Question

Length of the normal chord of the parabola $y^2=8x$ at the point where abscissa and ordinate are equal is:
A.13
B.8
C.$10\sqrt{5}$
D.4

Answer 1

Hint: Take the focal chord as PQ. Their coordinates are of the form $(a{t}^2,2at)$ . From the given equation of parabola and focal chord concept, we need to get the value of a, $t_1$ and $t_2$ . As abscissa of ordinate are equal, y=x puts that in the equation of parabola. Get value of P & Q, using distance formula. Find length or PQ.

Complete step-by-step answer:
We have been given the equation of parabola as $$y^2=8x$$ .
First let us draw the parabola $$y^2=8x$$

We know the general equation of a parabola as $y^2=4ax$ , Now let us compare both the general equation and the given equation of parabola.
From that we get, latus rectum $4a=8$
$\begin{array}{rl}& a={\displaystyle \frac{8}{4}}=2\\ & \text{i}\text{.e}\text{. }a=2\end{array}$
Let us take the two points on the parabola as $P\left(t_1\right)$ and $Q\left(t_2\right)$ . we know the relation $t_1{t}_2=-1$ .
Similarly we know that $t_2=-t_1-{\displaystyle \frac{2}{t_1}}$ .
It is said that the abscissa and ordinates are equal, the coordinates are equal so put y=x.
$y^2=8x$
So, $x^2=8x\Rightarrow x^2-8x=0$ .
$x(x-8)=0$ Hence $x=0$ or $x-8=0$
i.e. x can be either 0 or 8. So when x=8, y=8
Thus we got the coordinate of $(x,y)$ as $(8,8)$ .
We can take the coordinate of $P(a{t}_1^2,2a{t}_1)$ and $Q(a{t}_2^2,2a{t}_2)$ .
Thus we got a=2, So, $$P(2t_1^2,4t_1)$$ and $Q(2t_2^2,4a{t}_2)$ .
Now we get the coordinate of $(x,y)=(8,8)$ and coordinate $$(2t_1^2,4t_1)$$ .
Now let us equate the x and y coordinates of both.
Thus,
$\begin{array}{rl}& 4t_1=8\\ & t_1={\displaystyle \frac{8}{4}}=2\end{array}$
Thus we got $t_1=2$
We know that $t_2=-t_1-{\displaystyle \frac{2}{t_1}}=-2-{\displaystyle \frac{2}{2}}=-2-1=-3$
Thus $t_2=-3$ .
Thus let us put $t_1=2$ and $t_2=-3$ in the coordinate of P and Q
$$\begin{array}{rl}& P(2t_1^2,4t_1)=P(2\times 2^2,4\times 2)=P(8,8)\\ & Q(2t_2^2,4a{t}_2)=Q(2\times {(-3)}^2,4\times -3)Q(18,-12)\end{array}$$
Now, let us find the length of normal chord PQ using distance formula
 distance$\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
Put $(x_1,y_1)=(8,8)$ and $(x_2,y_2)=(18,12)$
Length of normal chord PQ $=\sqrt{{(18-8)}^2+{(-12-8)}^2}$
$$=\sqrt{{10}^2+{(-20)}^2}=\sqrt{100+400}=\sqrt{500}=\sqrt{100\times 5}=10\sqrt{5}$$ .
Hence we got the length of a normal chord as $10\sqrt{5}$ .
Therefore, option (C) is the correct answer.
Note: The normal at the point $$(2t_1^2,4t_1)$$ meets the parabola again in the point $(2t_2^2,4a{t}_2)$ , thus $t_2=-t_1-{\displaystyle \frac{2}{t_1}}$ . We may sometimes take $t_1{t}_2=-1$ , but that is not concept to use and it will give us wrong values of $t_1$ and $t_2$ .