Solveeit Logo
Login

Question

Question: The extremities of latus rectum of a parabola \(\left( {1,1} \right)\) and \(\left( {1, - 1} \right)...

The extremities of latus rectum of a parabola (1,1)\left( {1,1} \right) and (1,−1)\left( {1, - 1} \right), then the equation of the parabola can be
A.y2=2x−1{y^2} = 2x - 1
B.y2=1−2x{y^2} = 1 - 2x
C.y2=2x−3{y^2} = 2x - 3
D.y2=2x−4{y^2} = 2x - 4

Explanation

Solution

Hint: We will first find the distance of latus rectum and equate it to 4a4a and find the value of aa. We will then find the coordinates of focus using the given coordinates of the latus rectum. We will then find the possible coordinates of the vertex. We will substitute the values in the general equation of parabola, a(y−h)2+k=0a{\left( {y - h} \right)^2} + k = 0, where (h,k)\left( {h,k} \right) is the vertex of the parabola.

Complete step-by-step answer:
We are given the extremities of latus rectum of a parabola (1,1)\left( {1,1} \right) and (1,−1)\left( {1, - 1} \right)
Length of the latus rectum can be calculated using the distance formula between two pints, which is, (x1−x2)2+(y1−y2)2\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}}
Hence, the length of the latus rectum is (1−1)2+(1−(−1))2=0+22=2 units\sqrt {{{\left( {1 - 1} \right)}^2} + {{\left( {1 - \left( { - 1} \right)} \right)}^2}} = \sqrt {0 + {2^2}} = 2{\text{ units}}
We also that the length of the latus rectum is given by 4a4a
Hence,
4a=2 a=12  4a = 2 \\\ a = \dfrac{1}{2} \\\
Since, latus rectum is the line passing through the focus and touching the parabola.
From the given coordinates of the latus rectum we can observe that the xx coordinate is fixed.
Therefore, the parabola is symmetric to the xxaxis.
And also, the length of the latus rectum is 2 units, which gives us that it is 1 unit left from the focus and 1 unit right from the focus.
Therefore, the focus of the parabola is (1,0)\left( {1,0} \right)
The vertex of the parabola will be (1−12,0)\left( {1 - \dfrac{1}{2},0} \right) or (1+12,0)\left( {1 + \dfrac{1}{2},0} \right)
Then, the vertex of the parabola can be (12,0)\left( {\dfrac{1}{2},0} \right) or (32,0)\left( {\dfrac{3}{2},0} \right).
The general equation of the parabola symmetric to xx axis is a(y−h)2+k=0a{\left( {y - h} \right)^2} + k = 0, where (h,k)\left( {h,k} \right) is the coordinate of vertex of the parabola.

We can write the equation of parabola as, y2=2(x−12){y^2} = 2\left( {x - \dfrac{1}{2}} \right) or y2=−2(x−32){y^2} = - 2\left( {x - \dfrac{3}{2}} \right),
Which can be rewritten as, y2=2x−1{y^2} = 2x - 1 or y2=−2x+3{y^2} = - 2x + 3

Note: The distance between two points can be calculated using the formula, (x1−x2)2+(y1−y2)2\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} . The length of the latus rectum of a parabola is given by 4a4a. The general equation of the parabola symmetric to xx axis is a(y−h)2+k=0a{\left( {y - h} \right)^2} + k = 0, where (h,k)\left( {h,k} \right) is the coordinate of vertex of the parabola.