Question

The ends of the latus rectum of the conic ${x^2} + 10x - 16y + 25 = 0$ are
A. \[\left( {3,{}-{}4} \right),{}\left( {13,{}4} \right)\]
B. \[\left( { - 3,{} - 4} \right),{}\left( {13,{} - 4} \right)\]
C. \[\left( {3,{}4} \right),{}\left( { - 13,{}4} \right)\]
D. \[\left( {5,{} - 8} \right),{}\left( { - 5,{}8} \right)\]

Answer 1

Hint: We have to find the points of the Latus Rectum of the given conic equation . We solve this equation and determine the required general equation of the conic shape . Using the information about the shape obtained we obtain the end points of the Latus Rectum .

Complete step-by-step solution:
Given : ${x^2} + 10x - 16y + 25 = 0$
Firstly , we will make the equation in its general form by using the method of completing the square .
The formula of completing the square method is given as :
If a equation is given as ${x^2} + ax = 0$ , then by using the completing the square method , we get
Adding the square of half of the coefficient of $x$ both sides . So , in this equation we get
${x^2} + ax + {(\dfrac{a}{2})^2} = {(\dfrac{a}{2})^2}$
This becomes ,
${(x + \dfrac{a}{2})^2} = {(\dfrac{a}{2})^2}$
So , using the method we get
Adding ${5^2}$ both side
${x^2} + 10x + {5^2} = 16y - 25 + {5^2}$
On solving , we get
${(x + 5)^2} = 16y$
The equation obtained is of a parabola .
The general equation of parabola is given as :
${(x - h)^2} = 4a(y - k)$
Where \[\left( {h,k} \right)\] is the centre point or the point of the vertex .
So , on comparing the two equations , we get values as
\[h{} = {} - 5{},{}k{} = 0\]and \[a{} = {}4\]
Now , we know that the total length of the Latus Rectum is \[4a\] for a parabola .
So , using the parabola formed we can obtain the points at the end of Latus Rectum .
Also , the length of the latus rectum on one side is \[2a\] .
So , the end points of the latus rectum are given by \[\left( 2a,a \right)\] and \[\left( - 2a,a \right)\] . But these are the points when the parabola has a vertex \[\left( 0,0 \right)\] . In the given question the vertices are at point \[\left( - 5,0 \right)\] so the end points of Latus Rectum also shift .
Now , the end point becomes \[\left( 2a+h,a \right)\] and \[\left( - 2a + h,a \right)\]
So , finding the value\[\;\left( {{}{x_1} - {}h{} = {}2{}a{}} \right)\]and \[\left( {{}{x_2}{} - {}h{} = {} - 2a{}} \right)\]
On solving , we get
\[{x_1}{} = {}8{} - {}5\] and \[{x_2}{} = {} - 8{} - 5\]
\[{x_1}{} = {}3\] and \[{x_2}{} = {} - 13\]
Therefore, the end points of the Latus rectums are \[\left( { - 13{},{}4{}} \right)\] and \[\left( {{}3{},{}4} \right)\]. Hence, option (C) is correct.

Note: The equation of the parabola with focus at \[\left( {{}a{},{}0{}} \right){},{}a{} > {}0\] and directrix \[x{} = {} - a\] is ${y^2} = 4ax$ .
The Latus Rectum of a parabola is a line segment perpendicular to the axis of the parabola , through the focus and whose endpoints lie on the hyperbola.