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Question: The ends of the latus rectum of the conic \({x^2} + 10x - 16y + 25 = 0\) are A. \[\left( {3,{}-{}4...

The ends of the latus rectum of the conic x2+10x−16y+25=0{x^2} + 10x - 16y + 25 = 0 are
A. (3,−4),(13,4)\left( {3,{}-{}4} \right),{}\left( {13,{}4} \right)
B. (−3,−4),(13,−4)\left( { - 3,{} - 4} \right),{}\left( {13,{} - 4} \right)
C. (3,4),(−13,4)\left( {3,{}4} \right),{}\left( { - 13,{}4} \right)
D. (5,−8),(−5,8)\left( {5,{} - 8} \right),{}\left( { - 5,{}8} \right)

Explanation

Solution

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 : x2+10x−16y+25=0{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 x2+ax=0{x^2} + ax = 0 , then by using the completing the square method , we get
Adding the square of half of the coefficient of xx both sides . So , in this equation we get
x2+ax+(a2)2=(a2)2{x^2} + ax + {(\dfrac{a}{2})^2} = {(\dfrac{a}{2})^2}
This becomes ,
(x+a2)2=(a2)2{(x + \dfrac{a}{2})^2} = {(\dfrac{a}{2})^2}
So , using the method we get
Adding 52{5^2} both side
x2+10x+52=16y−25+52{x^2} + 10x + {5^2} = 16y - 25 + {5^2}
On solving , we get
(x+5)2=16y{(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){(x - h)^2} = 4a(y - k)
Where (h,k)\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=0h{} = {} - 5{},{}k{} = 0and a=4a{} = {}4
Now , we know that the total length of the Latus Rectum is 4a4a 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 2a2a .
So , the end points of the latus rectum are given by (2a,a)\left( 2a,a \right) and (−2a,a)\left( - 2a,a \right) . But these are the points when the parabola has a vertex (0,0)\left( 0,0 \right) . In the given question the vertices are at point (−5,0)\left( - 5,0 \right) so the end points of Latus Rectum also shift .
Now , the end point becomes (2a+h,a)\left( 2a+h,a \right) and (−2a+h,a)\left( - 2a + h,a \right)
So , finding the value  (x1−h=2a)\;\left( {{}{x_1} - {}h{} = {}2{}a{}} \right)and (x2−h=−2a)\left( {{}{x_2}{} - {}h{} = {} - 2a{}} \right)
On solving , we get
x1=8−5{x_1}{} = {}8{} - {}5 and x2=−8−5{x_2}{} = {} - 8{} - 5
x1=3{x_1}{} = {}3 and x2=−13{x_2}{} = {} - 13
Therefore, the end points of the Latus rectums are (−13,4)\left( { - 13{},{}4{}} \right) and (3,4)\left( {{}3{},{}4} \right). Hence, option (C) is correct.

Note: The equation of the parabola with focus at (a,0),a>0\left( {{}a{},{}0{}} \right){},{}a{} > {}0 and directrix x=−ax{} = {} - a is y2=4ax{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.