
The focal distance of a point on the parabola ${y^2} = 4x$ and above its axis, is 10 units. Its coordinates are
A). (9, 6)
B). (25, 10)
C). (25, -10)
D). None of these
Answer
591.3k+ views
Hint: Before attempting this question one must have prior knowledge of parabola, the equation which represents the parabola is ${y^2} = 4ax$, using this information will help you to approach towards the solution of the problem
Complete step-by-step solution -
According to the given information we have a parabola ${y^2} = 4x$ whose focal distance is 10
We know that the general equation of parabola is ${y^2} = 4ax$
Taking the given equation of parabola i.e. ${y^2} = 4x$ as equation 1
Comparing the general equation of parabola with the given equation of parabola we get
$4x = 4ax$
$ \Rightarrow $a = 1
So the coordinates of focus of parabola will be (1, 0)
We have focus distance = 10
So focal distance = 10 =$\sqrt {{{\left( {x - 1} \right)}^2} + {{\left( {y - 0} \right)}^2}} $
Squaring both sides we get
$ \Rightarrow $\[{\left( {10} \right)^2} = {\left( {\sqrt {{{\left( {x - 1} \right)}^2} + {{\left( {y - 0} \right)}^2}} } \right)^2}\]
$ \Rightarrow $\[100 = {\left( {x - 1} \right)^2} + {\left( {y - 0} \right)^2}\]
$ \Rightarrow $\[100 = {x^2} + 1 - 2x + {y^2}\]
Substituting the value of ${y^2}$equation 1
We get \[100 = {x^2} + 1 - 2x + 4x\]
$ \Rightarrow $\[99 = {x^2} + 2x\]
$ \Rightarrow $\[{x^2} + 2x - 99 = 0\]
By the method of splitting the middle term method
We get \[{x^2} + \left( {11 - 9} \right)x - 99 = 0\]
$ \Rightarrow $\[{x^2} + 11x - 9x - 99 = 0\]
$ \Rightarrow $\[x\left( {x + 11} \right) - 9\left( {x + 11} \right) = 0\]
So $x = -11, 9$
Since x can’t be negative
Now substituting the value of x in equation 1
For x = 9
${y^2} = 4\left( 9 \right)$
$ \Rightarrow $${y^2} = 36$
$ \Rightarrow $\[y = \sqrt {36} \]
So $y = 6, -6$
Therefore the coordinates are (9, 6)
Hence option A is the correct option.
Note: In the above solution we came across the terms parabola and focal distance which can be explained as a curve which consists of a set of all points that exist at equal distance from a fixed point (focus) this curve is named as a parabola. The distance from the vertex to focus which is measured along the symmetry of the axis is called focal distance.
Complete step-by-step solution -
According to the given information we have a parabola ${y^2} = 4x$ whose focal distance is 10
We know that the general equation of parabola is ${y^2} = 4ax$
Taking the given equation of parabola i.e. ${y^2} = 4x$ as equation 1
Comparing the general equation of parabola with the given equation of parabola we get
$4x = 4ax$
$ \Rightarrow $a = 1
So the coordinates of focus of parabola will be (1, 0)
We have focus distance = 10
So focal distance = 10 =$\sqrt {{{\left( {x - 1} \right)}^2} + {{\left( {y - 0} \right)}^2}} $
Squaring both sides we get
$ \Rightarrow $\[{\left( {10} \right)^2} = {\left( {\sqrt {{{\left( {x - 1} \right)}^2} + {{\left( {y - 0} \right)}^2}} } \right)^2}\]
$ \Rightarrow $\[100 = {\left( {x - 1} \right)^2} + {\left( {y - 0} \right)^2}\]
$ \Rightarrow $\[100 = {x^2} + 1 - 2x + {y^2}\]
Substituting the value of ${y^2}$equation 1
We get \[100 = {x^2} + 1 - 2x + 4x\]
$ \Rightarrow $\[99 = {x^2} + 2x\]
$ \Rightarrow $\[{x^2} + 2x - 99 = 0\]
By the method of splitting the middle term method
We get \[{x^2} + \left( {11 - 9} \right)x - 99 = 0\]
$ \Rightarrow $\[{x^2} + 11x - 9x - 99 = 0\]
$ \Rightarrow $\[x\left( {x + 11} \right) - 9\left( {x + 11} \right) = 0\]
So $x = -11, 9$
Since x can’t be negative
Now substituting the value of x in equation 1
For x = 9
${y^2} = 4\left( 9 \right)$
$ \Rightarrow $${y^2} = 36$
$ \Rightarrow $\[y = \sqrt {36} \]
So $y = 6, -6$
Therefore the coordinates are (9, 6)
Hence option A is the correct option.
Note: In the above solution we came across the terms parabola and focal distance which can be explained as a curve which consists of a set of all points that exist at equal distance from a fixed point (focus) this curve is named as a parabola. The distance from the vertex to focus which is measured along the symmetry of the axis is called focal distance.
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