Question

If a double ordinate of the parabola $y^2=4ax$ be of length $8a$, then the angle between the lines joining the vertex of the parabola to the ends of this double ordinate is

• A.
• B.
• C.
• D.

Answer 1

Answer: (C)

Given parabola is $y^2=4ax$

In parametric form,

Let us assume that the ends of double ordinate are $P(a{t}^2,2at$) and$P^{\prime }(a{t}^2,-2at$))

It is given that $P{P}^{\prime }=8a$

$\Rightarrow \sqrt{(a{t}^2-a{t}^2{)}^2+(2at+2at{)}^2}=8a$

$\Rightarrow 16a^2{t}^2=64a^2=0$

$\Rightarrow t^2=4$

$\Rightarrow t=2$

Hence $P(a{t}^2,2at$) $=P(4a,4a)$ and $P^{\prime }(a{t}^2,2at$) $=P^{\prime }(4a,-4a)$

Now, the vertex is $O(0,0)$ for the parabola.

slope of $OP=m_1=\frac{4a-0}{4a-0}=1$

slope of $O{P}^{\prime }=m_2=\frac{4a-0}{-4a-0}=-1$

Angle between the lines $OP$ and $O{P}^{\prime }$ is,

$\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}$

$\Rightarrow \tan \theta =\frac{1-(-1)}{1+1(-1)}$

$\Rightarrow \tan \theta =\frac{2}{0}$

$\Rightarrow \theta =\frac{\pi }{2}$

Hence, the line joining the origin to ends of the double ordinate will be perpendicular to each other.

It is clear from figure.

hence, Option C is correct.