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The abscissa of the point on the curve ay2 = x3, the normal at which cuts off equal intercepts from the coordinate axes is
- a)
- b)
- c)–
- d)–
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The abscissa of the point on the curve ay2= x3, the normal at which cu...
(That should be "axes" -- plural.)
Find the normal; find the intercepts -- for a point (u, v) say.
To get there, differentiate to find the gradient - of the tangent, hence of the normal.
2 a y dy/dx = 3 x^2
So dy/dx = 3 x^2 / (2 a y) .
At the point (u, v), this is 3 u^2 / (2 a v)
So the gradient of the normal there is (-2 a v) / (3 u^2)
So the equation of the normal is
y - v = (-2 a v) / (3 u^2) * (x - u)
When x = 0,
y = v - 2 a v / (3 u^2) (-u)
= v (1 + 2 a / (3 u) )
When y = 0,
x = u + 3 u^2 / (2 a v) * v
= u (1 + 3 u / (2 a) )
Now, "equal intercepts" could mean just by size, or by size and sign. Take the latter:
Then
v (2a + 3u) / (3u) = u (2a + 3u) / (2a)
So either u = -3 a / 2,
or v / (3u) = u / (2a),
2 a v = 3 u^2 --
each to be taken with a v^2 = u^3
One possibility, which fits all three equations, is (u, v) = (0, 0).; and if either u = 0 or v = 0, the so is the other.
If u ≠ 0, v ≠ 0 then
first case:
u = -3 a / 2
so a v^2 = -27 a^3 / 8
so v^2 = -27 a^2 / 8 -- not possible
second case:
2 a v = 3 u^2
and a v^2 = u^3 divide
So v / 2 = u /3 substitute
so a 4u^2 / 9 = u^3,
(u = 0 or) u = 4 a / 9 substitute, again:
a v^2 = 64 a^3 / 729
v = (-/)+ 8 a / 27
The (only) point that cuts off equal intercepts on the axes is (4a/9, 8a/27).
(if we allow equal distances, we get one other point -- y changes sign.)
Most Upvoted Answer
The abscissa of the point on the curve ay2= x3, the normal at which cu...
Slope of such normal is 1.⇒ dy/dx=1
ay2 = x3
⇒2ay dy/dx = 3x2
⇒ y=(3x2)/2a
⇒ a(3x2/2a)2
=x3
⇒x = 4a/9
ay2 = x3
⇒2ay dy/dx = 3x2
⇒ y=(3x2)/2a
⇒ a(3x2/2a)2
=x3
⇒x = 4a/9
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Community Answer
The abscissa of the point on the curve ay2= x3, the normal at which cu...
Slope of such normal is 1.⇒ dy/dx=1
ay2 = x3
⇒2ay dy/dx = 3x2
⇒ y=(3x2)/2a
⇒ a(3x2/2a)2
=x3
⇒x = 4a/9
ay2 = x3
⇒2ay dy/dx = 3x2
⇒ y=(3x2)/2a
⇒ a(3x2/2a)2
=x3
⇒x = 4a/9
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The abscissa of the point on the curve ay2= x3, the normal at which cuts off equal intercepts from the coordinate axes isa)b)c)–d)–Correct answer is option 'B'. Can you explain this answer?
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