Analytical Geometry> The points of intersection of the circle ...

The points of intersection of the circle x^2_+y2=a² with the parabolas y²=4ax and y²=-4ax form a rectangle whose area is ?

The point of intersection of the tangents to the parabola y²=4x at the points where the circle (x-3)²+y²=9 meets the parabola ,other than the origin is ?

priya , 9 Years ago

Grade 11

1 Answers

Divyarth

Last Activity: 7 Years ago

x² + y² = a² ----> y² = a² − x²

Now use this value of y in each of the 2 equations of parabolas to find points of intersection. Since we have both parabolas y² = 4ax and y² = −4ax, we can assume a > 0

y² = 4ax
a² − x² = 4ax
x² + 4ax = a²
x² + 4ax + 4a² = a² + 4a²
(x + 2a)² = 5a²
x + 2a = ± a√5
x = (−2±√5)a

Any point on circle has x-coordinate between −a and a
x = (√5−2)a

y² = a² − x² = a² − (9−4√5)a² = 4(√5−2)a²
y = ± 2√(√5−2)a

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y² = −4ax
a² − x² = −4ax
x² − 4ax = a²
x² − 4ax + 4a² = a² + 4a²
(x − 2a)² = 5a²
x − 2a = ± a√5
x = (2±√5)a

Any point on circle has x-coordinate between −a and a
x = (2−√5)a

y² = a² − x² = a² − (9−4√5)a² = 4(√5−2)a²
y = ± 2√(√5−2)a

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Width of rectangle = difference between x-coordinates:
(√5−2)a − (2−√5)a = 2(√5−2)a

Height of rectangle = difference between y-coordinates:
2√(√5−2)a − (−2√(√5−2)a) = 4√(√5−2)a

Area of rectangle
2(√5−2)a * 4√(√5−2)a
= (2*4) (√5−2)√(√5−2) * a²
= 8 (√5−2)^(3/2) a²