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Solving this question and explain please help me

The curve for which the normal at any point $(x,y)$ and line joining origin to that point form an isosceles with the $x$ axis as base ,is

  1. an ellipse
  2. a rectangular hyperbola
  3. a circle
  4. none of these
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    You are more likely to get help here if you edit the question to show what you have tried and where you are stuck. – 2017-01-24
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    I have not understand this question please solve it – 2017-01-24
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    To understand it, proceed in the reverse and next try to write the suitable equations. So is, draw an isosceles triangle with the base on the x-axis and a vertex on the origin, then, the third vertex, that out of the x axis, belong to the curve you are searching for. From here, distances are an important thing to consider. – 2017-01-24
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    Which part of the question do you not understand? If there are any words you do not know, you should look them up in your textbook. – 2017-01-24

1 Answers 1

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That won't do you any good- to understand it, you need to make the effort yourself. Can we, at least, assume you know what "normal to a curve" and "isosceles triangle" mean? Do you know what an "ellipse", "rectangular hyperbola", and "circle" are?

Suppose the curve is given by y= f(x). The "normal" to the curve at $(x_0, f(x_0)$ is $y= -\frac{1}{f'(x_0)}(x- x_0)+ f(x_0)$. That line will cross the x-axis when y= 0: at (x, 0) where x satisfies $-\frac{1}{f'(x_0)}(x- x_0)+ f(x_0)= 0$. You should easily be able to solve that for x. This will form an isosceles triangle if the two sides are of equal length. One side, from (0, 0) to $(x_0, f(x_0)$, has length $\sqrt{x_0^2+ f(x_0)^2}$. The other, from $(x_0, f(x_0))$ to (x, 0) (x being the solution to the equation above) has length $\sqrt{(x- x_0)^2+ f(x_0)^2}$. Set those equal and solve for $f(x)$.

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    Thank you sir for solving this i understand after read – 2017-01-25