The Equation of porabola having vertex at (h,k) and axis is paralle to y axisis (x-h)^2= 4a (y-k).
Answers
Step-by-step explanation:
Let A (h, k) be the vertex of the parabola, AM is the axis of the parabola which is parallel to y-axis. The distance between the vertex and focus is AS = a and let P (x, y) be any point on the required parabola.
Now we shift the origin of co-ordinate system at A. Draw two mutually perpendicular straight lines AM and AN through the point A as y and x-axes respectively.
According to the new co-ordinate axes (x', y ') be the co-ordinates of P. Therefore, the equation of the parabola is (x’)2 = 4ay' (a > 0) …………….. (i)
Therefore, we get,
Therefore, we get,AM = y' and PM = x'
Therefore, we get,AM = y' and PM = x'Also, OR = k, AR = h, OQ = y, PQ = x
Therefore, we get,AM = y' and PM = x'Also, OR = k, AR = h, OQ = y, PQ = xAgain, x = PQ
Therefore, we get,AM = y' and PM = x'Also, OR = k, AR = h, OQ = y, PQ = xAgain, x = PQ= PM + MQ
Therefore, we get,AM = y' and PM = x'Also, OR = k, AR = h, OQ = y, PQ = xAgain, x = PQ= PM + MQ= PM + AR
Therefore, we get,AM = y' and PM = x'Also, OR = k, AR = h, OQ = y, PQ = xAgain, x = PQ= PM + MQ= PM + AR = x' + h
Therefore, we get,AM = y' and PM = x'Also, OR = k, AR = h, OQ = y, PQ = xAgain, x = PQ= PM + MQ= PM + AR = x' + h Therefore, x' = x - h
Therefore, we get,AM = y' and PM = x'Also, OR = k, AR = h, OQ = y, PQ = xAgain, x = PQ= PM + MQ= PM + AR = x' + h Therefore, x' = x - hAnd, y = OQ = OR + RQ
Therefore, we get,AM = y' and PM = x'Also, OR = k, AR = h, OQ = y, PQ = xAgain, x = PQ= PM + MQ= PM + AR = x' + h Therefore, x' = x - hAnd, y = OQ = OR + RQ= OR + AM
Therefore, we get,AM = y' and PM = x'Also, OR = k, AR = h, OQ = y, PQ = xAgain, x = PQ= PM + MQ= PM + AR = x' + h Therefore, x' = x - hAnd, y = OQ = OR + RQ= OR + AM = k + y'
Therefore, we get,AM = y' and PM = x'Also, OR = k, AR = h, OQ = y, PQ = xAgain, x = PQ= PM + MQ= PM + AR = x' + h Therefore, x' = x - hAnd, y = OQ = OR + RQ= OR + AM = k + y' Therefore, y' = y - k
Now putting the value of x' and y' in (i) we get
Now putting the value of x' and y' in (i) we get(x - h)2 = 4a(y - k), which is the equation of the required parabola.
Now putting the value of x' and y' in (i) we get(x - h)2 = 4a(y - k), which is the equation of the required parabola.The equation (x - h)2 = 4a(y - k) represents the equation of a parabola whose co-ordinate of the vertex is at (h, k), the co-ordinates of the focus are (h, a + k), the distance between its vertex and focus is a, the equation of directrix is y - k = - a or, y + a = k, the equation of the axis is x = h, the axis is parallel to positive y-axis, the length of its latus rectum = 4a, co-ordinates of the extremity of the latus rectum are (h + 2a, k + a) and (h - 2a, k + a) and the equation of tangent at the vertex is y = k.
