Consider the parabola with vertex [( 1 2,3 4 )] and the directrix [y = 1 2].

Question

Consider the parabola with vertex $$\left( {{1 \over 2},{3 \over 4}} \right)$$ and the directrix $$y = {1 \over 2}$$. Let P be the point where the parabola meets the line $$x =  - {1 \over 2}$$. If the normal to the parabola at P intersects the parabola again at the point Q, then (PQ)2 is equal to :

Accepted Answer

$$\left( {y - {3 \over 4}} \right) = {\left( {x - {1 \over 2}} \right)^2}$$ .... (1)

For $$x =  - {1 \over 2}$$

$$y - {3 \over 4} = 1 \Rightarrow y = {7 \over 4} \Rightarrow P\left( { - {1 \over 2},{7 \over 4}} \right)$$

Now, $$y' = 2\left( {x - {1 \over 2}} \right)$$

At $$x =  - {1 \over 2}$$

$$ \Rightarrow {m_T} =  - 2,{m_N} = {1 \over 2}$$

Equation of normal is

$$y - {7 \over 4} = {1 \over 2}\left( {x + {1 \over 2}} \right)$$

$$y = {x \over 2} + 2$$

Now put y in equation (1)

$${x \over 2} + 2 - {3 \over 4} = {\left( {x - {1 \over 2}} \right)^2}$$

$$ \Rightarrow x = 2$$ \& $$ - {1 \over 2}$$

$$\Rightarrow$$ Q(2, 3)

Now, $${(PQ)^2} = {{125} \over {16}}$$

Option (b)

Solutions · 1