Solveeit Logo

Question

Question: The equation of the line joining the vertex of the parabola \({y^2} = 6x\) to the points on it whose...

The equation of the line joining the vertex of the parabola y2=6x{y^2} = 6x to the points on it whose abscissa is 24, is:

a. y±2x=0y \pm 2x = 0

b. 2y±x=02y \pm x = 0

c. x±2y=0x \pm 2y = 0

d. 2x±y=02x \pm y = 0

Explanation

Solution

If two points are known then find the equation of line using two point form. Abscissa is the x - coordinate of a point and ordinate is the y - coordinate of a point.

Complete step by step answer:

Given Equation of parabola is y2=6x{y^2} = 6x

y2=6x (1)\Rightarrow {y^2} = 6x{\text{ }}\left( 1 \right)

Now, we have first find ordinate of points with abscissa 24 that lie on the given parabola

Let the ordinate be y

So, (24,y) should satisfy the given equation of parabola

putting value of point in equation 1 we get,

y2=144\Rightarrow {y^2} = 144

y=±12\Rightarrow y = \pm 12

So, there will be two points on the given equation with abscissa as 24. Let these points be

P=(24,12)and Q=(24,12)\Rightarrow P = \left( {24,12} \right){\text{and }}Q = \left( {24, - 12} \right)

So, vertex of the equation 1 will be

vertex =(0,0)\Rightarrow {\text{vertex }} = \left( {0,0} \right)

So, equation of line joining vertex and point P will be,

Finding equation of line using two point form where points are vertex (0,0)(x1,y1) and P(24,12)(x2,y2)\equiv \left( {0,0} \right) \equiv \left( {{x_1},{y_1}} \right){\text{ and P}} \equiv \left( {24,12} \right) \equiv \left( {{x_2},{y_2}} \right)

(yy1)=y2y1x2x1(xx1)y=x22yx=0 (2)\Rightarrow \left( {y - {y_1}} \right) = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right) \Rightarrow y = \frac{x}{2} \Rightarrow 2y - x = 0{\text{ }}\left( 2 \right)

Now, equation of line joining vertex and point Q will be,

Finding equation of line using two point form where points are vertex (0,0)(x1,y1) and Q(24,12)(x2,y2)\equiv \left( {0,0} \right) \equiv \left( {{x_1},{y_1}} \right){\text{ and Q}} \equiv \left( {24, - 12} \right) \equiv \left( {{x_2},{y_2}} \right)

(yy1)=y2y1x2x1(xx1)y=x22y+x=0 (3)\Rightarrow \left( {y - {y_1}} \right) = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right) \Rightarrow y = \frac{{ - x}}{2} \Rightarrow 2y + x = 0{\text{ }}\left( 3 \right)

From equations 2 and 3 we get, 2y±x=02y \pm x = 0 is the equation of line required.

Correct option for the question will be (b).

Note: Understand the diagram properly whenever you are facing these kinds of problems and also never neglect signs otherwise you will get only one solution. A better knowledge of formulas will be an added advantage.