Find the equations to the tangent and normal at the ends of the latus rectum of the parabola $y^2=12 x$
Answer (2)
Find the focus of the parabola y 2 = 12 x , which is at ( 3 , 0 ) .
Determine the endpoints of the latus rectum by substituting x = 3 into the parabola equation, yielding ( 3 , 6 ) and ( 3 , − 6 ) .
Calculate the tangent at ( 3 , 6 ) using the formula y y 1 = 2 a ( x + x 1 ) , resulting in y = x + 3 .
Calculate the normal at ( 3 , 6 ) using the formula y − y 1 = − 2 a y 1 ( x − x 1 ) , resulting in y = − x + 9 .
Calculate the tangent at ( 3 , − 6 ) using the formula y y 1 = 2 a ( x + x 1 ) , resulting in y = − x − 3 .
Calculate the normal at ( 3 , − 6 ) using the formula y − y 1 = − 2 a y 1 ( x − x 1 ) , resulting in y = x − 9 .
Tangent at ( 3 , 6 ) : y = x + 3 ; Normal at ( 3 , 6 ) : y = − x + 9 ; Tangent at ( 3 , − 6 ) : y = − x − 3 ; Normal at ( 3 , − 6 ) : y = x − 9
Explanation
Problem Analysis We are given the equation of a parabola y 2 = 12 x . Our goal is to find the equations of the tangent and normal lines at the endpoints of the latus rectum. Let's break this down step by step.
Finding the Focus First, we need to find the coordinates of the focus and the endpoints of the latus rectum. The general form of a parabola is y 2 = 4 a x . Comparing this with y 2 = 12 x , we find that 4 a = 12 , which means a = 3 . The focus of the parabola is at the point ( a , 0 ) , so the focus is at ( 3 , 0 ) .
Endpoints of Latus Rectum The latus rectum is a line segment passing through the focus and perpendicular to the axis of the parabola. Since the parabola opens to the right, the latus rectum is a vertical line x = 3 . To find the endpoints of the latus rectum, we substitute x = 3 into the equation of the parabola: y 2 = 12 ( 3 ) = 36 . Thus, y = ± 6 . The endpoints of the latus rectum are ( 3 , 6 ) and ( 3 , − 6 ) .
Tangent at (3, 6) Now, we find the equation of the tangent at the point ( 3 , 6 ) . The equation of the tangent to the parabola y 2 = 4 a x at the point ( x 1 , y 1 ) is given by y y 1 = 2 a ( x + x 1 ) . In our case, ( x 1 , y 1 ) = ( 3 , 6 ) and a = 3 . Substituting these values, we get y ( 6 ) = 2 ( 3 ) ( x + 3 ) , which simplifies to 6 y = 6 x + 18 . Dividing by 6, we get y = x + 3 .
Normal at (3, 6) Next, we find the equation of the normal at the point ( 3 , 6 ) . The equation of the normal to the parabola y 2 = 4 a x at the point ( x 1 , y 1 ) is given by y − y 1 = − 2 a y 1 ( x − x 1 ) . Substituting ( x 1 , y 1 ) = ( 3 , 6 ) and a = 3 , we get y − 6 = − 2 ( 3 ) 6 ( x − 3 ) , which simplifies to y − 6 = − 1 ( x − 3 ) . Thus, y − 6 = − x + 3 , and y = − x + 9 .
Tangent at (3, -6) Now, we find the equation of the tangent at the point ( 3 , − 6 ) . Using the same formula y y 1 = 2 a ( x + x 1 ) with ( x 1 , y 1 ) = ( 3 , − 6 ) and a = 3 , we get y ( − 6 ) = 2 ( 3 ) ( x + 3 ) , which simplifies to − 6 y = 6 x + 18 . Dividing by -6, we get y = − x − 3 .
Normal at (3, -6) Finally, we find the equation of the normal at the point ( 3 , − 6 ) . Using the formula y − y 1 = − 2 a y 1 ( x − x 1 ) with ( x 1 , y 1 ) = ( 3 , − 6 ) and a = 3 , we get y − ( − 6 ) = − 2 ( 3 ) − 6 ( x − 3 ) , which simplifies to y + 6 = 1 ( x − 3 ) . Thus, y + 6 = x − 3 , and y = x − 9 .
Final Equations In summary, the equations of the tangent and normal at the ends of the latus rectum are:
Tangent at ( 3 , 6 ) : y = x + 3 Normal at ( 3 , 6 ) : y = − x + 9 Tangent at ( 3 , − 6 ) : y = − x − 3 Normal at ( 3 , − 6 ) : y = x − 9
Examples
Understanding tangents and normals to parabolas is crucial in various fields, such as optics and antenna design. For instance, parabolic reflectors used in satellite dishes and radio telescopes rely on the property that incoming parallel rays are focused at a single point. The tangent at any point on the parabola determines the angle of reflection, while the normal helps in aligning the receiver for optimal signal collection. By calculating these lines, engineers can precisely design and optimize these systems for maximum efficiency.
The tangent and normal lines at the endpoints of the latus rectum of the parabola y 2 = 12 x are y = x + 3 , y = − x + 9 at (3, 6) and y = − x − 3 , y = x − 9 at (3, -6).
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