Question

If $y=x\sqrt{2}-8\sqrt{2}$ is a normal chord to $y^2=8x$.Then its length (in units) is

• A. $16\sqrt{3}$
• B. $4\sqrt{3}$
• C. $12\sqrt{3}$
• D. $2\sqrt{3}$

Answer 1

The correct option is C $12\sqrt{3}$

For paarabola $y^2=4ax$ equation of normal at $P(a{t}_1^2,2a{t}_1)$ is $y=-t_{1}x+2a{t}_1+a{t}_1^3\cdots \left(1\right)$
here in $y^2=8x,a=2$
now equation $\left(1\right)$ will be $y=-t_{1}x+4t_1+2t_1^3\cdots \left(2\right)$
But given normal equation is $y=x\sqrt{2}-8\sqrt{2}\cdots \left(3\right)$
By comparing $\left(2\right)\&\left(3\right)$
$t_1=-\sqrt{2}$
So the point will be $P(4,-4\sqrt{2})$

as given that normal at $P(a{t}_1^2,2a{t}_1)$ passes through $Q(a{t}_2^2,2a{t}_2)$ so,
$t_2=-t_1-\frac{2}{t_1}\cdots \left(4\right)$

Now by equation $\left(4\right)$, $t_2=2\sqrt{2}$
So point $Q\equiv (16,8\sqrt{2})$
By distance formula we get, $PQ=12\sqrt{3}$ units