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Conic Section

Short Answer Type

61. Find the equation of the parabola that satisfying the following condition:
Vertex at (0,0), focus on the positive x-axis and length of latus rectum

16 over 9.

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62. Find the equation of the parabola that satisfying the following condition:
Vertex at (0, 0) focus on the negative side of y-axis and latus rectum equal to it.

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63. Find the equation of a parabola that satisfies the given condition:
Focus (6, 0), directrix is x = – 6

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64. Find the equation of a parabola that satisfies the given condition:
Focus (0 – 3); directrix y = 3

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65. Find the co-ordinates of a point on the parabola y² = l8x, where the ordinate is 3 times the abscissa.

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66. Find the area of the triangle formed by the lines joining the vertex of the parabola x² = 12y to the ends of the latus rectum.

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67.

An equilateral triangle is inscribed in the parabola $straight y squared equals 4 ax.$ , where one vertex is at the vertex of the parabola. Find (a) the length of the side of the triangle, (b) area of triangle ABC.

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68. If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

A parabolic reflection with diameter PR = 20 cm and OQ = 5 cm is shown below:
Here, vertex of the parabola is (0, 0)
Let the focus be S (a, 0)
Let the equation of the parabola be $straight y squared equals 4 ax.$
Now, PR = 20 cm $rightwards double arrow$ PQ = 10 cm
Also, OQ = 5 cm
∴ Point P is (5, 10)
Since it lies on the parabola $straight y squared equals 4 ax$
∴ $10 squared space equals 4 space left parenthesis straight a right parenthesis space 5 space rightwards double arrow space 100 space equals space 20 straight a space rightwards double arrow space straight a space equals space 5$
∴ Focus is S (5, 0) which is same as point Q.