Question

Let a focal chord of the parabola $y^2=16x$ cuts it at points $(f,g)$ and $(h,k)$. Then $f.h$ is:
A. -8
B. -16
C. 16
D. None of these

Answer 1

Hint: We here need to find $f.h$ , i.e. the product of the x-coordinates of the extremities of the focal chord of the given parabola. For this, we will first write the given coordinates in the parametric form and, i.e. $(a{t_1}^2,2a{t}_1)$ and $(a{t_2}^2,2a{t}_2)$ and hence write the product in the parametric form. Then we will use the property of $t_1{t}_2=-1$ where $t_1$ and $t_2$ are the parameters for the extremities of the focal chord. We will then put this value in the required product and hence we will get our result.

Complete step-by-step solution
Now, we have been given the two points the focal chord of the parabola $y^2=16x$ passes through. A focal chord of a parabola is the chord, which passes through its focus.

The parabola $y^2=16x$ is in the form of $y^2=4ax$ where the focus is $(a,0)$ . thus, we can write the given parabola as:
$\begin{array}{rl}& y^2=16x\\ & \Rightarrow y^2=4.4x\end{array}$
Thus, for the given parabola, $a=4$
Now, any point on a parabola $y^2=4ax$ is given by $(a{t}^2,2at)$ where ‘t’ is a parameter.
As mentioned above, here $a=4$
Thus, we can take the given points $(f,g)$ and $(h,k)$ as $$(4{t_1}^2,8t_1)$$ and $(4{t_2}^2,8t_2)$ respectively where ${}^{\prime }{t_1}^{\prime }$ and ${}^{\prime }{t_2}^{\prime }$ are two different parameters.
Thus, we get:
$f=4{t_1}^2$
$h=4{t_2}^2$
So we can get $f.h$ as by multiplying these both.
Thus, we get:
$\begin{array}{rl}& f.h=4{t_1}^2.4{t_2}^2\\ & \Rightarrow f.h=16{t_1}^2{t_2}^2\end{array}$
$\Rightarrow f.h=16{\left(t_1{t}_2\right)}^2$ ……(i)
Now, we know that if the focal chord of a parabola of the type $y^2=4ax$ passes through the points $(a{t_1}^2,2a{t}_1)$ and $(a{t_2}^2,2a{t}_2)$ then the relation between the points is given by:
$t_1{t}_2=-1$
Thus, by putting this value, we can find the value of $f.h$
Now, putting this value in equation (i) we get:
$\begin{array}{rl}& f.h=16{\left(t_1{t}_2\right)}^2\\ & \Rightarrow f.h=16{(-1)}^2\\ & \Rightarrow f.h=16\left(1\right)\\ & \Rightarrow f.h=16\end{array}$
Thus option (C) is the correct option.

Note: We have obtained the result $t_1{t}_2=-1$ by the following method:
In the given figure, we have a parabola $y^2=4ax$ with the focus at S. The extremities of the chord are P and Q.

Let the point P be $(a{t_1}^2,2a{t}_1)$ and let the point Q be $(a{t_2}^2,2a{t}_2)$.
Hence, the equation through point P and Q will be:
$\begin{array}{rl}& y-2a{t}_1={\displaystyle \frac{2a{t}_2-2a{t}_1}{a{t_2}^2-a{t_1}^2}}(x-a{t_1}^2)\\ & y-2a{t}_1={\displaystyle \frac{2}{t_2+t_1}}(x-a{t_1}^2)\end{array}$
Now, this line passes through the focus, i.e. S (a,0)
Putting this point in the equation of the chord we get:
$$\begin{array}{rl}& y-2a{t}_1={\displaystyle \frac{2}{t_2+t_1}}(x-a{t_1}^2)\\ & \Rightarrow 0-2a{t}_1={\displaystyle \frac{2}{t_2+t_1}}(a-a{t_1}^2)\\ & \Rightarrow -2a{t}_1={\displaystyle \frac{2a}{t_2+t_1}}(1-{t_1}^2)\\ & \Rightarrow -t_1={\displaystyle \frac{1-{t_1}^2}{t_2+t_1}}\\ & \Rightarrow -t_1{t}_2-{t_1}^2=1-{t_1}^2\\ & \Rightarrow -t_1{t}_2=1-{t_1}^2+{t_1}^2\\ & \Rightarrow -t_1{t}_2=1\\ & \Rightarrow t_1{t}_2=-1\end{array}$$