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The focal distance of a point on the parabola ${y^2} = 16x$ whose ordinate is twice as abscissa,
a. 6
b. 10
c. 8
d. 12
Hint: Compare the given equation with the standard equation of parabola. Take a point P on parabola, and find out the coordinates of it. Now, by using distance formulas, find the focal length.
Complete step by step answer:
As, we know that the standard equation of parabola is $(y - {y_0})^2 = 4a(x - {x_0})$.
In which,
$\Rightarrow$ Vertex = $\left( {x{}_0,{y_0}} \right)$ and,
$\Rightarrow$ Focus = $\left( {{x_0} + a,{y_0}} \right)$
Given Equation of parabola is ${y^2} = 16x$
$\Rightarrow {y^2} = 16x$ - (Eq 1)
Comparing equation 1with standard equation of parabola we get,
$\Rightarrow$ ${x_0} = 0,{y_0} = 0$ and $a = 4$
So, focus of the equation 1 will be,
$\Rightarrow$ focus = $\left( {4,0} \right)$
Let there be a point P on parabola, whose abscissa be t,
Then the ordinate of the point P will be 2t (According to question)
$\Rightarrow$ P = $\left( {{\text{t,2t}}} \right)$
According to the question point P lies on the parabola given (equation 1)
So, point P must satisfy equation 1
Putting value of P in equation 1 we get,
$\Rightarrow {(2t)^2} = 16(t)$
$\Rightarrow 4{t^2} - 16t = 0$
$\Rightarrow 4t(t - 4) = 0$
So, the value of t will be 4.
Hence point P = (4,8)
Now, as we know focal distance is the distance between focus and a point on a curve, focal distance will be the distance between focus and P.
Now, calculating distance between them as,
focal distance = $\sqrt {\left( {{{\left( {4 - 4} \right)}^2} + {{\left( {8 - 0} \right)}^2}} \right)} {\text{ }} = 8$
focal distance = 8
Hence the correct option for the question will be (c).
Note: - Point on a conic always satisfies the equation of conic. Abscissa is the x - coordinate of a point and ordinate is the y - coordinate of a point.
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