$ is equal to

Accepted Answer

Given, parabola

${y^2} = 8x$ ...... (i)

Equation of tangent at $P(2, - 4)$ is

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

or, $x + y + 2 = 0$ ..... (ii)

and Equation of normal to the parabola is

$x - y + C = 0$

$$\therefore$$ Normal passes through $(2, - 4)$

$$\therefore$$ $C =  - 6$

Normal : $x - y = 6$ ..... (iii)

Equation of directrix of parabola

$x =  - 2$ ..... (iv)

Point of intersection of tangent and normal with directrix are $x =  - 2$ at $A( - 2,0)$ and $B( - 2, - 8)$ respectively.

$Q(a,b)$ and $P(2, - 4)$ are given and AQBP is a square.

Mid-point of AB = Mid-point of PQ

$$ \Rightarrow ( - 2, - 4) = \left( {{{a + 2} \over 2},{{b - 4} \over 2}} \right) \Rightarrow a =  - 6,b =  - 4$$

$$ \Rightarrow 2a + b =  - 16$$

Solutions · 1